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@ebidel
Created April 18, 2012 03:23
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Uploading files using xhr.send(FormData) to PHP server
<?php
$fileName = $_FILES['afile']['name'];
$fileType = $_FILES['afile']['type'];
$fileContent = file_get_contents($_FILES['afile']['tmp_name']);
$dataUrl = 'data:' . $fileType . ';base64,' . base64_encode($fileContent);
$json = json_encode(array(
'name' => $fileName,
'type' => $fileType,
'dataUrl' => $dataUrl,
'username' => $_REQUEST['username'],
'accountnum' => $_REQUEST['accountnum']
));
echo $json;
?>
<!DOCTYPE html>
<!--
Copyright 2012 Google Inc.
Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at
http://www.apache.org/licenses/LICENSE-2.0
Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
Author: Eric Bidelman ([email protected])
-->
<html>
<head>
<meta charset="utf-8" />
<meta http-equiv="X-UA-Compatible" content="IE=Edge,chrome=1" />
<title>xhr.send(FormData) Example</title>
</head>
<body>
<input type="file" name="afile" id="afile" accept="image/*"/>
<script>
document.querySelector('#afile').addEventListener('change', function(e) {
var file = this.files[0];
var fd = new FormData();
fd.append("afile", file);
// These extra params aren't necessary but show that you can include other data.
fd.append("username", "Groucho");
fd.append("accountnum", 123456);
var xhr = new XMLHttpRequest();
xhr.open('POST', 'handle_file_upload.php', true);
xhr.upload.onprogress = function(e) {
if (e.lengthComputable) {
var percentComplete = (e.loaded / e.total) * 100;
console.log(percentComplete + '% uploaded');
}
};
xhr.onload = function() {
if (this.status == 200) {
var resp = JSON.parse(this.response);
console.log('Server got:', resp);
var image = document.createElement('img');
image.src = resp.dataUrl;
document.body.appendChild(image);
};
};
xhr.send(fd);
}, false);
</script>
<!--[if IE]>
<script src="http://ajax.googleapis.com/ajax/libs/chrome-frame/1/CFInstall.min.js"></script>
<script>CFInstall.check({mode: 'overlay'});</script>
<![endif]-->
</body>
</html>
@galicialuis
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and if i want to save the image in a carpet or path

@ahxxan
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ahxxan commented Mar 17, 2016

{"name":null,"type":null,"dataUrl":"data:;base64,","username":null,"accountnum":null} as output
i am receiving this . can some one help to resolve my issue @chrisp22 could you find solution

@ankurgupta1401
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You can find a working implementation of uploading file using XMLHttpRequest here.

@rod101
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rod101 commented Nov 16, 2016

Hi All,
Is there a way I can append the image path without using the file input
I have tried formData.append('avatar', avatar.files[0], 'cc.jpg'); any help please

@JuJussel
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Awesome, have been looking for this all day!
Thanks!

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ghost commented Oct 10, 2017

Thank you very much. Your example help me so much. I looking long time how to upload files to server via xhr. And again, thank you so much.

@AngeloAlamia
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AngeloAlamia commented Mar 12, 2018

Thanks a lot, after looking all day long I finally found the right script!!!

@navyjax2
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navyjax2 commented Apr 6, 2018

Note to anyone finding this via Google and working with MVC: To use FormData, you'd have to post using xhr.setRequestHeader("Content-Type","multipart/form-data"); - not included above - but then your server-side will throw an exception if you use

if (!Request.Content.IsMimeMultipartContent() { throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType); }

which is something it should never do, so I'm convinced FormData can only be used with PHP. Better to get the file into a base64 string (I have an example here: https://stackoverflow.com/questions/37134433/convert-input-file-to-byte-array/49676679#49676679 ), create a JSON object:

var dataObj = { afile: myBase64file, username: "Groucho", accountnum: 123456, };,

add in this after the xhr.open():
xhr.setRequestHeader("Content-Type","application/json;charset=UTF-8");

then do xhr.send(JSON.stringify(dataObj)); at the end.

On the MVC side, you need:

[System.Web.Http.HttpPost]
public virtual ActionResult Post([FromBody]MyModel myModelObject) { ... }

You'll need to have a model:
public class MyModel {
public string afile { get; set; }
public string username { get; set; }
public int accountnum { get; set; }
}

Inside the Post, the parameters should be accessible: string file = myModelObject.afile;

@luischicllarosas
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Hey! you are great!! thnks !!!

@Gataquadrada
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Note to anyone finding this via Google Search about Cordova File Upload:
You can use this code on input:file with position:absolute; opacity:0; trick too. Works like a charm.

@dicksonmundia
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can send(fd) plus other such as image size?

@Gataquadrada
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Yes. You can still populate the data: {} property as you wish.
What I said just triggers the usual use case for the plugin.

And it still works to this day!

@PaYo90
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PaYo90 commented Jul 16, 2020

I can access rsp.dataUrl my object all the time refuses to read into it, i have it undefined, pls help.

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