繁体   English   中英

如何使用 JS fetch API 上传文件?

[英]How do I upload a file with the JS fetch API?

我仍然试图绕过它。

我可以让用户使用文件输入选择文件(甚至多个):

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

我是这样做的:

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
})

这是一个带有注释的基本示例。 upload功能正是您要找的:

// 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);

使用 Fetch API 发送文件的重要说明

需要省略 Fetch 请求的content-type标头。 然后浏览器会自动添加Content type标题,包括表单边界,看起来像

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

表单边界是表单数据的分隔符

如果你想要多个文件,你可以使用这个

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
})

要提交单个文件,您可以直接使用input.files数组中的File对象作为body:在您的fetch()初始值设定项中的值:

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File继承自Blob ,而Blob是 Fetch Standard 中定义的允许的BodyInit类型之一。

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);
});

从 Alex Montoya 的多个文件输入元素的方法中跳出来

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 })

添加 php 端点示例会很好。 所以这是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);

    }
}

那是php:

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

我的问题是我正在使用 response.blob() 来填充表单数据。 显然你至少不能用 react native 做到这一点,所以我最终使用了

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

Fetch 似乎可以识别该格式并将文件发送到 uri 指向的位置。

这是我的代码:

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>

如何使用 HTML5 fetch<\/code>在选择时上传单个文件<\/h4>
<label role="button"> Upload a picture <input accept="image\/*" type="file" hidden \/> <\/label><\/code><\/pre>
 const input = document.querySelector(`input[type="file"]`); function upload() { fetch(uplaodURL, { method: "PUT", body: input.files[0] }); } input.addEventListener("change", upload);<\/code><\/pre>"

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM