javascript 为什么 XmlHttpRequest readyState = 2 on 200 HTTP 响应代码
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Why XmlHttpRequest readyState = 2 on 200 HTTP response code
提问by snowbound
So I'm using plain javascript (no jquery), to send a file to the server. Server script PHP returns status code 200 at the end, but instead javascript is getting readyState == 2.
所以我使用普通的javascript（没有jquery）将文件发送到服务器。服务器脚本 PHP 在最后返回状态代码 200，但 javascript 变得 readyState == 2。
The PHP code sends back status code 200:
PHP 代码发回状态代码 200：
header('X-PHP-Response-Code: 200', true, 200);
exit;
The javascript is doing:
javascript正在做：
request.onreadystatechange = function() {
if (request.readyState == 4) {
var message;
switch(request.status) {
case '200':
message = "Data uploaded successfully.";
break;
case '406':
message = "Incorrect file format. Please try again.";
break;
case '410':
message = "Unexpected error. Please contact support.";
break;
default:
break;
}
status_message_container.innerHTML = message;
submit_button.disabled = false;
}
else {
alert( "Unexpected error: " + this.statusText + ".\nPlease try again");
}
};
request.send(formData);
Even know the HTTP 200 status code comes back correctly (I get 'OK') on frontend. The JS script is seeing readyState==2(i.e. else block always hit)
甚至知道 HTTP 200 状态代码在前端正确返回（我得到“OK”）。JS脚本看到readyState==2（即else块总是命中）
My understanding is that a server status code of 200 should give readyState == 4??
我的理解是服务器状态代码 200 应该给出readyState == 4??
回答by CodingIntrigue
Firstly, onreadystateisn't just fired once. It's fired multiple times, you need to be able to handle that. These are the codes you need to handle:
首先，onreadystate不只是被解雇一次。它被多次触发，你需要能够处理它。这些是您需要处理的代码：
0 UNSENT - open()has not been called yet
1 OPENED - send()has not been called yet
2 HEADERS_RECEIVED - send() has been called, and headers and status are available
3 LOADING Downloading; - responseText holds partial data
4 - The operation is complete
0 UNSENT - open() 还没有被调用
1 OPENED - send() 还没有被调用
2 HEADERS_RECEIVED - send() 已经被调用，并且头部和状态可用
3 LOADING 下载；- responseText 保存部分数据
4 -操作完成
Your code is hitting the else block on readyState == 2(headers received) and assuming that is an error status when it's not.
您的代码遇到了 else 块readyState == 2（收到的标头），并假设它不是错误状态。
You error check should be inside the request.readyState == 4check. That way, the request is complete but there could also have been an error:
你的错误检查应该在request.readyState == 4检查里面。这样，请求就完成了，但也可能出现错误：
if (request.readyState == 4) {
switch(request.status) {
case '200':
message = "Data uploaded successfully.";
break;
// Error handling here
default: alert( "Unexpected error: " + this.statusText + ".\nPlease try again"); break;
}
}
https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest
https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest
