我还在试着接受这件事。
我可以让用户选择文件(甚至多个)与文件输入:
<form>
<div>
<label>Select file to upload</label>
<input type="file">
</div>
<button type="submit">Convert</button>
</form>
我可以用<填充事件处理程序>来捕获提交事件。但是一旦我这样做了，我如何使用fetch发送文件?
fetch('/files', {
method: 'post',
// what goes here? What is the "body" for this? content-type header?
}).then(/* whatever */);
当前回答
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 })
其他回答
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>
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 })
添加php端点示例会很好。 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);
}
}
这就是php:
$file = $_FILES['nameusedinFormData'];
$temp = $file['tmp_name'];
$target_file = './targetfilename.jpg';
move_uploaded_file($_FILES["image"]["tmp_name"], $target_file);
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);
});