JavaScript: Upload file

Let's say I have this element on the page:

<input id="image-file" type="file" />

This will create a button that allows the users of the web page to select a file via an OS "File open..." dialog in the browser.

Let's say the user clicks said button, selects a file in the dialog, then clicks the "Ok" button to close the dialog.

The selected file name is now stored in:

document.getElementById("image-file").value

Now, let's say that the server handles multi-part POSTs at the URL "/upload/image".

How do I send the file to "/upload/image"?

Also, how do I listen for notification that the file is finished uploading?

Answers:

Answer

Unless you're trying to upload the file using ajax, just submit the form to /upload/image.

<form enctype="multipart/form-data" action="/upload/image" method="post">
    <input id="image-file" type="file" />
</form>

If you do want to upload the image in the background (e.g. without submitting the whole form), you can use ajax:

Answer

Pure JS

You can use fetch optionally with await-try-catch

let photo = document.getElementById("image-file").files[0];
let formData = new FormData();

formData.append("photo", photo);
fetch('/upload/image', {method: "POST", body: formData});

async function SavePhoto(inp) 
{
    let user = { name:'john', age:34 };
    let formData = new FormData();
    let photo = inp.files[0];      
         
    formData.append("photo", photo);
    formData.append("user", JSON.stringify(user));  
    
    try {
       let r = await fetch('/upload/image', {method: "POST", body: formData}); 
       console.log('HTTP response code:',r.status); 
    } catch(e) {
       console.log('Huston we have problem...:', e);
    }
    
}
<input id="image-file" type="file" onchange="SavePhoto(this)" >
<br><br>
Before selecting the file open chrome console > network tab to see the request details.
<br><br>
<small>Because in this example we send request to https://stacksnippets.net/upload/image the response code will be 404 ofcourse...</small>

<br><br>
(in stack overflow snippets there is problem with error handling, however in <a href="https://jsfiddle.net/Lamik/b8ed5x3y/5/">jsfiddle version</a> for 404 errors 4xx/5xx are <a href="https://stackoverflow.com/a/33355142/860099">not throwing</a> at all but we can read response status which contains code)

Old school approach - xhr

let photo = document.getElementById("image-file").files[0];  // file from input
let req = new XMLHttpRequest();
let formData = new FormData();

formData.append("photo", photo);                                
req.open("POST", '/upload/image');
req.send(formData);

function SavePhoto(e) 
{
    let user = { name:'john', age:34 };
    let xhr = new XMLHttpRequest();
    let formData = new FormData();
    let photo = e.files[0];      
    
    formData.append("user", JSON.stringify(user));   
    formData.append("photo", photo);
    
    xhr.onreadystatechange = state => { console.log(xhr.status); } // err handling
    xhr.open("POST", '/upload/image');    
    xhr.send(formData);
}
<input id="image-file" type="file" onchange="SavePhoto(this)" >
<br><br>
Choose file and open chrome console > network tab to see the request details.
<br><br>
<small>Because in this example we send request to https://stacksnippets.net/upload/image the response code will be 404 ofcourse...</small>

<br><br>
(the stack overflow snippets, has some problem with error handling - the xhr.status is zero (instead of 404) which is similar to situation when we run script from file on <a href="https://stackoverflow.com/a/10173639/860099">local disc</a> - so I provide also js fiddle version which shows proper http error code <a href="https://jsfiddle.net/Lamik/k6jtq3uh/2/">here</a>)

SUMMARY

  • In server side you can read original file name (and other info) which is automatically included to request by browser in filename formData parameter.
  • You do NOT need to set request header Content-Type to multipart/form-data - this will be set automatically by browser.
  • Instead of /upload/image you can use full address like http://.../upload/image.
  • If you want to send many files in single request use multiple attribute: <input multiple type=... />, and attach all chosen files to formData in similar way (e.g. photo2=...files[2];... formData.append("photo2", photo2);)
  • You can include additional data (json) to request e.g. let user = {name:'john', age:34} in this way: formData.append("user", JSON.stringify(user));
  • This solutions should work on all major browsers.
Answer

As its creator I'm biased ;) but you could also consider using something like https://uppy.io. It does file uploading without navigating away from the page and offers a few bonuses like drag & drop, resuming uploads in case of browser crashes/flaky networks, and importing from e.g. Instagram. It's also open source and does not rely on jQuery or anything like that.

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