Find substring containing the escaped form of a delimiter (Regexp)












4















Hi All!



I am playing with markdown, dealing with inline markers and escaped characters.



Problem:



I want to transform this: some text *some number * other number* more text



Into this: some text <strong>some number * other number</strong> more text



My current pattern is: /((?!\)*)(.*?)((?!\)*)/g



But the (.*?) group seems to capture the character, so the third group finds the second * character and stops looking for the third one, which should be its target.



Possible solution:



I can solve this problem using negative lookbehind: /((?<!\)*)(.*?)((?<!\)*)/g, but I'd like to avoid it, if it is possible.



Can I modify my other pattern to make it work?






javascript regex







share|improve this question


















  • 2





    regex101.com/r/afdKgi/2 ?

    – splash58
    Jan 2 at 12:21











  • @splash58 Would you post it as an answer?

    – Nekomajin42
    Jan 2 at 12:32











  • /(^|[^\])*(.*?)($|[^\])*/g does not work if the * is at the start of the string. Even if you fix that, you won't match \*some number * other number* more text that should be since the first \ defines a backslash.

    – Wiktor Stribiżew
    Jan 2 at 12:38








  • 1





    @WiktorStribiżew regex101.com/r/afdKgi/6

    – splash58
    Jan 2 at 12:54













  • @splash58 Still won't work, \*some number * other number* more text starts with a backslash and an escaped *, but there is a match. This kind of task cannot be solved with .*? and lookarounds.

    – Wiktor Stribiżew
    Jan 2 at 13:24


















4















Hi All!



I am playing with markdown, dealing with inline markers and escaped characters.



Problem:



I want to transform this: some text *some number * other number* more text



Into this: some text <strong>some number * other number</strong> more text



My current pattern is: /((?!\)*)(.*?)((?!\)*)/g



But the (.*?) group seems to capture the character, so the third group finds the second * character and stops looking for the third one, which should be its target.



Possible solution:



I can solve this problem using negative lookbehind: /((?<!\)*)(.*?)((?<!\)*)/g, but I'd like to avoid it, if it is possible.



Can I modify my other pattern to make it work?






javascript regex







share|improve this question


















  • 2





    regex101.com/r/afdKgi/2 ?

    – splash58
    Jan 2 at 12:21











  • @splash58 Would you post it as an answer?

    – Nekomajin42
    Jan 2 at 12:32











  • /(^|[^\])*(.*?)($|[^\])*/g does not work if the * is at the start of the string. Even if you fix that, you won't match \*some number * other number* more text that should be since the first \ defines a backslash.

    – Wiktor Stribiżew
    Jan 2 at 12:38








  • 1





    @WiktorStribiżew regex101.com/r/afdKgi/6

    – splash58
    Jan 2 at 12:54













  • @splash58 Still won't work, \*some number * other number* more text starts with a backslash and an escaped *, but there is a match. This kind of task cannot be solved with .*? and lookarounds.

    – Wiktor Stribiżew
    Jan 2 at 13:24
















4












4








4








Hi All!



I am playing with markdown, dealing with inline markers and escaped characters.



Problem:



I want to transform this: some text *some number * other number* more text



Into this: some text <strong>some number * other number</strong> more text



My current pattern is: /((?!\)*)(.*?)((?!\)*)/g



But the (.*?) group seems to capture the character, so the third group finds the second * character and stops looking for the third one, which should be its target.



Possible solution:



I can solve this problem using negative lookbehind: /((?<!\)*)(.*?)((?<!\)*)/g, but I'd like to avoid it, if it is possible.



Can I modify my other pattern to make it work?






javascript regex







share|improve this question














Hi All!



I am playing with markdown, dealing with inline markers and escaped characters.



Problem:



I want to transform this: some text *some number * other number* more text



Into this: some text <strong>some number * other number</strong> more text



My current pattern is: /((?!\)*)(.*?)((?!\)*)/g



But the (.*?) group seems to capture the character, so the third group finds the second * character and stops looking for the third one, which should be its target.



Possible solution:



I can solve this problem using negative lookbehind: /((?<!\)*)(.*?)((?<!\)*)/g, but I'd like to avoid it, if it is possible.



