Uploading both data and files in one form using POST method, AJAX and jQuery (Error)





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1















I have a problem. formData.append("id_element",1) doesn`t add the field into formData. I have found a lot of solves but in my case it doesn`t work right.



PHP code here:



var_dump($_FILES);





$("#hw-upload_image-form").submit(function(e) {
e.preventDefault();

var formData = new FormData(this);
formData.append("id_element",1); // doesn`t work here
$.ajax({
type:"POST",
processData: false,
contentType: false,
cache: false,
url:$(this).prop('action'),
data:formData,
success:function (data) {
console.log(data); // show returned data from php
}
});

});

<form action="action.php"  method="post" enctype="multipart/form-data" id="hw-upload_image-form">
<input type="file" name="hwImage"> <!-- hw = homework (just for you :) -->
</form>







Result (from the console)



array(1) {
["hwImage"]=>
array(5) {
["name"]=>
string(70) "73b38ef5d1f5849ea800c18990acde94_ce_1920x1200x0x0_cropped_800x427.jpeg"
["type"]=>
string(10) "image/jpeg"
["tmp_name"]=>
string(36) "D:OSPaneluserdatatempphp9F48.tmp"
["error"]=>
int(0)
["size"]=>
int(68411)
}
}









share|improve this question

























  • Possible duplicate of Uploading both data and files in one form using Ajax?

    – ttrasn
    Jan 3 at 21:10











  • You removed (stealth edit) a console.log() line that would actually be quite helpful in debugging. What was the result/output of that line?

    – Patrick Q
    Jan 3 at 21:10











  • I removed it, because the output was: ibb.co/KN3bMyC (random site for share images)

    – Nekear
    Jan 4 at 13:05




















1















I have a problem. formData.append("id_element",1) doesn`t add the field into formData. I have found a lot of solves but in my case it doesn`t work right.



PHP code here:



var_dump($_FILES);





$("#hw-upload_image-form").submit(function(e) {
e.preventDefault();

var formData = new FormData(this);
formData.append("id_element",1); // doesn`t work here
$.ajax({
type:"POST",
processData: false,
contentType: false,
cache: false,
url:$(this).prop('action'),
data:formData,
success:function (data) {
console.log(data); // show returned data from php
}
});

});

<form action="action.php"  method="post" enctype="multipart/form-data" id="hw-upload_image-form">
<input type="file" name="hwImage"> <!-- hw = homework (just for you :) -->
</form>







Result (from the console)



array(1) {
["hwImage"]=>
array(5) {
["name"]=>
string(70) "73b38ef5d1f5849ea800c18990acde94_ce_1920x1200x0x0_cropped_800x427.jpeg"
["type"]=>
string(10) "image/jpeg"
["tmp_name"]=>
string(36) "D:OSPaneluserdatatempphp9F48.tmp"
["error"]=>
int(0)
["size"]=>
int(68411)
}
}









share|improve this question

























  • Possible duplicate of Uploading both data and files in one form using Ajax?

    – ttrasn
    Jan 3 at 21:10











  • You removed (stealth edit) a console.log() line that would actually be quite helpful in debugging. What was the result/output of that line?

    – Patrick Q
    Jan 3 at 21:10











  • I removed it, because the output was: ibb.co/KN3bMyC (random site for share images)

    – Nekear
    Jan 4 at 13:05
















1












1








1








I have a problem. formData.append("id_element",1) doesn`t add the field into formData. I have found a lot of solves but in my case it doesn`t work right.



PHP code here:



var_dump($_FILES);





$("#hw-upload_image-form").submit(function(e) {
e.preventDefault();

var formData = new FormData(this);
formData.append("id_element",1); // doesn`t work here
$.ajax({
type:"POST",
processData: false,
contentType: false,
cache: false,
url:$(this).prop('action'),
data:formData,
success:function (data) {
console.log(data); // show returned data from php
}
});

});

<form action="action.php"  method="post" enctype="multipart/form-data" id="hw-upload_image-form">
<input type="file" name="hwImage"> <!-- hw = homework (just for you :) -->
</form>







Result (from the console)



array(1) {
["hwImage"]=>
array(5) {
["name"]=>
string(70) "73b38ef5d1f5849ea800c18990acde94_ce_1920x1200x0x0_cropped_800x427.jpeg"
["type"]=>
string(10) "image/jpeg"
["tmp_name"]=>
string(36) "D:OSPaneluserdatatempphp9F48.tmp"
["error"]=>
int(0)
["size"]=>
int(68411)
}
}









share|improve this question
















I have a problem. formData.append("id_element",1) doesn`t add the field into formData. I have found a lot of solves but in my case it doesn`t work right.



