Mixing
Upload File With Ajax XmlHttpRequest
Hi i am trying to send file with xmlhttprequest with this code.
<script>
var url= "http://localhost:80/....";
$(document).ready(function(){
document.getElementById('upload').addEventListener('change', function(e) {
var file = this.files[0];
var xhr = new XMLHttpRequest();
xhr.file = file; // not necessary if you create scopes like this
xhr.addEventListener('progress', function(e) {
var done = e.position || e.loaded, total = e.totalSize || e.total;
console.log('xhr progress: ' + (Math.floor(done/total*1000)/10) + '%');
}, false);
if ( xhr.upload ) {
xhr.upload.onprogress = function(e) {
var done = e.position || e.loaded, total = e.totalSize || e.total;
console.log('xhr.upload progress: ' + done + ' / ' + total + ' = ' + (Math.floor(done/total*1000)/10) + '%');
};
}
xhr.onreadystatechange = function(e) {
if ( 4 == this.readyState ) {
console.log(['xhr upload complete', e]);
}
};
xhr.open('post', url, true);
xhr.setRequestHeader("Content-Type","multipart/form-data");
xhr.send(file);
}, false);
});
</script>
but i got this error : the request was rejected because no multipart boundary was found
help me pls..
javascript ajax jquery file-upload xmlhttprequest
asked Jun 2 '11 at 6:19
Sedat BaşarSedat Başar
2,30121828
add a comment |
Hi i am trying to send file with xmlhttprequest with this code.
<script>
var url= "http://localhost:80/....";
$(document).ready(function(){
document.getElementById('upload').addEventListener('change', function(e) {
var file = this.files[0];
var xhr = new XMLHttpRequest();
xhr.file = file; // not necessary if you create scopes like this
xhr.addEventListener('progress', function(e) {
var done = e.position || e.loaded, total = e.totalSize || e.total;
console.log('xhr progress: ' + (Math.floor(done/total*1000)/10) + '%');
}, false);
if ( xhr.upload ) {
xhr.upload.onprogress = function(e) {
var done = e.position || e.loaded, total = e.totalSize || e.total;
console.log('xhr.upload progress: ' + done + ' / ' + total + ' = ' + (Math.floor(done/total*1000)/10) + '%');
};
}
xhr.onreadystatechange = function(e) {
if ( 4 == this.readyState ) {
console.log(['xhr upload complete', e]);
}
};
xhr.open('post', url, true);
xhr.setRequestHeader("Content-Type","multipart/form-data");
xhr.send(file);
}, false);
});
</script>
but i got this error : the request was rejected because no multipart boundary was found
help me pls..
javascript ajax jquery file-upload xmlhttprequest
asked Jun 2 '11 at 6:19
Sedat BaşarSedat Başar
2,30121828
add a comment |
Hi i am trying to send file with xmlhttprequest with this code.
<script>
var url= "http://localhost:80/....";
$(document).ready(function(){
document.getElementById('upload').addEventListener('change', function(e) {
var file = this.files[0];
var xhr = new XMLHttpRequest();
xhr.file = file; // not necessary if you create scopes like this
xhr.addEventListener('progress', function(e) {
var done = e.position || e.loaded, total = e.totalSize || e.total;
console.log('xhr progress: ' + (Math.floor(done/total*1000)/10) + '%');
}, false);
if ( xhr.upload ) {
xhr.upload.onprogress = function(e) {
var done = e.position || e.loaded, total = e.totalSize || e.total;
console.log('xhr.upload progress: ' + done + ' / ' + total + ' = ' + (Math.floor(done/total*1000)/10) + '%');
};
}
xhr.onreadystatechange = function(e) {
if ( 4 == this.readyState ) {
console.log(['xhr upload complete', e]);
}
};
xhr.open('post', url, true);
xhr.setRequestHeader("Content-Type","multipart/form-data");
xhr.send(file);
}, false);
});
</script>
but i got this error : the request was rejected because no multipart boundary was found
help me pls..
javascript ajax jquery file-upload xmlhttprequest
asked Jun 2 '11 at 6:19
Sedat BaşarSedat Başar
2,30121828
Hi i am trying to send file with xmlhttprequest with this code.
