How to implement this by RegularExpression
my $s1 = 'a1a2a3a4a5';
my $pat = '(ad(?=ad)ad)';
while($s1 =~ m/$pat/g)
{
print "$1n"
}
I want to get output:
a1a2a3
a2a3a4
a3a4a5
But output is:
a1a2
a3a4
By my knowledge, next match in loop is started from ?=, but by my practice, it is not work in this way. Who can point out the problem.
Thanks
perl
asked Jan 2 at 3:42
user2799433user2799433
234
add a comment |
my $s1 = 'a1a2a3a4a5';
my $pat = '(ad(?=ad)ad)';
while($s1 =~ m/$pat/g)
{
print "$1n"
}
I want to get output:
a1a2a3
a2a3a4
a3a4a5
But output is:
a1a2
a3a4
By my knowledge, next match in loop is started from ?=, but by my practice, it is not work in this way. Who can point out the problem.
Thanks
perl
asked Jan 2 at 3:42
user2799433user2799433
234
add a comment |
my $s1 = 'a1a2a3a4a5';
my $pat = '(ad(?=ad)ad)';
while($s1 =~ m/$pat/g)
{
print "$1n"
}
I want to get output:
a1a2a3
a2a3a4
a3a4a5
But output is:
a1a2
a3a4
By my knowledge, next match in loop is started from ?=, but by my practice, it is not work in this way. Who can point out the problem.
Thanks
perl
asked Jan 2 at 3:42
user2799433user2799433
234
my $s1 = 'a1a2a3a4a5';
my $pat = '(ad(?=ad)ad)';
while($s1 =~ m/$pat/g)
{
print "$1n"
}
I want to get output:
a1a2a3
a2a3a4
a3a4a5
But output is:
a1a2
a3a4
By my knowledge, next match in loop is started from ?=, but by my practice, it is not work in this way. Who can point out the problem.
Thanks
perl
perl
asked Jan 2 at 3:42
user2799433user2799433
234
asked Jan 2 at 3:42
user2799433user2799433
234
asked Jan 2 at 3:42
user2799433user2799433
234
asked Jan 2 at 3:42
user2799433user2799433
234
asked Jan 2 at 3:42
user2799433user2799433
234
234
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
.... next match in loop is started from ?=
The next match within //g is started where the last match left off. A (?=...) does not magically set the position where the next match will start; it simply checks if the part inside (?=...) would match at the current position without advancing the position.
If your regex would be /(ad(?=ad))/ then the match would be done after the first ad although it makes sure that there is a second ad behind it. But your regex is /(ad(?=ad)ad)/ which means simplified regarding the match /adad/, i.e. it is done after the second ad.
What you could do to achieve what you want is for example the following:
my $s1 = 'a1a2a3a4a5';
while($s1 =~ m/(ad(?=(adad)))/g)
{
print "$1$2n"
}
This will put the first ad into $1, capture the second adad into $2 but still finish the match after the first ad. Only you need to print "$1$2" then instead of only "$1".
edited Jan 2 at 17:51
ikegami
265k11178401
answered Jan 2 at 4:14
Steffen UllrichSteffen Ullrich
61.6k358100
By my understand, (?=pat1)pat2 only take a look whether there is a pat1 at current matching position but do not match it to move current matching position. Then continue to match pat2 at current match position and then advance the position. Right?
– user2799433
Jan 3 at 7:45
@user2799433: right: the(?=pat1)only looks if the pattern exists but does not move the end of the match. But thepat2then actually matches and also moves the end of the match. Since you essentially used the same pattern forpat1andpat2(i.e.(?=pat)pat) the positive lookahead(?=pat)was essentially irrelevant since it was followed by exactly the same pattern with a "real" match, i.e. the lookahead and the real match started and ended at the same position.
– Steffen Ullrich
Jan 3 at 8:19
add a comment |
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1 Answer
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1 Answer
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active
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active
oldest
votes
.... next match in loop is started from ?=
The next match within //g is started where the last match left off. A (?=...) does not magically set the position where the next match will start; it simply checks if the part inside (?=...) would match at the current position without advancing the position.
