File Uploading with JavaScript

Question

File Uploading with JavaScript

Imagine you have an element on your webpage like this:

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

With this element, users can click a button to select a file through their browser's "File open..." dialog.

If a user selects a file and clicks "Ok" to close the dialog, the selected file name will be stored in:

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

Assuming the server accepts multi-part POSTs at "/upload/image", how do you send the file there?

And how can you listen for notification that the upload process is complete?

javascript file-upload

Answer 1

Answer №1

Vanilla JavaScript

If you prefer using pure JavaScript, you have the option to utilize the Fetch API, which can be combined with await-try-catch for error handling.

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)

Traditional Approach - XMLHttpRequest (xhr)

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

Vanilla JavaScript

If you prefer using pure JavaScript, you have the option to utilize the Fetch API, which can be combined with await-try-catch for error handling.

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)

Traditional Approach - XMLHttpRequest (xhr)

let photo = document.getElementById("image-file").files[0];  
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 (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));
```
You can set timeout: for fetch using AbortController, for old approach by xhr.timeout= milisec
This solutions should work on all major browsers.

Answer 3

Answer №2

To upload the file, simply 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 prefer to upload the image in the background without submitting the entire form, you can utilize ajax:

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

Answer 4

To upload the file, simply 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 prefer to upload the image in the background without submitting the entire form, you can utilize ajax:

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

Answer 5

Answer №3

After attempting various solutions without success, I finally found a method that worked for me in uploading files.

The setup involved having a file selection input along with a submit button:

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

Subsequently, I added this JavaScript function to handle the upload process:

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 successfully uploaded');
    location.reload();
};

If you're a fan of Stack Snippets...

The PUT method differs slightly from the POST method. For instance, the POST method might not be supported on certain web servers such as in Chrome.

I tested this functionality using the Web Server for Chrome extension - Link Here

Remember - With the Web Server for Chrome, ensure you enable 'file upload' in the advanced settings to prevent any errors related to permissions.

Answer 6

After attempting various solutions without success, I finally found a method that worked for me in uploading files.

The setup involved having a file selection input along with a submit button:

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

Subsequently, I added this JavaScript function to handle the upload process:

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 successfully uploaded');
    location.reload();
};

If you're a fan of Stack Snippets...

The PUT method differs slightly from the POST method. For instance, the POST method might not be supported on certain web servers such as in Chrome.

I tested this functionality using the Web Server for Chrome extension - Link Here

Remember - With the Web Server for Chrome, ensure you enable 'file upload' in the advanced settings to prevent any errors related to permissions.

Answer 7

Answer №4

This code snippet handles form submissions with files that are not directly in the form:

const formData = new FormData();
const files = event.target.files;

for (const file of files) {
   formData.append('files[]', file);
}

$.ajax({
   type: "POST",
   url: urlString,
   data: formData,
   error : function (result) {
       console.log('error occurred:', result);
   },
   success : function (result) {
      console.log('success!', result)
   }
});

Answer 8

This code snippet handles form submissions with files that are not directly in the form:

const formData = new FormData();
const files = event.target.files;

for (const file of files) {
   formData.append('files[]', file);
}

$.ajax({
   type: "POST",
   url: urlString,
   data: formData,
   error : function (result) {
       console.log('error occurred:', result);
   },
   success : function (result) {
      console.log('success!', result)
   }
});

File Uploading with JavaScript

Answer №1

Vanilla JavaScript

Answer №2

Answer №3

Answer №4

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