Mixing
Passing an Image with Ajax
0
HTML form for uploading an Image:
<form method="post" enctype="multipart/form-data">
<div>
<input type="file" id="image">
<button type="button" class="ImgSubmitButton" onclick="uploadImages();">UPLOAD IMAGE</button>
</div>
</form>
Javascript/Ajax for sending the data.
var RequestObject = false;
if (window.XMLHttpRequest) {
RequestObject = new XMLHttpRequest();
} else if (window.ActiveXObject) {
RequestObject = new ActiveXObject("Microsoft.XMLHTTP");
}
function uploadImages() {
if (RequestObject) {
var formData = new FormData();
var myfile = document.getElementById('image').files[0];
formData.append('file', myfile);
RequestObject.open("POST", "processFileA.php");
RequestObject.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
RequestObject.onreadystatechange = function() {
if (RequestObject.readyState == 4 && RequestObject.status == 200) {
document.getElementById('err').innerHTML = RequestObject.responseText;
}
}
RequestObject.send("data=" + formData);
}
return false;
}
PHP is simple just to check if the data is set.
if(isset($_POST['data'])){
echo $_POST['data'];
echo "data is set";
} else {
echo "data is not set";
}
I've checked 3 header requests.
First:the data isn't set.
RequestObject.setRequestHeader('Content-Type','multipart/form-data');
Second:the data isn't set.
RequestObject.setRequestHeader('Content-Type', "multipart/form-data; charset=utf-8; boundary=" + Math.random().toString().substr(2));
Third:returns this [object FormData].
RequestObject.setRequestHeader('Content-Type','application/x-www-form-urlencoded');
I have also tried no header requests and data isn't set.
I know how to safely process the form through regular form submit with PHP, but not sure how to handle [object FormData] as it is passed via Ajax. If their is a better method or something I'm doing wrong please let me know. My question is how do I properly send the image file Via Ajax to process it as you would in a regular form submit to properly process it in PHP.
Please no JQuery.
javascript php ajax ajaxform
edited Jan 1 at 8:31
adiga
10.7k62444
asked Jan 1 at 8:25
JonnyJonny
9301720
add a comment |
0
HTML form for uploading an Image:
<form method="post" enctype="multipart/form-data">
<div>
<input type="file" id="image">
<button type="button" class="ImgSubmitButton" onclick="uploadImages();">UPLOAD IMAGE</button>
</div>
</form>
Javascript/Ajax for sending the data.
var RequestObject = false;
if (window.XMLHttpRequest) {
RequestObject = new XMLHttpRequest();
} else if (window.ActiveXObject) {
RequestObject = new ActiveXObject("Microsoft.XMLHTTP");
}
function uploadImages() {
if (RequestObject) {
var formData = new FormData();
var myfile = document.getElementById('image').files[0];
formData.append('file', myfile);
RequestObject.open("POST", "processFileA.php");
RequestObject.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
RequestObject.onreadystatechange = function() {
if (RequestObject.readyState == 4 && RequestObject.status == 200) {
document.getElementById('err').innerHTML = RequestObject.responseText;
}
}
RequestObject.send("data=" + formData);
}
return false;
}
PHP is simple just to check if the data is set.
if(isset($_POST['data'])){
echo $_POST['data'];
echo "data is set";
} else {
echo "data is not set";
}
I've checked 3 header requests.
First:the data isn't set.
RequestObject.setRequestHeader('Content-Type','multipart/form-data');
Second:the data isn't set.
RequestObject.setRequestHeader('Content-Type', "multipart/form-data; charset=utf-8; boundary=" + Math.random().toString().substr(2));
Third:returns this [object FormData].
RequestObject.setRequestHeader('Content-Type','application/x-www-form-urlencoded');
I have also tried no header requests and data isn't set.
I know how to safely process the form through regular form submit with PHP, but not sure how to handle [object FormData] as it is passed via Ajax. If their is a better method or something I'm doing wrong please let me know. My question is how do I properly send the image file Via Ajax to process it as you would in a regular form submit to properly process it in PHP.
