JavaScript：上传文件

Question

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?

Answer 1

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)); const ctrl = new AbortController() // timeout setTimeout(() => ctrl.abort(), 5000); try { let r = await fetch('/upload/image', {method: "POST", body: formData, signal: ctrl.signal}); 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.timeout = 5000; 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 (which will include the mandatory boundary parameter ).
• Instead of /upload/image you can use full address like http://.../upload/image (of course both addresses are arbitrary and depends on server - and same situation with param method - usually on servers "POST" is used for file upload but sometimes "PUT" or other can be used).
• 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 (eg photo2=...files[2]; ... formData.append("photo2", photo2); )
• You can include additional data (json) to request eg let user = {name:'john', age:34} in this way: formData.append("user", JSON.stringify(user));
• You can set timeout: for fetch using AbortController , for old approach by xhr.timeout= milisec
• This solutions should work on all major browsers.

Answer 2

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 (eg without submitting the whole form), you can use ajax:

• Asynchronous file upload (AJAX file upload) using jsp and javascript
• jQuery Ajax File Upload
• Ajax using file upload

Answer 3

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 eg Instagram. It's also open source and does not rely on jQuery or anything like that.

Answer 4

I have been trying to do this for a while and none of these answers worked for me. This is how I did it.

I had a select file and a submit button

<input type="file" name="file" id="file">
<button onclick="doupload()" name="submit">Upload File</button>

Then in my javascript code I put this

function doupload() {
    let data = document.getElementById("file").files[0];
    let entry = document.getElementById("file").files[0];
    console.log('doupload',entry,data)
    fetch('uploads/' + encodeURIComponent(entry.name), {method:'PUT',body:data});
    alert('your file has been uploaded');
    location.reload();
};

If you like StackSnippets...

 function doupload() { let data = document.getElementById("file").files[0]; let entry = document.getElementById("file").files[0]; console.log('doupload',entry,data) fetch('uploads/' + encodeURIComponent(entry.name), {method:'PUT',body:data}); alert('your file has been uploaded'); };

 <input type="file" name="file" id="file"> <button onclick="doupload()" name="submit">Upload File</button>

The PUT method is slightly different than the POST method. In this case, in web server for chrome, the POST method is not implemented.

Tested with web server for chrome - https://chrome.google.com/webstore/detail/web-server-for-chrome/ofhbbkphhbklhfoeikjpcbhemlocgigb?hl=en

Note- When using web server for chrome you need to go into advanced options and check the option 'enable file upload'. If you do not, you will get an error for not allowed.