Uploading both data and files in one form using POST method, AJAX and jQuery (Error)
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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)
}
}
javascript php jquery ajax
add a comment |
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)
}
}
javascript php jquery ajax
Possible duplicate of Uploading both data and files in one form using Ajax?
– ttrasn
Jan 3 at 21:10
You removed (stealth edit) aconsole.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
add a comment |
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)
}
}
javascript php jquery ajax
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
javascript php jquery ajax
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) aconsole.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
add a comment |
Possible duplicate of Uploading both data and files in one form using Ajax?
– ttrasn
Jan 3 at 21:10
You removed (stealth edit) aconsole.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
add a comment |
1 Answer
1
active
oldest
votes
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"
}
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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"
}
add a comment |
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"
}
add a comment |
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"
}
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"
}
answered Jan 4 at 13:17
NekearNekear
62
62
add a comment |
add a comment |
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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