Please no JQuery.
javascript php ajax ajaxform
edited Jan 1 at 8:31
adiga
10.7k62444
asked Jan 1 at 8:25
JonnyJonny
9301720
try use print_r($_FILES) and see if you file image received
– HamzaNig
Jan 1 at 8:33
I get Array ( ) 1 with print_r($_FILES) using this RequestObject.setRequestHeader('Content-Type','application/x-www-form-urlencoded'); Still isnt set with the other 2 header requests
– Jonny
Jan 1 at 8:34
add a comment |
0
0
0
HTML form for uploading an Image:
<form method="post" enctype="multipart/form-data">
<div>
<input type="file" id="image">
<button type="button" class="ImgSubmitButton" onclick="uploadImages();">UPLOAD IMAGE</button>
</div>
</form>
Javascript/Ajax for sending the data.
var RequestObject = false;
if (window.XMLHttpRequest) {
RequestObject = new XMLHttpRequest();
} else if (window.ActiveXObject) {
RequestObject = new ActiveXObject("Microsoft.XMLHTTP");
}
function uploadImages() {
if (RequestObject) {
var formData = new FormData();
var myfile = document.getElementById('image').files[0];
formData.append('file', myfile);
RequestObject.open("POST", "processFileA.php");
RequestObject.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
RequestObject.onreadystatechange = function() {
if (RequestObject.readyState == 4 && RequestObject.status == 200) {
document.getElementById('err').innerHTML = RequestObject.responseText;
}
}
RequestObject.send("data=" + formData);
}
return false;
}
PHP is simple just to check if the data is set.
if(isset($_POST['data'])){
echo $_POST['data'];
echo "data is set";
} else {
echo "data is not set";
}
I've checked 3 header requests.
First:the data isn't set.
RequestObject.setRequestHeader('Content-Type','multipart/form-data');
Second:the data isn't set.
RequestObject.setRequestHeader('Content-Type', "multipart/form-data; charset=utf-8; boundary=" + Math.random().toString().substr(2));
Third:returns this [object FormData].
RequestObject.setRequestHeader('Content-Type','application/x-www-form-urlencoded');
I have also tried no header requests and data isn't set.
I know how to safely process the form through regular form submit with PHP, but not sure how to handle [object FormData] as it is passed via Ajax. If their is a better method or something I'm doing wrong please let me know. My question is how do I properly send the image file Via Ajax to process it as you would in a regular form submit to properly process it in PHP.
Please no JQuery.
javascript php ajax ajaxform
edited Jan 1 at 8:31
adiga
10.7k62444
asked Jan 1 at 8:25
JonnyJonny
9301720
HTML form for uploading an Image:
<form method="post" enctype="multipart/form-data">
<div>
<input type="file" id="image">
<button type="button" class="ImgSubmitButton" onclick="uploadImages();">UPLOAD IMAGE</button>
</div>
</form>
Javascript/Ajax for sending the data.
var RequestObject = false;
if (window.XMLHttpRequest) {
RequestObject = new XMLHttpRequest();
} else if (window.ActiveXObject) {
RequestObject = new ActiveXObject("Microsoft.XMLHTTP");
}
function uploadImages() {
if (RequestObject) {
var formData = new FormData();
var myfile = document.getElementById('image').files[0];
formData.append('file', myfile);
RequestObject.open("POST", "processFileA.php");
RequestObject.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
RequestObject.onreadystatechange = function() {
if (RequestObject.readyState == 4 && RequestObject.status == 200) {
document.getElementById('err').innerHTML = RequestObject.responseText;
}
}
RequestObject.send("data=" + formData);
}
return false;
}
PHP is simple just to check if the data is set.
if(isset($_POST['data'])){
echo $_POST['data'];
echo "data is set";
} else {
echo "data is not set";
}
I've checked 3 header requests.
First:the data isn't set.
RequestObject.setRequestHeader('Content-Type','multipart/form-data');
Second:the data isn't set.