<script>
var url= "http://localhost:80/....";
$(document).ready(function(){
document.getElementById('upload').addEventListener('change', function(e) {
var file = this.files[0];
var xhr = new XMLHttpRequest();
xhr.file = file; // not necessary if you create scopes like this
xhr.addEventListener('progress', function(e) {
var done = e.position || e.loaded, total = e.totalSize || e.total;
console.log('xhr progress: ' + (Math.floor(done/total*1000)/10) + '%');
}, false);
if ( xhr.upload ) {
xhr.upload.onprogress = function(e) {
var done = e.position || e.loaded, total = e.totalSize || e.total;
console.log('xhr.upload progress: ' + done + ' / ' + total + ' = ' + (Math.floor(done/total*1000)/10) + '%');
};
}
xhr.onreadystatechange = function(e) {
if ( 4 == this.readyState ) {
console.log(['xhr upload complete', e]);
}
};
xhr.open('post', url, true);
xhr.setRequestHeader("Content-Type","multipart/form-data");
xhr.send(file);
}, false);
});
</script>
but i got this error : the request was rejected because no multipart boundary was found
help me pls..
javascript ajax jquery file-upload xmlhttprequest
javascript ajax jquery file-upload xmlhttprequest
asked Jun 2 '11 at 6:19
Sedat BaşarSedat Başar
2,30121828
asked Jun 2 '11 at 6:19
Sedat BaşarSedat Başar
2,30121828
edited May 20 '13 at 2:58
user1823761
asked Jun 2 '11 at 6:19
Sedat BaşarSedat Başar
2,30121828
asked Jun 2 '11 at 6:19
Sedat BaşarSedat Başar
2,30121828
asked Jun 2 '11 at 6:19
Sedat BaşarSedat Başar
2,30121828
2,30121828
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
- There is no such thing as
xhr.file = file;; the file object is not supposed to be attached this way.
xhr.send(file)doesn't send the file. You have to use theFormDataobject to wrap the file into amultipart/form-datapost data object:
var formData = new FormData();
formData.append("thefile", file);
xhr.send(formData);
After that, the file can be access in $_FILES['thefile'] (if you are using PHP).
Remember, MDC and Mozilla Hack demos are your best friends.
EDIT: The (2) above was incorrect. It does send the file, but it would send it as raw post data. That means you would have to parse it yourself on the server (and it's often not possible, depend on server configuration). Read how to get raw post data in PHP here.
answered Jun 2 '11 at 9:09
timdreamtimdream
4,30831722
1
ty so much that helped me..
– Sedat Başar
Jun 2 '11 at 11:48
5
xhr.send(file) is based on XHR2 which is a new version of the XMLHttpRequest object only avalable in some browsers.
– nkassis
Apr 20 '12 at 19:19
1
@nkassis yeah, but it would send the file itself as the post body, instead of constructing a multipart postdata for server to parse.
– timdream
Apr 22 '12 at 18:19
How come jquery can do it and pure javascript cannot?
– John Smith
Oct 18 '12 at 22:48
2
This method requires at least IE10 or Android 3.0.
– andreszs
Aug 5 '14 at 15:14
|
show 4 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
- There is no such thing as
xhr.file = file;; the file object is not supposed to be attached this way.
xhr.send(file)doesn't send the file. You have to use theFormDataobject to wrap the file into amultipart/form-datapost data object:
var formData = new FormData();
formData.append("thefile", file);
xhr.send(formData);
After that, the file can be access in $_FILES['thefile'] (if you are using PHP).
Remember, MDC and Mozilla Hack demos are your best friends.
EDIT: The (2) above was incorrect. It does send the file, but it would send it as raw post data. That means you would have to parse it yourself on the server (and it's often not possible, depend on server configuration). Read how to get raw post data in PHP here.
answered Jun 2 '11 at 9:09
timdreamtimdream
4,30831722
1
ty so much that helped me..
– Sedat Başar
Jun 2 '11 at 11:48
5
xhr.send(file) is based on XHR2 which is a new version of the XMLHttpRequest object only avalable in some browsers.
– nkassis
Apr 20 '12 at 19:19
1
@nkassis yeah, but it would send the file itself as the post body, instead of constructing a multipart postdata for server to parse.
– timdream
Apr 22 '12 at 18:19
How come jquery can do it and pure javascript cannot?
– John Smith
Oct 18 '12 at 22:48
2
This method requires at least IE10 or Android 3.0.
– andreszs
Aug 5 '14 at 15:14
|
show 4 more comments
- There is no such thing as
xhr.file = file;; the file object is not supposed to be attached this way.
xhr.send(file)doesn't send the file. You have to use theFormDataobject to wrap the file into amultipart/form-datapost data object:
var formData = new FormData();
formData.append("thefile", file);
xhr.send(formData);
After that, the file can be access in $_FILES['thefile'] (if you are using PHP).
Remember, MDC and Mozilla Hack demos are your best friends.
EDIT: The (2) above was incorrect. It does send the file, but it would send it as raw post data. That means you would have to parse it yourself on the server (and it's often not possible, depend on server configuration). Read how to get raw post data in PHP here.
answered Jun 2 '11 at 9:09
timdreamtimdream
4,30831722
1
ty so much that helped me..
– Sedat Başar
Jun 2 '11 at 11:48
5
xhr.send(file) is based on XHR2 which is a new version of the XMLHttpRequest object only avalable in some browsers.