If your regex would be /(ad(?=ad))/ then the match would be done after the first ad although it makes sure that there is a second ad behind it. But your regex is /(ad(?=ad)ad)/ which means simplified regarding the match /adad/, i.e. it is done after the second ad.
What you could do to achieve what you want is for example the following:
my $s1 = 'a1a2a3a4a5';
while($s1 =~ m/(ad(?=(adad)))/g)
{
print "$1$2n"
}
This will put the first ad into $1, capture the second adad into $2 but still finish the match after the first ad. Only you need to print "$1$2" then instead of only "$1".
edited Jan 2 at 17:51
ikegami
265k11178401
answered Jan 2 at 4:14
Steffen UllrichSteffen Ullrich
61.6k358100
By my understand, (?=pat1)pat2 only take a look whether there is a pat1 at current matching position but do not match it to move current matching position. Then continue to match pat2 at current match position and then advance the position. Right?
– user2799433
Jan 3 at 7:45
@user2799433: right: the(?=pat1)only looks if the pattern exists but does not move the end of the match. But thepat2then actually matches and also moves the end of the match. Since you essentially used the same pattern forpat1andpat2(i.e.(?=pat)pat) the positive lookahead(?=pat)was essentially irrelevant since it was followed by exactly the same pattern with a "real" match, i.e. the lookahead and the real match started and ended at the same position.
– Steffen Ullrich
Jan 3 at 8:19
add a comment |
.... next match in loop is started from ?=
The next match within //g is started where the last match left off. A (?=...) does not magically set the position where the next match will start; it simply checks if the part inside (?=...) would match at the current position without advancing the position.
If your regex would be /(ad(?=ad))/ then the match would be done after the first ad although it makes sure that there is a second ad behind it. But your regex is /(ad(?=ad)ad)/ which means simplified regarding the match /adad/, i.e. it is done after the second ad.
What you could do to achieve what you want is for example the following:
my $s1 = 'a1a2a3a4a5';
while($s1 =~ m/(ad(?=(adad)))/g)
{
print "$1$2n"
}
This will put the first ad into $1, capture the second adad into $2 but still finish the match after the first ad. Only you need to print "$1$2" then instead of only "$1".
edited Jan 2 at 17:51
ikegami
265k11178401
answered Jan 2 at 4:14
Steffen UllrichSteffen Ullrich
61.6k358100
By my understand, (?=pat1)pat2 only take a look whether there is a pat1 at current matching position but do not match it to move current matching position. Then continue to match pat2 at current match position and then advance the position. Right?
– user2799433
Jan 3 at 7:45
@user2799433: right: the(?=pat1)only looks if the pattern exists but does not move the end of the match. But thepat2then actually matches and also moves the end of the match. Since you essentially used the same pattern forpat1andpat2(i.e.(?=pat)pat) the positive lookahead(?=pat)was essentially irrelevant since it was followed by exactly the same pattern with a "real" match, i.e. the lookahead and the real match started and ended at the same position.
– Steffen Ullrich
Jan 3 at 8:19
add a comment |
.... next match in loop is started from ?=
The next match within //g is started where the last match left off. A (?=...) does not magically set the position where the next match will start; it simply checks if the part inside (?=...) would match at the current position without advancing the position.
If your regex would be /(ad(?=ad))/ then the match would be done after the first ad although it makes sure that there is a second ad behind it. But your regex is /(ad(?=ad)ad)/ which means simplified regarding the match /adad/, i.e. it is done after the second ad.
What you could do to achieve what you want is for example the following:
my $s1 = 'a1a2a3a4a5';
while($s1 =~ m/(ad(?=(adad)))/g)
{
print "$1$2n"
}
This will put the first ad into $1, capture the second adad into $2 but still finish the match after the first ad. Only you need to print "$1$2" then instead of only "$1".
edited Jan 2 at 17:51
ikegami
265k11178401
answered Jan 2 at 4:14
Steffen UllrichSteffen Ullrich
61.6k358100
.... next match in loop is started from ?=
The next match within //g is started where the last match left off. A (?=...) does not magically set the position where the next match will start; it simply checks if the part inside (?=...) would match at the current position without advancing the position.