RequestObject.setRequestHeader('Content-Type', "multipart/form-data; charset=utf-8; boundary=" + Math.random().toString().substr(2));
Third:returns this [object FormData].
RequestObject.setRequestHeader('Content-Type','application/x-www-form-urlencoded');
I have also tried no header requests and data isn't set.
I know how to safely process the form through regular form submit with PHP, but not sure how to handle [object FormData] as it is passed via Ajax. If their is a better method or something I'm doing wrong please let me know. My question is how do I properly send the image file Via Ajax to process it as you would in a regular form submit to properly process it in PHP.
Please no JQuery.
javascript php ajax ajaxform
javascript php ajax ajaxform
edited Jan 1 at 8:31
adiga
10.7k62444
asked Jan 1 at 8:25
JonnyJonny
9301720
edited Jan 1 at 8:31
adiga
10.7k62444
asked Jan 1 at 8:25
JonnyJonny
9301720
edited Jan 1 at 8:31
adiga
10.7k62444
edited Jan 1 at 8:31
adiga
10.7k62444
edited Jan 1 at 8:31
adiga
10.7k62444
10.7k62444
asked Jan 1 at 8:25
JonnyJonny
9301720
asked Jan 1 at 8:25
JonnyJonny
9301720
asked Jan 1 at 8:25
JonnyJonny
9301720
9301720
try use print_r($_FILES) and see if you file image received
– HamzaNig
Jan 1 at 8:33
I get Array ( ) 1 with print_r($_FILES) using this RequestObject.setRequestHeader('Content-Type','application/x-www-form-urlencoded'); Still isnt set with the other 2 header requests
– Jonny
Jan 1 at 8:34
add a comment |
try use print_r($_FILES) and see if you file image received
– HamzaNig
Jan 1 at 8:33
I get Array ( ) 1 with print_r($_FILES) using this RequestObject.setRequestHeader('Content-Type','application/x-www-form-urlencoded'); Still isnt set with the other 2 header requests
– Jonny
Jan 1 at 8:34
try use print_r($_FILES) and see if you file image received
– HamzaNig
Jan 1 at 8:33
try use print_r($_FILES) and see if you file image received
– HamzaNig
Jan 1 at 8:33
I get Array ( ) 1 with print_r($_FILES) using this RequestObject.setRequestHeader('Content-Type','application/x-www-form-urlencoded'); Still isnt set with the other 2 header requests
– Jonny
Jan 1 at 8:34
I get Array ( ) 1 with print_r($_FILES) using this RequestObject.setRequestHeader('Content-Type','application/x-www-form-urlencoded'); Still isnt set with the other 2 header requests
– Jonny
Jan 1 at 8:34
add a comment |
2 Answers
2
active
oldest
votes
1
your problem in this string:
RequestObject.send("data=" + formData);
when you try String + formData, you did concatenation, and formData convert to String to, as we know formData it's object.
this is correct approach to send data:
RequestObject.send(formData);
just send a data like in this examples: https://developer.mozilla.org/en-US/docs/Learn/HTML/Forms/Sending_forms_through_JavaScript
answered Jan 1 at 8:41
Vadim HulevichVadim Hulevich
775111
I get this Warning: Unknown: Input variables exceeded 2500. To increase the limit change max_input_vars in php.ini. in Unknown on line 0 and it returned data is not set.