PHP code here:



var_dump($_FILES);





$("#hw-upload_image-form").submit(function(e) {
e.preventDefault();

var formData = new FormData(this);
formData.append("id_element",1); // doesn`t work here
$.ajax({
type:"POST",
processData: false,
contentType: false,
cache: false,
url:$(this).prop('action'),
data:formData,
success:function (data) {
console.log(data); // show returned data from php
}
});

});

<form action="action.php"  method="post" enctype="multipart/form-data" id="hw-upload_image-form">
<input type="file" name="hwImage"> <!-- hw = homework (just for you :) -->
</form>







Result (from the console)



array(1) {
["hwImage"]=>
array(5) {
["name"]=>
string(70) "73b38ef5d1f5849ea800c18990acde94_ce_1920x1200x0x0_cropped_800x427.jpeg"
["type"]=>
string(10) "image/jpeg"
["tmp_name"]=>
string(36) "D:OSPaneluserdatatempphp9F48.tmp"
["error"]=>
int(0)
["size"]=>
int(68411)
}
}





$("#hw-upload_image-form").submit(function(e) {
e.preventDefault();

var formData = new FormData(this);
formData.append("id_element",1); // doesn`t work here
$.ajax({
type:"POST",
processData: false,
contentType: false,
cache: false,
url:$(this).prop('action'),
data:formData,
success:function (data) {
console.log(data); // show returned data from php
}
});

});

<form action="action.php"  method="post" enctype="multipart/form-data" id="hw-upload_image-form">
<input type="file" name="hwImage"> <!-- hw = homework (just for you :) -->
</form>





$("#hw-upload_image-form").submit(function(e) {
e.preventDefault();

var formData = new FormData(this);
formData.append("id_element",1); // doesn`t work here
$.ajax({
type:"POST",
processData: false,
contentType: false,
cache: false,
url:$(this).prop('action'),
data:formData,
success:function (data) {
console.log(data); // show returned data from php
}
});

});

<form action="action.php"  method="post" enctype="multipart/form-data" id="hw-upload_image-form">
<input type="file" name="hwImage"> <!-- hw = homework (just for you :) -->
</form>






javascript php jquery ajax






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edited Jan 4 at 15:36









wanttobeprofessional

1,02931323




1,02931323










asked Jan 3 at 21:02









NekearNekear

62




62













  • Possible duplicate of Uploading both data and files in one form using Ajax?

    – ttrasn
    Jan 3 at 21:10











  • You removed (stealth edit) a console.log() line that would actually be quite helpful in debugging. What was the result/output of that line?

    – Patrick Q
    Jan 3 at 21:10











  • I removed it, because the output was: ibb.co/KN3bMyC (random site for share images)

    – Nekear
    Jan 4 at 13:05





















  • Possible duplicate of Uploading both data and files in one form using Ajax?

    – ttrasn
    Jan 3 at 21:10











  • You removed (stealth edit) a console.log() line that would actually be quite helpful in debugging. What was the result/output of that line?

    – Patrick Q
    Jan 3 at 21:10











  • I removed it, because the output was: ibb.co/KN3bMyC (random site for share images)

    – Nekear
    Jan 4 at 13:05



















Possible duplicate of Uploading both data and files in one form using Ajax?

– ttrasn
Jan 3 at 21:10





Possible duplicate of Uploading both data and files in one form using Ajax?

– ttrasn
Jan 3 at 21:10













You removed (stealth edit) a console.log() line that would actually be quite helpful in debugging. What was the result/output of that line?

– Patrick Q
Jan 3 at 21:10





You removed (stealth edit) a console.log() line that would actually be quite helpful in debugging. What was the result/output of that line?

– Patrick Q
Jan 3 at 21:10













I removed it, because the output was: ibb.co/KN3bMyC (random site for share images)

– Nekear
Jan 4 at 13:05







I removed it, because the output was: ibb.co/KN3bMyC (random site for share images)

– Nekear
Jan 4 at 13:05














1 Answer
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active

oldest

votes


















0














I have already undestood my mistake.



$_FILES will show me only file data, but if I want to see id_element I have to use $_POST.



The result is:



array(1) {
["id_element"]=>
string(1) "1"
}





share|improve this answer
























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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0














    I have already undestood my mistake.



    $_FILES will show me only file data, but if I want to see id_element I have to use $_POST.



    The result is:



    array(1) {
    ["id_element"]=>
    string(1) "1"
    }





    share|improve this answer




























      0














      I have already undestood my mistake.



      $_FILES will show me only file data, but if I want to see id_element I have to use $_POST.



      The result is:



      array(1) {
      ["id_element"]=>
      string(1) "1"
      }





      share|improve this answer


























        0












        0








        0







        I have already undestood my mistake.



        $_FILES will show me only file data, but if I want to see id_element I have to use $_POST.



        The result is:



        array(1) {
        ["id_element"]=>
        string(1) "1"
        }





        share|improve this answer













        I have already undestood my mistake.



        $_FILES will show me only file data, but if I want to see id_element I have to use $_POST.



        The result is:



        array(1) {
        ["id_element"]=>
        string(1) "1"
        }






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jan 4 at 13:17









        NekearNekear

        62




        62
































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