276

Display Below error in Safari.

Failed to execute 'createObjectURL' on 'URL': No function was found that matched the signature provided.

My Code is:

function createObjectURL(object) {
    return (window.URL) ? window.URL.createObjectURL(object) : window.webkitURL.createObjectURL(object);
}

This is my Code for image:

function myUploadOnChangeFunction() { 
    if (this.files.length) { 
       for (var i in this.files) { 
           if (this.files.hasOwnProperty(i)) { 
              var src = createObjectURL(this.files[i]); 
              var image = new Image(); 
              image.src = src; 
              imagSRC = src; 
              $('#img').attr('src', src); 
            }
       }           
   } 
} 
8
  • 7
    Hi Hardik, what are you passing to your createObjectURL(...) function when you get that error? Commented Nov 25, 2014 at 7:38
  • 4
    object must be a File object or a Blob object to create a object URL for.see devdocs.io/dom/window.url.createobjecturl Commented Nov 25, 2014 at 7:38
  • 1
    This is my Code for image: function myUploadOnChangeFunction() { if (this.files.length) { for (var i in this.files) { if (this.files.hasOwnProperty(i)) { var src = createObjectURL(this.files[i]); var image = new Image(); image.src = src; imagSRC = src; $('#img').attr('src', src); } } } } Commented Nov 25, 2014 at 7:46
  • @HardikMansaraa Go ahead and edit that in to your question. Commented Dec 5, 2014 at 3:16
  • window.URL.createObjectURL('broken') throws an error: Uncaught TypeError: Failed to execute 'createObjectURL' on 'URL': No function was found that matched the signature provided. Commented Nov 17, 2015 at 14:36

14 Answers 14

311

I experienced the same error, when I passed raw data to createObjectURL:

window.URL.createObjectURL(data)

It has to be a Blob, File or MediaSource object, not data itself. This worked for me:

var binaryData = [];
binaryData.push(data);
window.URL.createObjectURL(new Blob(binaryData, {type: "application/zip"}))

Check also the MDN for more info: https://developer.mozilla.org/en-US/docs/Web/API/URL/createObjectURL


UPDATE

Back in the day we could also use createObjectURL() method with MediaStream objects. This use has been dropped by the specs and by browsers.
If you need to set a MediaStream as the source of an HTMLMediaElement just attach the MediaStream object directly to the srcObject property of the HTMLMediaElement e.g. <video> element.

const mediaStream = new MediaStream();
const video = document.getElementById('video-player');
video.srcObject = mediaStream;

However, if you need to work with MediaSource, Blob or File, you still have to create a blob:// URL with URL.createObjectURL() and assign it to HTMLMediaElement.src.

Read more details here: https://developer.mozilla.org/en-US/docs/Web/API/HTMLMediaElement/srcObject

Sign up to request clarification or add additional context in comments.

3 Comments

Hi.. What to do if I am dealing with "application/pdf" ? I am getting the same error on console when I am dealing with PDF file
@mimo I am using same code to download file. But two files are getting downloaded. One is correct file and one extra file with same name but failed status is getting downloaded. Do you have any idea why it is happening?
I'm confused with this comment, in MDN it discourages the use of URL.createObjectURL() for media streams. However it doesn't state NOT to use it for a file input as stated in the initial question.
192

This error is caused because the function createObjectURL no longer accepts media stream object for Google Chrome

I changed this:

video.src=vendorUrl.createObjectURL(stream);
video.play();

to this:

video.srcObject=stream;
video.play();

This worked for me.

6 Comments

createObjectURL is not deprecated as shown here but no longer accepts media stream object
This should be the answer.
there is one another problem video.play() seems to be restricted : DOMException: play() can only be initiated by a user gesture.
@user889030 that just means you can't open a webpage and expect the webcam stream to start. You have to let the user explicitly start the stream. Just use a button to start the stream
|
38

My code was broken because I was using a deprecated technique. It used to be this:

video.src = window.URL.createObjectURL(localMediaStream);
video.play();

Then I replaced that with this:

video.srcObject = localMediaStream;
video.play();

That worked beautifully.