– Jonny
Jan 1 at 8:44
can you try with small image
– Vadim Hulevich
Jan 1 at 8:46
try in php var_dump($_POST) and see what inside
– Vadim Hulevich
Jan 1 at 8:47
yes i just did no warning but it returned data is not set
– Jonny
Jan 1 at 8:47
ok, what inside var_dump($_POST) and var_dump($_FILES)
– Vadim Hulevich
Jan 1 at 8:49
|
show 6 more comments
1
You can use diffrent way by ajax see example bellow :
THIS ANSWER WITH JUST JAVASCRIPT AJAX
<html>
<head>
<script >
function uploadImages(){
var xhr = new XMLHttpRequest();
var url = "processFileA.php";
xhr.open("POST", url, true);
//xhr.setRequestHeader("Content-Type", "application/json");
//xhr.setRequestHeader("accept", "application/json, text/plain, */*");
xhr.onreadystatechange = function () {
if (xhr.readyState === 4 && xhr.status === 200) {
var content= xhr.responseText;
console.log(content);
}
};
var datae=document.getElementById('uploadimage');
var data = new FormData(datae);
xhr.send(data);
}
</script>
</head>
<body>
<form id="uploadimage" method="post" enctype="multipart/form-data">
<div>
<input type="file" name="file" id="image">
<button type="button" class="ImgSubmitButton" onclick="uploadImages();">UPLOAD IMAGE</button>
</div>
</form>
</body>
</html>
THIS ANSWER USING JQUERY AJAX
<link href='http://fonts.googleapis.com/css?family=Roboto+Condensed|Open+Sans+Condensed:300' rel='stylesheet' type='text/css'>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script >
$(document).ready(function (e) {
$("#uploadimage").on('submit',(function(e) {
e.preventDefault();
$.ajax({
url: "processFileA.php", // Url to which the request is send
type: "POST", // Type of request to be send, called as method
data: new FormData(this), // Data sent to server, a set of key/value pairs (i.e. form fields and values)
contentType: false, // The content type used when sending data to the server.
cache: false, // To unable request pages to be cached
processData:false, // To send DOMDocument or non processed data file it is set to false
success: function(data) // A function to be called if request succeeds
{
alert('done')
}
});
}));
});
</script>
</head>
<body>
<form id="uploadimage" method="post" enctype="multipart/form-data">
<div>
<input type="file" name="file" id="image">
<button type="submit" class="ImgSubmitButton" >UPLOAD
IMAGE</button>
</div>
</form>
</body>
</html>
PHP FILE : processFileA.php
<?php
if(isset($_FILES["file"]["type"]))
{
$file=(file_get_contents($_FILES["file"]['tmp_name']));
file_put_contents('tmp_name.'.str_replace('image/','',$_FILES["file"]["type"]),$file);
}
?>
answered Jan 1 at 9:23
HamzaNigHamzaNig
810531
cant use jquery
– Jonny
Jan 1 at 9:24
@Jonny ah you using just javascript ok i updated my code check it
– HamzaNig
Jan 1 at 9:49
I get data not set with this solution.
– Jonny
Jan 1 at 23:35
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
your problem in this string:
RequestObject.send("data=" + formData);
when you try String + formData, you did concatenation, and formData convert to String to, as we know formData it's object.
this is correct approach to send data:
RequestObject.send(formData);
just send a data like in this examples: https://developer.mozilla.org/en-US/docs/Learn/HTML/Forms/Sending_forms_through_JavaScript
answered Jan 1 at 8:41
Vadim HulevichVadim Hulevich
775111
I get this Warning: Unknown: Input variables exceeded 2500. To increase the limit change max_input_vars in php.ini. in Unknown on line 0 and it returned data is not set.
– Jonny
Jan 1 at 8:44
can you try with small image
– Vadim Hulevich
Jan 1 at 8:46
try in php var_dump($_POST) and see what inside
– Vadim Hulevich
Jan 1 at 8:47
yes i just did no warning but it returned data is not set
– Jonny
Jan 1 at 8:47
ok, what inside var_dump($_POST) and var_dump($_FILES)
– Vadim Hulevich
Jan 1 at 8:49
|
show 6 more comments
1
your problem in this string:
RequestObject.send("data=" + formData);
when you try String + formData, you did concatenation, and formData convert to String to, as we know formData it's object.
this is correct approach to send data:
RequestObject.send(formData);
just send a data like in this examples: https://developer.mozilla.org/en-US/docs/Learn/HTML/Forms/Sending_forms_through_JavaScript
answered Jan 1 at 8:41
Vadim HulevichVadim Hulevich
775111
I get this Warning: Unknown: Input variables exceeded 2500. To increase the limit change max_input_vars in php.ini. in Unknown on line 0 and it returned data is not set.