Can I modify my other pattern to make it work?






javascript regex




javascript regex






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 2 at 12:11









Nekomajin42Nekomajin42

110139




110139








  • 2





    regex101.com/r/afdKgi/2 ?

    – splash58
    Jan 2 at 12:21











  • @splash58 Would you post it as an answer?

    – Nekomajin42
    Jan 2 at 12:32











  • /(^|[^\])*(.*?)($|[^\])*/g does not work if the * is at the start of the string. Even if you fix that, you won't match \*some number * other number* more text that should be since the first \ defines a backslash.

    – Wiktor Stribiżew
    Jan 2 at 12:38








  • 1





    @WiktorStribiżew regex101.com/r/afdKgi/6

    – splash58
    Jan 2 at 12:54













  • @splash58 Still won't work, \*some number * other number* more text starts with a backslash and an escaped *, but there is a match. This kind of task cannot be solved with .*? and lookarounds.

    – Wiktor Stribiżew
    Jan 2 at 13:24
















  • 2





    regex101.com/r/afdKgi/2 ?

    – splash58
    Jan 2 at 12:21











  • @splash58 Would you post it as an answer?

    – Nekomajin42
    Jan 2 at 12:32











  • /(^|[^\])*(.*?)($|[^\])*/g does not work if the * is at the start of the string. Even if you fix that, you won't match \*some number * other number* more text that should be since the first \ defines a backslash.

    – Wiktor Stribiżew
    Jan 2 at 12:38








  • 1





    @WiktorStribiżew regex101.com/r/afdKgi/6

    – splash58
    Jan 2 at 12:54













  • @splash58 Still won't work, \*some number * other number* more text starts with a backslash and an escaped *, but there is a match. This kind of task cannot be solved with .*? and lookarounds.

    – Wiktor Stribiżew
    Jan 2 at 13:24










2




2





regex101.com/r/afdKgi/2 ?

– splash58
Jan 2 at 12:21





regex101.com/r/afdKgi/2 ?

– splash58
Jan 2 at 12:21













@splash58 Would you post it as an answer?

– Nekomajin42
Jan 2 at 12:32





@splash58 Would you post it as an answer?

– Nekomajin42
Jan 2 at 12:32













/(^|[^\])*(.*?)($|[^\])*/g does not work if the * is at the start of the string. Even if you fix that, you won't match \*some number * other number* more text that should be since the first \ defines a backslash.

– Wiktor Stribiżew
Jan 2 at 12:38







/(^|[^\])*(.*?)($|[^\])*/g does not work if the * is at the start of the string. Even if you fix that, you won't match \*some number * other number* more text that should be since the first \ defines a backslash.

– Wiktor Stribiżew
Jan 2 at 12:38






1




1





@WiktorStribiżew regex101.com/r/afdKgi/6

– splash58
Jan 2 at 12:54







@WiktorStribiżew regex101.com/r/afdKgi/6

– splash58
Jan 2 at 12:54















@splash58 Still won't work, \*some number * other number* more text starts with a backslash and an escaped *, but there is a match. This kind of task cannot be solved with .*? and lookarounds.

– Wiktor Stribiżew
Jan 2 at 13:24







@splash58 Still won't work, \*some number * other number* more text starts with a backslash and an escaped *, but there is a match. This kind of task cannot be solved with .*? and lookarounds.

– Wiktor Stribiżew
Jan 2 at 13:24














3 Answers
3






active

oldest

votes


















2














You may use






var str = "some text *some number \* other number* more text";

console.log(

 str.replace(/((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g, 

   function($0, $1, $2) { return $1 + '<strong>' + $2.replace(/\([sS])/g, '$1') + '</strong>'; }

 )

)





The first /((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g regex matches all the strings within unescaped *:





  • ((?:^|[^\])(?:\{2})*) - Group 1:



    • (?:^|[^\]) - start of string or a non-backslash


    • (?:\{2})* - any 0+ occurrences of double backslash (this avoids matching escaped *)




  • * - a * char


  • ([^\*]*(?:\[sS][^*\]*)*) - Group 2:



    • [^\*]* - 0+ chars other than and *


    • (?:\[sS][^*\]*)* - 0+ sequences of



      • \[sS] - a and any char


      • [^*\]* - 0+ chars other than and *






  • * - a * char.