If your regex would be /(ad(?=ad))/ then the match would be done after the first ad although it makes sure that there is a second ad behind it. But your regex is /(ad(?=ad)ad)/ which means simplified regarding the match /adad/, i.e. it is done after the second ad.
What you could do to achieve what you want is for example the following:
my $s1 = 'a1a2a3a4a5';
while($s1 =~ m/(ad(?=(adad)))/g)
{
print "$1$2n"
}
This will put the first ad into $1, capture the second adad into $2 but still finish the match after the first ad. Only you need to print "$1$2" then instead of only "$1".
edited Jan 2 at 17:51
ikegami
265k11178401
answered Jan 2 at 4:14
Steffen UllrichSteffen Ullrich
61.6k358100
edited Jan 2 at 17:51
ikegami
265k11178401
edited Jan 2 at 17:51
ikegami
265k11178401
edited Jan 2 at 17:51
ikegami
265k11178401
265k11178401
answered Jan 2 at 4:14
Steffen UllrichSteffen Ullrich
61.6k358100
answered Jan 2 at 4:14
Steffen UllrichSteffen Ullrich
61.6k358100
answered Jan 2 at 4:14
Steffen UllrichSteffen Ullrich
61.6k358100
61.6k358100
By my understand, (?=pat1)pat2 only take a look whether there is a pat1 at current matching position but do not match it to move current matching position. Then continue to match pat2 at current match position and then advance the position. Right?
– user2799433
Jan 3 at 7:45
@user2799433: right: the(?=pat1)only looks if the pattern exists but does not move the end of the match. But thepat2then actually matches and also moves the end of the match. Since you essentially used the same pattern forpat1andpat2(i.e.(?=pat)pat) the positive lookahead(?=pat)was essentially irrelevant since it was followed by exactly the same pattern with a "real" match, i.e. the lookahead and the real match started and ended at the same position.
– Steffen Ullrich
Jan 3 at 8:19
add a comment |
By my understand, (?=pat1)pat2 only take a look whether there is a pat1 at current matching position but do not match it to move current matching position. Then continue to match pat2 at current match position and then advance the position. Right?
– user2799433
Jan 3 at 7:45
@user2799433: right: the(?=pat1)only looks if the pattern exists but does not move the end of the match. But thepat2then actually matches and also moves the end of the match. Since you essentially used the same pattern forpat1andpat2(i.e.(?=pat)pat) the positive lookahead(?=pat)was essentially irrelevant since it was followed by exactly the same pattern with a "real" match, i.e. the lookahead and the real match started and ended at the same position.
– Steffen Ullrich
Jan 3 at 8:19
By my understand, (?=pat1)pat2 only take a look whether there is a pat1 at current matching position but do not match it to move current matching position. Then continue to match pat2 at current match position and then advance the position. Right?
– user2799433
Jan 3 at 7:45
By my understand, (?=pat1)pat2 only take a look whether there is a pat1 at current matching position but do not match it to move current matching position. Then continue to match pat2 at current match position and then advance the position. Right?
– user2799433
Jan 3 at 7:45
@user2799433: right: the
(?=pat1) only looks if the pattern exists but does not move the end of the match. But the pat2 then actually matches and also moves the end of the match. Since you essentially used the same pattern for pat1 and pat2 (i.e. (?=pat)pat) the positive lookahead (?=pat) was essentially irrelevant since it was followed by exactly the same pattern with a "real" match, i.e. the lookahead and the real match started and ended at the same position.– Steffen Ullrich
Jan 3 at 8:19
@user2799433: right: the
(?=pat1) only looks if the pattern exists but does not move the end of the match. But the pat2 then actually matches and also moves the end of the match. Since you essentially used the same pattern for pat1 and pat2 (i.e. (?=pat)pat) the positive lookahead (?=pat) was essentially irrelevant since it was followed by exactly the same pattern with a "real" match, i.e. the lookahead and the real match started and ended at the same position.– Steffen Ullrich
Jan 3 at 8:19
add a comment |
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