– Jonny
Jan 1 at 8:44
can you try with small image
– Vadim Hulevich
Jan 1 at 8:46
try in php var_dump($_POST) and see what inside
– Vadim Hulevich
Jan 1 at 8:47
yes i just did no warning but it returned data is not set
– Jonny
Jan 1 at 8:47
ok, what inside var_dump($_POST) and var_dump($_FILES)
– Vadim Hulevich
Jan 1 at 8:49
|
show 6 more comments
1
1
1
your problem in this string:
RequestObject.send("data=" + formData);
when you try String + formData, you did concatenation, and formData convert to String to, as we know formData it's object.
this is correct approach to send data:
RequestObject.send(formData);
just send a data like in this examples: https://developer.mozilla.org/en-US/docs/Learn/HTML/Forms/Sending_forms_through_JavaScript
answered Jan 1 at 8:41
Vadim HulevichVadim Hulevich
775111
your problem in this string:
RequestObject.send("data=" + formData);
when you try String + formData, you did concatenation, and formData convert to String to, as we know formData it's object.
this is correct approach to send data:
RequestObject.send(formData);
just send a data like in this examples: https://developer.mozilla.org/en-US/docs/Learn/HTML/Forms/Sending_forms_through_JavaScript
answered Jan 1 at 8:41
Vadim HulevichVadim Hulevich
775111
answered Jan 1 at 8:41
Vadim HulevichVadim Hulevich
775111
answered Jan 1 at 8:41
Vadim HulevichVadim Hulevich
775111
answered Jan 1 at 8:41
Vadim HulevichVadim Hulevich
775111
775111
I get this Warning: Unknown: Input variables exceeded 2500. To increase the limit change max_input_vars in php.ini. in Unknown on line 0 and it returned data is not set.
– Jonny
Jan 1 at 8:44
can you try with small image
– Vadim Hulevich
Jan 1 at 8:46
try in php var_dump($_POST) and see what inside
– Vadim Hulevich
Jan 1 at 8:47
yes i just did no warning but it returned data is not set
– Jonny
Jan 1 at 8:47
ok, what inside var_dump($_POST) and var_dump($_FILES)
– Vadim Hulevich
Jan 1 at 8:49
|
show 6 more comments
I get this Warning: Unknown: Input variables exceeded 2500. To increase the limit change max_input_vars in php.ini. in Unknown on line 0 and it returned data is not set.
– Jonny
Jan 1 at 8:44
can you try with small image
– Vadim Hulevich
Jan 1 at 8:46
try in php var_dump($_POST) and see what inside
– Vadim Hulevich
Jan 1 at 8:47
yes i just did no warning but it returned data is not set
– Jonny
Jan 1 at 8:47
ok, what inside var_dump($_POST) and var_dump($_FILES)
– Vadim Hulevich
Jan 1 at 8:49
I get this Warning: Unknown: Input variables exceeded 2500. To increase the limit change max_input_vars in php.ini. in Unknown on line 0 and it returned data is not set.
– Jonny
Jan 1 at 8:44
I get this Warning: Unknown: Input variables exceeded 2500. To increase the limit change max_input_vars in php.ini. in Unknown on line 0 and it returned data is not set.