The match is passed to the anonymous method as the second argument to the replace method and the contents of Group 2 are processed to "unescape" any escape sequence with .replace(/\([sS])/g, '$1'): \ matches a backslash and ([sS]) matches and captures any char into Group 1, and this is what remains after the replacement with the group placeholder $1.






share|improve this answer



















  • 1





    The thing is, I need this stuff for educational purposes, and it's way more complicated than the ideal, but thank you for the detailed explanation.

    – Nekomajin42
    Jan 2 at 12:34



















1














You can use this 



*(.*)*


This uses above regex to find * up to the last *. And than with \(.) i am finding the escaped character and replacing it with captured group.






const regex = /*(.*)*/gm;

const str = `some text *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/,'$1')   //replacing escaped character here



console.log(finalResult);





UPDATE: For matching more than one substring






const regex = /*(.*?[^\])*/gm;

const str = `some text *some number \* other number* blah blah *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/g,'$1')   //replacing escaped character here



console.log(finalResult);








share|improve this answer


























  • Can you modify it to work with multiple instances of the substring within str?

    – Nekomajin42
    Jan 2 at 12:39











  • @Nekomajin42 check the updated one. let me know if something is missing

    – Code Maniac
    Jan 2 at 12:58











  • The updated pattern seems to find only the first and last * marker in the string, and the same markers are ignored between them.

    – Nekomajin42
    Jan 5 at 14:42



















0














There could be a simpler way to accomplish the same task using the following regex:



\.|*((\.|[^*])+)*


The idea is matching a desired string should occur after all escaped characters are consumed. We try to match all escaped characters using first side of alternation then at the second attempt we want to match our desired pattern if exists.



JS code:






var str = `some text *some number \* other number* more text`



console.log(str.replace(/\.|*((\.|[^*])+)*/g, function(match, $1) {

	return $1 ? '<strong>' + $1 + '</strong>' : match;

}));





Breakdown:





  • \. Match an escaped character


  • | Or


  • * Match a literal *


  • ( Start of first capturing group



    • ( Start of second capturing group



      • \. Match an escaped character


      • | Or


      • [^*]+ Match anything except *




    • )+ End of second capturing group, repeat one or more time


    • ) End of first capturing group




  • * Match a literal *






share|improve this answer
























  • This pattern seems to match * substrings outside of the * markers.

    – Nekomajin42
    Jan 5 at 14:40











  • It matches but doesn't replace. There is a callback for replace method. Even the answer you accepted has two consecutive calls to replace() which is complicating things. If you experience any problems please show an example.

    – revo
    Jan 5 at 18:19













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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














You may use






var str = "some text *some number \* other number* more text";

console.log(

 str.replace(/((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g, 

   function($0, $1, $2) { return $1 + '<strong>' + $2.replace(/\([sS])/g, '$1') + '</strong>'; }

 )

)





The first /((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g regex matches all the strings within unescaped *:





  • ((?:^|[^\])(?:\{2})*) - Group 1:



    • (?:^|[^\]) - start of string or a non-backslash


    • (?:\{2})* - any 0+ occurrences of double backslash (this avoids matching escaped *)




  • * - a * char


  • ([^\*]*(?:\[sS][^*\]*)*) - Group 2:



    • [^\*]* - 0+ chars other than and *


    • (?:\[sS][^*\]*)* - 0+ sequences of



      • \[sS] - a and any char


      • [^*\]* - 0+ chars other than and *






  • * - a * char.


The match is passed to the anonymous method as the second argument to the replace method and the contents of Group 2 are processed to "unescape" any escape sequence with .replace(/\([sS])/g, '$1'): \ matches a backslash and ([sS]) matches and captures any char into Group 1, and this is what remains after the replacement with the group placeholder $1.






share|improve this answer



















  • 1





    The thing is, I need this stuff for educational purposes, and it's way more complicated than the ideal, but thank you for the detailed explanation.