EDIT: Recently localMediaStream has been deprecated and replaced with MediaStream. The latest code looks like this:

video.srcObject = new MediaStream();

References:

  1. Deprecated technique: https://developer.mozilla.org/en-US/docs/Web/API/URL/createObjectURL
  2. Modern deprecated technique: https://developer.mozilla.org/en-US/docs/Web/API/HTMLMediaElement/srcObject
  3. Modern technique: https://developer.mozilla.org/en-US/docs/Web/API/MediaStream

Comments

10

Video with fall back:

try {
  video.srcObject = mediaSource;
} catch (error) {
  video.src = URL.createObjectURL(mediaSource);
}
video.play();

From: https://developer.mozilla.org/en-US/docs/Web/API/HTMLMediaElement/srcObject

Comments

9

I had the same error for the MediaStream. The solution is set a stream to the srcObject.

From the docs:

Important: If you still have code that relies on createObjectURL() to attach streams to media elements, you need to update your code to simply set srcObject to the MediaStream directly.

Comments

5

The problem is that the keys provided in the loop do not refer to the index of the file.

for (var i in this.files) {
    console.log(i);
}

The output of the above code is:

0
length
item

But what was expected was:

0
1
2
etc...

Then the error occurs when the browser tries to execute, for example:

window.URL.createObjectURL(this.files["length"])

I suggest implementation based on the following code:

var files = this.files;
for (var i = 0; i < files.length; i++) {
    var file = files[i],
        src = (window.URL || window.webkitURL).createObjectURL(file);
    ...
}

I hope this can help someone.

Greetings!

Comments

2

If you are using ajax, it is possible to add the options xhrFields: { responseType: 'blob' }:

$.ajax({
  url: 'yourURL',
  type: 'POST',
  data: yourData,
  xhrFields: { responseType: 'blob' },
  success: function (data, textStatus, jqXHR) {
    let src = window.URL.createObjectURL(data);
  }
});

Comments

2

If you are using angular this tutorial will be helpful: link. However you will need to replace this line:

this.video.src = window.URL.createObjectURL(stream);

with this, since createObjectURL() is deprecated on chrome for MediaStream.

this.video.srcObject = stream;

Comments

2
//my code was:

this._videoEl = videoEl;
        navigator.mediaDevices.getUserMedia({
            video : true
        }).then(stream => {
            this._videoEl.src = URL.createObjectURL(stream);
            this._videoEl.play();
        }).catch(err => {
            console.log(err);
        });

//and replace to this worked for me :

this._videoEl = videoEl;
        navigator.mediaDevices.getUserMedia({
            video : true
        }).then(stream => {
            this._videoEl.srcObject = stream;
            this._videoEl.play();
        }).catch(err => {
            console.log(err);
        });

2 Comments

Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.
This answer aligns with more recent MDN examples at developer.mozilla.org/en-US/docs/Web/API/MediaDevices/… which show video.srcObject = mediaStream; without using URL.createObjectURL().
2

I was able to fix this by checking if the object is null. {object ? URL.createObjectURL(object) : "default.png"}

This made me to conclude that the error occur when object is null.

Comments

0

I tried few things, but for me simply assigning stream to src worked.

video.srcObject=stream;

Comments

0

In jQuery - I had to add:

    xhrFields: {
        responseType: 'blob'
    }

to data: of my request

Comments

0

In case you're using Axios and noticed my comment, just add this line: responseType: 'blob'

the full code would be:

const response = await axios.get(`/example/image`, {
      responseType: 'blob' // This ensures the response is treated as binary data (Blob)
});

This tells Axios to handle the response as a Blob, which is useful when dealing with binary data such as images or file downloads.

GLHF :)

Comments

-12

I fixed it bydownloading the latest version from GitHub

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.