我如何上传一个文件与JS获取API?

这是一个带有注释的基本示例。upload函数就是你要找的:

// 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);

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
})

要提交单个文件，可以直接使用input的.files数组中的File对象作为fetch()初始化式中body: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],
});

这是因为File继承自Blob，而Blob是Fetch标准中定义的允许的BodyInit类型之一。

使用Fetch API发送文件的一个重要注意事项

我们需要忽略Fetch请求的content-type报头。然后浏览器会自动添加Content type头文件，包括表单边界，如下所示

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
})

从Alex Montoya的多文件输入元素方法出发

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 })

对我来说，问题是我正在使用response.blob()填充表单数据。显然你不能这样做，至少用react native，所以我最终使用

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>

这里公认的答案有点过时了。截至2020年4月，MDN网站上的推荐方法建议使用FormData，也不要求设置内容类型。https://developer.mozilla.org/en-US/docs/Web/API/Fetch_API/Using_Fetch

为了方便起见，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);
});

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);


}
}

这就是php:

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

如何上传一个文件选择使用HTML5 fetch

<label role="button">
Upload a picture
<input accept="image/*" type="file" hidden />
</label>

const input = document.querySelector(`input[type="file"]`);


function upload() {
fetch(uplaodURL, { method: "PUT", body: input.files[0] });
}


input.addEventListener("change", upload);