    – Nekomajin42
    Jan 2 at 12:34
















2














You may use






var str = "some text *some number \* other number* more text";

console.log(

 str.replace(/((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g, 

   function($0, $1, $2) { return $1 + '<strong>' + $2.replace(/\([sS])/g, '$1') + '</strong>'; }

 )

)





The first /((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g regex matches all the strings within unescaped *:





  • ((?:^|[^\])(?:\{2})*) - Group 1:



    • (?:^|[^\]) - start of string or a non-backslash


    • (?:\{2})* - any 0+ occurrences of double backslash (this avoids matching escaped *)




  • * - a * char


  • ([^\*]*(?:\[sS][^*\]*)*) - Group 2:



    • [^\*]* - 0+ chars other than and *


    • (?:\[sS][^*\]*)* - 0+ sequences of



      • \[sS] - a and any char


      • [^*\]* - 0+ chars other than and *






  • * - a * char.


The match is passed to the anonymous method as the second argument to the replace method and the contents of Group 2 are processed to "unescape" any escape sequence with .replace(/\([sS])/g, '$1'): \ matches a backslash and ([sS]) matches and captures any char into Group 1, and this is what remains after the replacement with the group placeholder $1.






share|improve this answer



















  • 1





    The thing is, I need this stuff for educational purposes, and it's way more complicated than the ideal, but thank you for the detailed explanation.

    – Nekomajin42
    Jan 2 at 12:34














2












2








2







You may use






var str = "some text *some number \* other number* more text";

console.log(

 str.replace(/((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g, 

   function($0, $1, $2) { return $1 + '<strong>' + $2.replace(/\([sS])/g, '$1') + '</strong>'; }

 )

)





The first /((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g regex matches all the strings within unescaped *:





  • ((?:^|[^\])(?:\{2})*) - Group 1:



    • (?:^|[^\]) - start of string or a non-backslash


    • (?:\{2})* - any 0+ occurrences of double backslash (this avoids matching escaped *)




  • * - a * char


  • ([^\*]*(?:\[sS][^*\]*)*) - Group 2:



    • [^\*]* - 0+ chars other than and *


    • (?:\[sS][^*\]*)* - 0+ sequences of



      • \[sS] - a and any char


      • [^*\]* - 0+ chars other than and *






  • * - a * char.


The match is passed to the anonymous method as the second argument to the replace method and the contents of Group 2 are processed to "unescape" any escape sequence with .replace(/\([sS])/g, '$1'): \ matches a backslash and ([sS]) matches and captures any char into Group 1, and this is what remains after the replacement with the group placeholder $1.






share|improve this answer













You may use






var str = "some text *some number \* other number* more text";

console.log(

 str.replace(/((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g, 

   function($0, $1, $2) { return $1 + '<strong>' + $2.replace(/\([sS])/g, '$1') + '</strong>'; }

 )

)





The first /((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g regex matches all the strings within unescaped *:





  • ((?:^|[^\])(?:\{2})*) - Group 1:



    • (?:^|[^\]) - start of string or a non-backslash


    • (?:\{2})* - any 0+ occurrences of double backslash (this avoids matching escaped *)




  • * - a * char


  • ([^\*]*(?:\[sS][^*\]*)*) - Group 2:



    • [^\*]* - 0+ chars other than and *


    • (?:\[sS][^*\]*)* - 0+ sequences of



      • \[sS] - a and any char


      • [^*\]* - 0+ chars other than and *






  • * - a * char.


The match is passed to the anonymous method as the second argument to the replace method and the contents of Group 2 are processed to "unescape" any escape sequence with .replace(/\([sS])/g, '$1'): \ matches a backslash and ([sS]) matches and captures any char into Group 1, and this is what remains after the replacement with the group placeholder $1.






var str = "some text *some number \* other number* more text";

console.log(

 str.replace(/((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g, 

   function($0, $1, $2) { return $1 + '<strong>' + $2.replace(/\([sS])/g, '$1') + '</strong>'; }

 )

)





var str = "some text *some number \* other number* more text";

console.log(

 str.replace(/((?:^|[^\])(?:\{2})*)*([^\*]*(?:\[sS][^*\]*)*)*/g, 

   function($0, $1, $2) { return $1 + '<strong>' + $2.replace(/\([sS])/g, '$1') + '</strong>'; }

 )

)






share|improve this answer












share|improve this answer



share|improve this answer










answered Jan 2 at 12:21









Wiktor StribiżewWiktor Stribiżew

322k16143224




322k16143224








  • 1





    The thing is, I need this stuff for educational purposes, and it's way more complicated than the ideal, but thank you for the detailed explanation.