– Jonny
Jan 1 at 8:44
can you try with small image
– Vadim Hulevich
Jan 1 at 8:46
can you try with small image
– Vadim Hulevich
Jan 1 at 8:46
try in php var_dump($_POST) and see what inside
– Vadim Hulevich
Jan 1 at 8:47
try in php var_dump($_POST) and see what inside
– Vadim Hulevich
Jan 1 at 8:47
yes i just did no warning but it returned data is not set
– Jonny
Jan 1 at 8:47
yes i just did no warning but it returned data is not set
– Jonny
Jan 1 at 8:47
ok, what inside var_dump($_POST) and var_dump($_FILES)
– Vadim Hulevich
Jan 1 at 8:49
ok, what inside var_dump($_POST) and var_dump($_FILES)
– Vadim Hulevich
Jan 1 at 8:49
|
show 6 more comments
1
You can use diffrent way by ajax see example bellow :
THIS ANSWER WITH JUST JAVASCRIPT AJAX
<html>
<head>
<script >
function uploadImages(){
var xhr = new XMLHttpRequest();
var url = "processFileA.php";
xhr.open("POST", url, true);
//xhr.setRequestHeader("Content-Type", "application/json");
//xhr.setRequestHeader("accept", "application/json, text/plain, */*");
xhr.onreadystatechange = function () {
if (xhr.readyState === 4 && xhr.status === 200) {
var content= xhr.responseText;
console.log(content);
}
};
var datae=document.getElementById('uploadimage');
var data = new FormData(datae);
xhr.send(data);
}
</script>
</head>
<body>
<form id="uploadimage" method="post" enctype="multipart/form-data">
<div>
<input type="file" name="file" id="image">
<button type="button" class="ImgSubmitButton" onclick="uploadImages();">UPLOAD IMAGE</button>
</div>
</form>
</body>
</html>
THIS ANSWER USING JQUERY AJAX
<link href='http://fonts.googleapis.com/css?family=Roboto+Condensed|Open+Sans+Condensed:300' rel='stylesheet' type='text/css'>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script >
$(document).ready(function (e) {
$("#uploadimage").on('submit',(function(e) {
e.preventDefault();
$.ajax({
url: "processFileA.php", // Url to which the request is send
type: "POST", // Type of request to be send, called as method
data: new FormData(this), // Data sent to server, a set of key/value pairs (i.e. form fields and values)
contentType: false, // The content type used when sending data to the server.
cache: false, // To unable request pages to be cached
processData:false, // To send DOMDocument or non processed data file it is set to false
success: function(data) // A function to be called if request succeeds
{
alert('done')
}
});
}));
});
</script>
</head>
<body>
<form id="uploadimage" method="post" enctype="multipart/form-data">
<div>
<input type="file" name="file" id="image">
<button type="submit" class="ImgSubmitButton" >UPLOAD
IMAGE</button>
</div>
</form>
</body>
</html>
PHP FILE : processFileA.php
<?php
if(isset($_FILES["file"]["type"]))
{
$file=(file_get_contents($_FILES["file"]['tmp_name']));
file_put_contents('tmp_name.'.str_replace('image/','',$_FILES["file"]["type"]),$file);
}
?>
answered Jan 1 at 9:23
HamzaNigHamzaNig
810531
cant use jquery
– Jonny
Jan 1 at 9:24
@Jonny ah you using just javascript ok i updated my code check it
– HamzaNig
Jan 1 at 9:49
I get data not set with this solution.
– Jonny
Jan 1 at 23:35
add a comment |
1
You can use diffrent way by ajax see example bellow :
THIS ANSWER WITH JUST JAVASCRIPT AJAX
<html>
<head>
<script >
function uploadImages(){
var xhr = new XMLHttpRequest();
var url = "processFileA.php";
xhr.open("POST", url, true);
//xhr.setRequestHeader("Content-Type", "application/json");
//xhr.setRequestHeader("accept", "application/json, text/plain, */*");
xhr.onreadystatechange = function () {
if (xhr.readyState === 4 && xhr.status === 200) {
var content= xhr.responseText;
console.log(content);
}
};
var datae=document.getElementById('uploadimage');
var data = new FormData(datae);
xhr.send(data);
}
</script>
</head>
<body>
<form id="uploadimage" method="post" enctype="multipart/form-data">
<div>
<input type="file" name="file" id="image">
<button type="button" class="ImgSubmitButton" onclick="uploadImages();">UPLOAD IMAGE</button>
</div>
</form>
</body>
</html>
THIS ANSWER USING JQUERY AJAX
<link href='http://fonts.googleapis.com/css?family=Roboto+Condensed|Open+Sans+Condensed:300' rel='stylesheet' type='text/css'>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script >
$(document).ready(function (e) {
$("#uploadimage").on('submit',(function(e) {
e.preventDefault();
$.ajax({
url: "processFileA.php", // Url to which the request is send
type: "POST", // Type of request to be send, called as method
data: new FormData(this), // Data sent to server, a set of key/value pairs (i.e. form fields and values)
contentType: false, // The content type used when sending data to the server.
