226
votes

I am still trying to wrap my head around it.

I can have the user select the file (or even multiple) with the file input:

<form>
  <div>
    <label>Select file to upload</label>
    <input type="file">
  </div>
  <button type="submit">Convert</button>
</form>

And I can catch the submit event using <fill in your event handler here>. But once I do, how do I send the file using fetch?

fetch('/files', {
  method: 'post',
  // what goes here? What is the "body" for this? content-type header?
}).then(/* whatever */);
10
official document works for me after trying some answers failed: developer.mozilla.org/en-US/docs/Web/API/Fetch_API/… , something can confirm: 1. need wrap file in FromData; 2. doesn't need to declare Content-Type: multipart/form-data in request header - Spark.Bao

10 Answers

174
votes

This is a basic example with comments. The upload function is what you are looking for:

// Select your input type file and store it in a variable
const input = document.getElementById('fileinput');

// This will upload the file after having read it
const upload = (file) => {
  fetch('http://www.example.net', { // Your POST endpoint
    method: 'POST',
    headers: {
      // Content-Type may need to be completely **omitted**
      // or you may need something
      "Content-Type": "You will perhaps need to define a content-type here"
    },
    body: file // This is your file object
  }).then(
    response => response.json() // if the response is a JSON object
  ).then(
    success => console.log(success) // Handle the success response object
  ).catch(
    error => console.log(error) // Handle the error response object
  );
};

// Event handler executed when a file is selected
const onSelectFile = () => upload(input.files[0]);

// Add a listener on your input
// It will be triggered when a file will be selected
input.addEventListener('change', onSelectFile, false);
272
votes

I've done it like this:

var input = document.querySelector('input[type="file"]')

var data = new FormData()
data.append('file', input.files[0])
data.append('user', 'hubot')

fetch('/avatars', {
  method: 'POST',
  body: data
})
106
votes

An important note for sending Files with Fetch API

One needs to omit content-type header for the Fetch request. Then the browser will automatically add the Content type header including the Form Boundary which looks like

Content-Type: multipart/form-data; boundary=—-WebKitFormBoundaryfgtsKTYLsT7PNUVD

Form boundary is the delimiter for the form data

42
votes

If you want multiple files, you can use this

var input = document.querySelector('input[type="file"]')

var data = new FormData()
for (const file of input.files) {
  data.append('files',file,file.name)
}

fetch('/avatars', {
  method: 'POST',
  body: data
})
24
votes

To submit a single file, you can simply use the File object from the input's .files array directly as the value of body: in your fetch() initializer:

const myInput = document.getElementById('my-input');

// Later, perhaps in a form 'submit' handler or the input's 'change' handler:
fetch('https://example.com/some_endpoint', {
  method: 'POST',
  body: myInput.files[0],
});

This works because File inherits from Blob, and Blob is one of the permissible BodyInit types defined in the Fetch Standard.

13
votes

The accepted answer here is a bit dated. As of April 2020, a recommended approach seen on the MDN website suggests using FormData and also does not ask to set the content type. https://developer.mozilla.org/en-US/docs/Web/API/Fetch_API/Using_Fetch

I'm quoting the code snippet for convenience:

const formData = new FormData();
const fileField = document.querySelector('input[type="file"]');

formData.append('username', 'abc123');
formData.append('avatar', fileField.files[0]);

fetch('https://example.com/profile/avatar', {
  method: 'PUT',
  body: formData
})
.then((response) => response.json())
.then((result) => {
  console.log('Success:', result);
})
.catch((error) => {
  console.error('Error:', error);
});
4
votes

Jumping off from Alex Montoya's approach for multiple file input elements

const inputFiles = document.querySelectorAll('input[type="file"]');
const formData = new FormData();

for (const file of inputFiles) {
    formData.append(file.name, file.files[0]);
}

fetch(url, {
    method: 'POST',
    body: formData })
2
votes

It would be nice to add php endpoint example. So that is js:

const uploadinput = document.querySelector('#uploadinputid');
const uploadBtn = document.querySelector('#uploadBtnid');
uploadBtn.addEventListener('click',uploadFile);

async function uploadFile(){
    const formData = new FormData();
    formData.append('nameusedinFormData',uploadinput.files[0]);    
    try{
        const response = await fetch('server.php',{
            method:'POST',
            body:formData
        } );
        const result = await response.json();
        console.log(result);
    }catch(e){
        console.log(e);

    }
}

That is php:

$file = $_FILES['nameusedinFormData'];
$temp = $file['tmp_name'];
$target_file = './targetfilename.jpg';
move_uploaded_file($_FILES["image"]["tmp_name"], $target_file);
1
votes

The problem for me was that I was using a response.blob() to populate the form data. Apparently you can't do that at least with react native so I ended up using

data.append('fileData', {
  uri : pickerResponse.uri,
  type: pickerResponse.type,
  name: pickerResponse.fileName
 });

Fetch seems to recognize that format and send the file where the uri is pointing.

1
votes

Here is my code:

html:

const upload = (file) => {
    console.log(file);

    

    fetch('http://localhost:8080/files/uploadFile', { 
    method: 'POST',
    // headers: {
    //   //"Content-Disposition": "attachment; name='file'; filename='xml2.txt'",
    //   "Content-Type": "multipart/form-data; boundary=BbC04y " //"multipart/mixed;boundary=gc0p4Jq0M2Yt08jU534c0p" //  ή // multipart/form-data 
    // },
    body: file // This is your file object
  }).then(
    response => response.json() // if the response is a JSON object
  ).then(
    success => console.log(success) // Handle the success response object
  ).catch(
    error => console.log(error) // Handle the error response object
  );

  //cvForm.submit();
};

const onSelectFile = () => upload(uploadCvInput.files[0]);

uploadCvInput.addEventListener('change', onSelectFile, false);
<form id="cv_form" style="display: none;"
										enctype="multipart/form-data">
										<input id="uploadCV" type="file" name="file"/>
										<button type="submit" id="upload_btn">upload</button>
</form>
<ul class="dropdown-menu">
<li class="nav-item"><a class="nav-link" href="#" id="upload">UPLOAD CV</a></li>
<li class="nav-item"><a class="nav-link" href="#" id="download">DOWNLOAD CV</a></li>
</ul>