    – Nekomajin42
    Jan 2 at 12:34














  • 1





    The thing is, I need this stuff for educational purposes, and it's way more complicated than the ideal, but thank you for the detailed explanation.

    – Nekomajin42
    Jan 2 at 12:34








1




1





The thing is, I need this stuff for educational purposes, and it's way more complicated than the ideal, but thank you for the detailed explanation.

– Nekomajin42
Jan 2 at 12:34





The thing is, I need this stuff for educational purposes, and it's way more complicated than the ideal, but thank you for the detailed explanation.

– Nekomajin42
Jan 2 at 12:34













1














You can use this 



*(.*)*


This uses above regex to find * up to the last *. And than with \(.) i am finding the escaped character and replacing it with captured group.






const regex = /*(.*)*/gm;

const str = `some text *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/,'$1')   //replacing escaped character here



console.log(finalResult);





UPDATE: For matching more than one substring






const regex = /*(.*?[^\])*/gm;

const str = `some text *some number \* other number* blah blah *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/g,'$1')   //replacing escaped character here



console.log(finalResult);








share|improve this answer


























  • Can you modify it to work with multiple instances of the substring within str?

    – Nekomajin42
    Jan 2 at 12:39











  • @Nekomajin42 check the updated one. let me know if something is missing

    – Code Maniac
    Jan 2 at 12:58











  • The updated pattern seems to find only the first and last * marker in the string, and the same markers are ignored between them.

    – Nekomajin42
    Jan 5 at 14:42
















1














You can use this 



*(.*)*


This uses above regex to find * up to the last *. And than with \(.) i am finding the escaped character and replacing it with captured group.






const regex = /*(.*)*/gm;

const str = `some text *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/,'$1')   //replacing escaped character here



console.log(finalResult);





UPDATE: For matching more than one substring






const regex = /*(.*?[^\])*/gm;

const str = `some text *some number \* other number* blah blah *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/g,'$1')   //replacing escaped character here



console.log(finalResult);








share|improve this answer


























  • Can you modify it to work with multiple instances of the substring within str?

    – Nekomajin42
    Jan 2 at 12:39











  • @Nekomajin42 check the updated one. let me know if something is missing

    – Code Maniac
    Jan 2 at 12:58











  • The updated pattern seems to find only the first and last * marker in the string, and the same markers are ignored between them.

    – Nekomajin42
    Jan 5 at 14:42














1












1








1







You can use this 



*(.*)*


This uses above regex to find * up to the last *. And than with \(.) i am finding the escaped character and replacing it with captured group.






const regex = /*(.*)*/gm;

const str = `some text *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/,'$1')   //replacing escaped character here



console.log(finalResult);





UPDATE: For matching more than one substring






const regex = /*(.*?[^\])*/gm;

const str = `some text *some number \* other number* blah blah *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/g,'$1')   //replacing escaped character here



console.log(finalResult);








share|improve this answer















You can use this 



*(.*)*


This uses above regex to find * up to the last *. And than with \(.) i am finding the escaped character and replacing it with captured group.






const regex = /*(.*)*/gm;

const str = `some text *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/,'$1')   //replacing escaped character here



console.log(finalResult);





