如何使用JS fetch API上传文件？

Question

我仍然试图绕过它.

我可以让用户通过文件输入选择文件(甚至多个):

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

我可以submit使用这个事件<fill in your event handler here>.但是一旦我这样做,我该如何使用该文件发送fetch？

fetch('/files', {
  method: 'post',
  // what goes here? What is the "body" for this? content-type header?
}).then(/* whatever */);

Answer 1

我这样做了:

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
})

Answer 2

这是带注释的基本示例.该upload功能正是您所需要的:

// 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);

Answer 3

使用Fetch API发送文件的重要注意事项

需要省略 content-typeFetch请求的标头.然后浏览器将自动添加Content type包含表格边界的标题

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

表单边界是表单数据的分隔符

Answer 4

如果你想要多个文件,你可以使用它

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
})

Answer 5

要提交一个文件,你可以简单地使用File对象从input的.files阵列作为直接的价值body:在你的fetch()初始化:

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],
});

这是有效的,因为它File继承自Fetch标准中定义的允许类型Blob,并且Blob是其中一种BodyInit.

Answer 6

这里接受的答案有点过时了。截至 2020 年 4 月，在 MDN 网站上看到的推荐方法建议使用FormData并且也不要求设置内容类型。https://developer.mozilla.org/en-US/docs/Web/API/Fetch_API/Using_Fetch

为方便起见，我引用了代码片段：

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);
});

Answer 7

最好添加 php 端点示例。所以这就是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);

    }
}

那就是php：

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

Answer 8

借鉴 Alex Montoya 的多文件输入元素方法

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 })