– nkassis
Apr 20 '12 at 19:19
1
@nkassis yeah, but it would send the file itself as the post body, instead of constructing a multipart postdata for server to parse.
– timdream
Apr 22 '12 at 18:19
How come jquery can do it and pure javascript cannot?
– John Smith
Oct 18 '12 at 22:48
2
This method requires at least IE10 or Android 3.0.
– andreszs
Aug 5 '14 at 15:14
|
show 4 more comments
- There is no such thing as
xhr.file = file;; the file object is not supposed to be attached this way.
xhr.send(file)doesn't send the file. You have to use theFormDataobject to wrap the file into amultipart/form-datapost data object:
var formData = new FormData();
formData.append("thefile", file);
xhr.send(formData);
After that, the file can be access in $_FILES['thefile'] (if you are using PHP).
Remember, MDC and Mozilla Hack demos are your best friends.
EDIT: The (2) above was incorrect. It does send the file, but it would send it as raw post data. That means you would have to parse it yourself on the server (and it's often not possible, depend on server configuration). Read how to get raw post data in PHP here.
answered Jun 2 '11 at 9:09
timdreamtimdream
4,30831722
- There is no such thing as
xhr.file = file;; the file object is not supposed to be attached this way.
xhr.send(file)doesn't send the file. You have to use theFormDataobject to wrap the file into amultipart/form-datapost data object:
var formData = new FormData();
formData.append("thefile", file);
xhr.send(formData);
After that, the file can be access in $_FILES['thefile'] (if you are using PHP).
Remember, MDC and Mozilla Hack demos are your best friends.
EDIT: The (2) above was incorrect. It does send the file, but it would send it as raw post data. That means you would have to parse it yourself on the server (and it's often not possible, depend on server configuration). Read how to get raw post data in PHP here.
answered Jun 2 '11 at 9:09
timdreamtimdream
4,30831722
edited May 20 '13 at 2:43
answered Jun 2 '11 at 9:09
timdreamtimdream
4,30831722
answered Jun 2 '11 at 9:09
timdreamtimdream
4,30831722
answered Jun 2 '11 at 9:09
timdreamtimdream
4,30831722
4,30831722
1
ty so much that helped me..
– Sedat Başar
Jun 2 '11 at 11:48
5
xhr.send(file) is based on XHR2 which is a new version of the XMLHttpRequest object only avalable in some browsers.
– nkassis
Apr 20 '12 at 19:19
1
@nkassis yeah, but it would send the file itself as the post body, instead of constructing a multipart postdata for server to parse.
– timdream
Apr 22 '12 at 18:19
How come jquery can do it and pure javascript cannot?
– John Smith
Oct 18 '12 at 22:48
2
This method requires at least IE10 or Android 3.0.
– andreszs
Aug 5 '14 at 15:14
|
show 4 more comments
1
ty so much that helped me..
– Sedat Başar
Jun 2 '11 at 11:48
5
xhr.send(file) is based on XHR2 which is a new version of the XMLHttpRequest object only avalable in some browsers.
– nkassis
Apr 20 '12 at 19:19
1
@nkassis yeah, but it would send the file itself as the post body, instead of constructing a multipart postdata for server to parse.
– timdream
Apr 22 '12 at 18:19
How come jquery can do it and pure javascript cannot?
– John Smith
Oct 18 '12 at 22:48
2
This method requires at least IE10 or Android 3.0.
– andreszs
Aug 5 '14 at 15:14
1
1
ty so much that helped me..
– Sedat Başar
Jun 2 '11 at 11:48
ty so much that helped me..
– Sedat Başar
Jun 2 '11 at 11:48
5
5
xhr.send(file) is based on XHR2 which is a new version of the XMLHttpRequest object only avalable in some browsers.
– nkassis
Apr 20 '12 at 19:19
xhr.send(file) is based on XHR2 which is a new version of the XMLHttpRequest object only avalable in some browsers.
– nkassis
Apr 20 '12 at 19:19
1
1
@nkassis yeah, but it would send the file itself as the post body, instead of constructing a multipart postdata for server to parse.
– timdream
Apr 22 '12 at 18:19
@nkassis yeah, but it would send the file itself as the post body, instead of constructing a multipart postdata for server to parse.
– timdream
Apr 22 '12 at 18:19
How come jquery can do it and pure javascript cannot?
– John Smith
Oct 18 '12 at 22:48
How come jquery can do it and pure javascript cannot?
– John Smith
Oct 18 '12 at 22:48
2
2
This method requires at least IE10 or Android 3.0.
– andreszs
Aug 5 '14 at 15:14
This method requires at least IE10 or Android 3.0.
– andreszs
Aug 5 '14 at 15:14
|
show 4 more comments
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