cache: false, // To unable request pages to be cached
processData:false, // To send DOMDocument or non processed data file it is set to false
success: function(data) // A function to be called if request succeeds
{
alert('done')
}
});
}));
});
</script>
</head>
<body>
<form id="uploadimage" method="post" enctype="multipart/form-data">
<div>
<input type="file" name="file" id="image">
<button type="submit" class="ImgSubmitButton" >UPLOAD
IMAGE</button>
</div>
</form>
</body>
</html>
PHP FILE : processFileA.php
<?php
if(isset($_FILES["file"]["type"]))
{
$file=(file_get_contents($_FILES["file"]['tmp_name']));
file_put_contents('tmp_name.'.str_replace('image/','',$_FILES["file"]["type"]),$file);
}
?>
answered Jan 1 at 9:23
HamzaNigHamzaNig
810531
cant use jquery
– Jonny
Jan 1 at 9:24
@Jonny ah you using just javascript ok i updated my code check it
– HamzaNig
Jan 1 at 9:49
I get data not set with this solution.
– Jonny
Jan 1 at 23:35
add a comment |
1
1
1
You can use diffrent way by ajax see example bellow :
THIS ANSWER WITH JUST JAVASCRIPT AJAX
<html>
<head>
<script >
function uploadImages(){
var xhr = new XMLHttpRequest();
var url = "processFileA.php";
xhr.open("POST", url, true);
//xhr.setRequestHeader("Content-Type", "application/json");
//xhr.setRequestHeader("accept", "application/json, text/plain, */*");
xhr.onreadystatechange = function () {
if (xhr.readyState === 4 && xhr.status === 200) {
var content= xhr.responseText;
console.log(content);
}
};
var datae=document.getElementById('uploadimage');
var data = new FormData(datae);
xhr.send(data);
}
</script>
</head>
<body>
<form id="uploadimage" method="post" enctype="multipart/form-data">
<div>
<input type="file" name="file" id="image">
<button type="button" class="ImgSubmitButton" onclick="uploadImages();">UPLOAD IMAGE</button>
</div>
</form>
</body>
</html>
THIS ANSWER USING JQUERY AJAX
<link href='http://fonts.googleapis.com/css?family=Roboto+Condensed|Open+Sans+Condensed:300' rel='stylesheet' type='text/css'>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script >
$(document).ready(function (e) {
$("#uploadimage").on('submit',(function(e) {
e.preventDefault();
$.ajax({
url: "processFileA.php", // Url to which the request is send
type: "POST", // Type of request to be send, called as method
data: new FormData(this), // Data sent to server, a set of key/value pairs (i.e. form fields and values)
contentType: false, // The content type used when sending data to the server.
