纯 JS
您可以选择将抓取与 await-try-catch 结合使用
let photo = document.getElementById("image-file").files[0];
let formData = new FormData();
formData.append("photo", photo);
fetch('/upload/image', {method: "POST", body: formData});
async function SavePhoto(inp)
{
let user = { name:'john', age:34 };
let formData = new FormData();
let photo = inp.files[0];
formData.append("photo", photo);
formData.append("user", JSON.stringify(user));
const ctrl = new AbortController() // timeout
setTimeout(() => ctrl.abort(), 5000);
try {
let r = await fetch('/upload/image',
{method: "POST", body: formData, signal: ctrl.signal});
console.log('HTTP response code:',r.status);
} catch(e) {
console.log('Huston we have problem...:', e);
}
}
<input id="image-file" type="file" onchange="SavePhoto(this)" >
<br><br>
Before selecting the file open chrome console > network tab to see the request details.
<br><br>
<small>Because in this example we send request to https://stacksnippets.net/upload/image the response code will be 404 ofcourse...</small>
<br><br>
(in stack overflow snippets there is problem with error handling, however in <a href="https://jsfiddle.net/Lamik/b8ed5x3y/5/">jsfiddle version</a> for 404 errors 4xx/5xx are <a href="https://stackoverflow.com/a/33355142/860099">not throwing</a> at all but we can read response status which contains code)
老派方法 - xhr
let photo = document.getElementById("image-file").files[0]; // file from input
let req = new XMLHttpRequest();
let formData = new FormData();
formData.append("photo", photo);
req.open("POST", '/upload/image');
req.send(formData);
function SavePhoto(e)
{
let user = { name:'john', age:34 };
let xhr = new XMLHttpRequest();
let formData = new FormData();
let photo = e.files[0];
formData.append("user", JSON.stringify(user));
formData.append("photo", photo);
xhr.onreadystatechange = state => { console.log(xhr.status); } // err handling
xhr.timeout = 5000;
xhr.open("POST", '/upload/image');
xhr.send(formData);
}
<input id="image-file" type="file" onchange="SavePhoto(this)" >
<br><br>
Choose file and open chrome console > network tab to see the request details.
<br><br>
<small>Because in this example we send request to https://stacksnippets.net/upload/image the response code will be 404 ofcourse...</small>
<br><br>
(the stack overflow snippets, has some problem with error handling - the xhr.status is zero (instead of 404) which is similar to situation when we run script from file on <a href="https://stackoverflow.com/a/10173639/860099">local disc</a> - so I provide also js fiddle version which shows proper http error code <a href="https://jsfiddle.net/Lamik/k6jtq3uh/2/">here</a>)
总结
- 在服务器端,您可以读取原始文件名(和其他信息),这些信息会自动包含在formData参数中,由浏览器请求。
filename
- 您不需要将请求标头设置为 - 这将由浏览器自动设置(这将包括必需的
边界
参数)。Content-Type
multipart/form-data
- 而不是你可以使用完整的地址,如(当然这两个地址都是任意的,取决于服务器 - 以及参数的情况相同 - 通常在服务器上“POST”用于文件上传,但有时可以使用“PUT”或其他)。
/upload/image
http://.../upload/image
method
- 如果要在单个请求中发送多个文件,请使用属性:,并以类似的方式将所有选定的文件附加到formData(例如...
multiple
<input multiple type=... />
photo2=...files[2];
formData.append("photo2", photo2);
)
- 您可以包含其他数据(json)来请求,例如 通过这种方式:
let user = {name:'john', age:34}
formData.append("user", JSON.stringify(user));
- 您可以设置超时:用于使用,用于旧方法
fetch
AbortController
xhr.timeout= milisec
- 此解决方案应该适用于所有主流浏览器。