UPDATE: For matching more than one substring






const regex = /*(.*?[^\])*/gm;

const str = `some text *some number \* other number* blah blah *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/g,'$1')   //replacing escaped character here



console.log(finalResult);








const regex = /*(.*)*/gm;

const str = `some text *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/,'$1')   //replacing escaped character here



console.log(finalResult);





const regex = /*(.*)*/gm;

const str = `some text *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/,'$1')   //replacing escaped character here



console.log(finalResult);





const regex = /*(.*?[^\])*/gm;

const str = `some text *some number \* other number* blah blah *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/g,'$1')   //replacing escaped character here



console.log(finalResult);





const regex = /*(.*?[^\])*/gm;

const str = `some text *some number \* other number* blah blah *some number \* other number* more text`;

const subst = `<strong>$1</strong>`;



// The substituted value will be contained in the result variable

const result = str.replace(regex, subst);

const finalResult = result.replace(/\(.)/g,'$1')   //replacing escaped character here



console.log(finalResult);






share|improve this answer














share|improve this answer



share|improve this answer








edited Jan 2 at 12:57

























answered Jan 2 at 12:19









Code ManiacCode Maniac

8,9202630




8,9202630













  • Can you modify it to work with multiple instances of the substring within str?

    – Nekomajin42
    Jan 2 at 12:39











  • @Nekomajin42 check the updated one. let me know if something is missing

    – Code Maniac
    Jan 2 at 12:58











  • The updated pattern seems to find only the first and last * marker in the string, and the same markers are ignored between them.

    – Nekomajin42
    Jan 5 at 14:42



















  • Can you modify it to work with multiple instances of the substring within str?

    – Nekomajin42
    Jan 2 at 12:39











  • @Nekomajin42 check the updated one. let me know if something is missing

    – Code Maniac
    Jan 2 at 12:58











  • The updated pattern seems to find only the first and last * marker in the string, and the same markers are ignored between them.

    – Nekomajin42
    Jan 5 at 14:42

















Can you modify it to work with multiple instances of the substring within str?

– Nekomajin42
Jan 2 at 12:39





Can you modify it to work with multiple instances of the substring within str?

– Nekomajin42
Jan 2 at 12:39













@Nekomajin42 check the updated one. let me know if something is missing

– Code Maniac
Jan 2 at 12:58





@Nekomajin42 check the updated one. let me know if something is missing

– Code Maniac
Jan 2 at 12:58













The updated pattern seems to find only the first and last * marker in the string, and the same markers are ignored between them.

– Nekomajin42
Jan 5 at 14:42





The updated pattern seems to find only the first and last * marker in the string, and the same markers are ignored between them.

– Nekomajin42
Jan 5 at 14:42











0














There could be a simpler way to accomplish the same task using the following regex:



\.|*((\.|[^*])+)*


The idea is matching a desired string should occur after all escaped characters are consumed. We try to match all escaped characters using first side of alternation then at the second attempt we want to match our desired pattern if exists.



JS code:






var str = `some text *some number \* other number* more text`



console.log(str.replace(/\.|*((\.|[^*])+)*/g, function(match, $1) {

	return $1 ? '<strong>' + $1 + '</strong>' : match;

}));





Breakdown:





  • \. Match an escaped character


  • | Or


  • * Match a literal *


  • ( Start of first capturing group



    • ( Start of second capturing group



      • \. Match an escaped character


      • | Or


      • [^*]+ Match anything except *




    • )+ End of second capturing group, repeat one or more time


    • ) End of first capturing group




  • * Match a literal *






share|improve this answer
























  • This pattern seems to match * substrings outside of the * markers.

    – Nekomajin42
    Jan 5 at 14:40











  • It matches but doesn't replace. There is a callback for replace method. Even the answer you accepted has two consecutive calls to replace() which is complicating things. If you experience any problems please show an example.

    – revo
    Jan 5 at 18:19


















0














There could be a simpler way to accomplish the same task using the following regex:



\.|*((\.|[^*])+)*


The idea is matching a desired string should occur after all escaped characters are consumed. We try to match all escaped characters using first side of alternation then at the second attempt we want to match our desired pattern if exists.



JS code:






var str = `some text *some number \* other number* more text`



console.log(str.replace(/\.|*((\.|[^*])+)*/g, function(match, $1) {

	return $1 ? '<strong>' + $1 + '</strong>' : match;

}));





Breakdown:





  • \. Match an escaped character


  • | Or


  • * Match a literal *


  • ( Start of first capturing group



    • ( Start of second capturing group



      • \. Match an escaped character


      • | Or


      • [^*]+ Match anything except *




    • )+ End of second capturing group, repeat one or more time


    • ) End of first capturing group




  • * Match a literal *






share|improve this answer
























  • This pattern seems to match * substrings outside of the * markers.