cache: false, // To unable request pages to be cached
processData:false, // To send DOMDocument or non processed data file it is set to false
success: function(data) // A function to be called if request succeeds
{
alert('done')
}
});
}));
});
</script>
</head>
<body>
<form id="uploadimage" method="post" enctype="multipart/form-data">
<div>
<input type="file" name="file" id="image">
<button type="submit" class="ImgSubmitButton" >UPLOAD
IMAGE</button>
</div>
</form>
</body>
</html>
PHP FILE : processFileA.php
<?php
if(isset($_FILES["file"]["type"]))
{
$file=(file_get_contents($_FILES["file"]['tmp_name']));
file_put_contents('tmp_name.'.str_replace('image/','',$_FILES["file"]["type"]),$file);
}
?>
answered Jan 1 at 9:23
HamzaNigHamzaNig
810531
You can use diffrent way by ajax see example bellow :
THIS ANSWER WITH JUST JAVASCRIPT AJAX
<html>
<head>
<script >
function uploadImages(){
var xhr = new XMLHttpRequest();
var url = "processFileA.php";
xhr.open("POST", url, true);
//xhr.setRequestHeader("Content-Type", "application/json");
//xhr.setRequestHeader("accept", "application/json, text/plain, */*");
xhr.onreadystatechange = function () {
if (xhr.readyState === 4 && xhr.status === 200) {
var content= xhr.responseText;
console.log(content);
}
};
var datae=document.getElementById('uploadimage');
var data = new FormData(datae);
xhr.send(data);
}
</script>
</head>
<body>
<form id="uploadimage" method="post" enctype="multipart/form-data">
<div>
<input type="file" name="file" id="image">
<button type="button" class="ImgSubmitButton" onclick="uploadImages();">UPLOAD IMAGE</button>
</div>
</form>
</body>
</html>
THIS ANSWER USING JQUERY AJAX
<link href='http://fonts.googleapis.com/css?family=Roboto+Condensed|Open+Sans+Condensed:300' rel='stylesheet' type='text/css'>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script >
$(document).ready(function (e) {
$("#uploadimage").on('submit',(function(e) {
e.preventDefault();
$.ajax({
url: "processFileA.php", // Url to which the request is send
type: "POST", // Type of request to be send, called as method
data: new FormData(this), // Data sent to server, a set of key/value pairs (i.e. form fields and values)
contentType: false, // The content type used when sending data to the server.
cache: false, // To unable request pages to be cached
processData:false, // To send DOMDocument or non processed data file it is set to false
success: function(data) // A function to be called if request succeeds
{
alert('done')
}
});
}));
});
</script>
</head>
<body>
<form id="uploadimage" method="post" enctype="multipart/form-data">
<div>
<input type="file" name="file" id="image">
<button type="submit" class="ImgSubmitButton" >UPLOAD
IMAGE</button>
</div>
</form>
</body>
</html>
PHP FILE : processFileA.php
<?php
if(isset($_FILES["file"]["type"]))
{
$file=(file_get_contents($_FILES["file"]['tmp_name']));
file_put_contents('tmp_name.'.str_replace('image/','',$_FILES["file"]["type"]),$file);
}
?>
answered Jan 1 at 9:23
HamzaNigHamzaNig
810531
edited Jan 1 at 9:48
answered Jan 1 at 9:23
HamzaNigHamzaNig
810531
answered Jan 1 at 9:23
HamzaNigHamzaNig
810531
answered Jan 1 at 9:23
HamzaNigHamzaNig
810531
810531
cant use jquery
– Jonny
Jan 1 at 9:24
@Jonny ah you using just javascript ok i updated my code check it
– HamzaNig
Jan 1 at 9:49
I get data not set with this solution.
– Jonny
Jan 1 at 23:35
add a comment |
cant use jquery
– Jonny
Jan 1 at 9:24
@Jonny ah you using just javascript ok i updated my code check it
– HamzaNig
Jan 1 at 9:49
I get data not set with this solution.
– Jonny
Jan 1 at 23:35
cant use jquery
– Jonny
Jan 1 at 9:24
cant use jquery
– Jonny
Jan 1 at 9:24
@Jonny ah you using just javascript ok i updated my code check it
– HamzaNig
Jan 1 at 9:49
@Jonny ah you using just javascript ok i updated my code check it
– HamzaNig
Jan 1 at 9:49
I get data not set with this solution.
– Jonny
Jan 1 at 23:35
I get data not set with this solution.
– Jonny
Jan 1 at 23:35
add a comment |
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try use print_r($_FILES) and see if you file image received
– HamzaNig
Jan 1 at 8:33
I get Array ( ) 1 with print_r($_FILES) using this RequestObject.setRequestHeader('Content-Type','application/x-www-form-urlencoded'); Still isnt set with the other 2 header requests
– Jonny
Jan 1 at 8:34