    – Nekomajin42
    Jan 5 at 14:40











  • It matches but doesn't replace. There is a callback for replace method. Even the answer you accepted has two consecutive calls to replace() which is complicating things. If you experience any problems please show an example.

    – revo
    Jan 5 at 18:19
















0












0








0







There could be a simpler way to accomplish the same task using the following regex:



\.|*((\.|[^*])+)*


The idea is matching a desired string should occur after all escaped characters are consumed. We try to match all escaped characters using first side of alternation then at the second attempt we want to match our desired pattern if exists.



JS code:






var str = `some text *some number \* other number* more text`



console.log(str.replace(/\.|*((\.|[^*])+)*/g, function(match, $1) {

	return $1 ? '<strong>' + $1 + '</strong>' : match;

}));





Breakdown:





  • \. Match an escaped character


  • | Or


  • * Match a literal *


  • ( Start of first capturing group



    • ( Start of second capturing group



      • \. Match an escaped character


      • | Or


      • [^*]+ Match anything except *




    • )+ End of second capturing group, repeat one or more time


    • ) End of first capturing group




  • * Match a literal *






share|improve this answer













There could be a simpler way to accomplish the same task using the following regex:



\.|*((\.|[^*])+)*


The idea is matching a desired string should occur after all escaped characters are consumed. We try to match all escaped characters using first side of alternation then at the second attempt we want to match our desired pattern if exists.



JS code:






var str = `some text *some number \* other number* more text`



console.log(str.replace(/\.|*((\.|[^*])+)*/g, function(match, $1) {

	return $1 ? '<strong>' + $1 + '</strong>' : match;

}));





Breakdown:





  • \. Match an escaped character


  • | Or


  • * Match a literal *


  • ( Start of first capturing group



    • ( Start of second capturing group



      • \. Match an escaped character


      • | Or


      • [^*]+ Match anything except *




    • )+ End of second capturing group, repeat one or more time


    • ) End of first capturing group




  • * Match a literal *






var str = `some text *some number \* other number* more text`



console.log(str.replace(/\.|*((\.|[^*])+)*/g, function(match, $1) {

	return $1 ? '<strong>' + $1 + '</strong>' : match;

}));





var str = `some text *some number \* other number* more text`



console.log(str.replace(/\.|*((\.|[^*])+)*/g, function(match, $1) {

	return $1 ? '<strong>' + $1 + '</strong>' : match;

}));






share|improve this answer












share|improve this answer



share|improve this answer










answered Jan 2 at 13:08









revorevo

33.6k135086




33.6k135086













  • This pattern seems to match * substrings outside of the * markers.

    – Nekomajin42
    Jan 5 at 14:40











  • It matches but doesn't replace. There is a callback for replace method. Even the answer you accepted has two consecutive calls to replace() which is complicating things. If you experience any problems please show an example.

    – revo
    Jan 5 at 18:19





















  • This pattern seems to match * substrings outside of the * markers.

    – Nekomajin42
    Jan 5 at 14:40











  • It matches but doesn't replace. There is a callback for replace method. Even the answer you accepted has two consecutive calls to replace() which is complicating things. If you experience any problems please show an example.

    – revo
    Jan 5 at 18:19



















This pattern seems to match * substrings outside of the * markers.

– Nekomajin42
Jan 5 at 14:40





This pattern seems to match * substrings outside of the * markers.

– Nekomajin42
Jan 5 at 14:40













It matches but doesn't replace. There is a callback for replace method. Even the answer you accepted has two consecutive calls to replace() which is complicating things. If you experience any problems please show an example.

– revo
Jan 5 at 18:19







It matches but doesn't replace. There is a callback for replace method. Even the answer you accepted has two consecutive calls to replace() which is complicating things. If you experience any problems please show an example.

– revo
Jan 5 at 18:19




















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