问题描述
我正在尝试根据我的 readystate
的 xhr
运行条件语句。 readystate
在调用 1
后切换到 open()
(第 2 行),但此后不再更改。它跳过条件,什么也不发送。
我很想知道为什么 readystate
没有改变?
var xhr = new XMLHttpRequest();
xhr.open('POST',submitUrl,true);
xhr.setRequestHeader('Content-Type','application/json');
xhr.onreadystatechange = function () {
if (xhr.readyState === 4) {
function1();
function2();
} else if (xhr.status === 400) {
function3();
function2();
} else if (xhr.status != 400 && xhr.status != 200) {
function5();
function6();
}
}
xhr.send(body);
})
解决方法
根据您的代码、您的观察以及您在评论中提供的上下文,您有两个问题:
- 发送表单数据
- 评估响应
让我们假设一些像这样的基本形式:
<form action="endpoint.php" method="post">
<input type="hidden" name="token" value="value">
<input type="submit" name="submit" value="submit">
</form>
为了能够自己发送此表单的数据,我们需要确保我们拦截浏览器的默认行为,即在点击提交按钮后立即提交(参见 epascarello 的评论):
// Intercept the onsubmit event
document.querySelector('form').onsubmit = function (evt) {
// Make sure to prevent the form from being submitted by
// the browser,which is the default behaviour.
evt.preventDefault();
// Get the form's data
let form = new FormData(evt.target);
// We're going to explicitly submitting our data
// as JSON,so make sure it actually is JSON.
let data = JSON.stringify(Object.fromEntries(form)); // https://stackoverflow.com/a/55874235/3323348
sendRequest(data); // Our submit function,which we'll define next (see below)
};
现在,我们将能够实际发送数据,并正确处理服务器发回的消息和状态代码。但首先,让我们快速浏览一下您的 if
子句,因为它们可能不会按您期望的方式工作。尤其是因为 state 和 status 不是互斥的 - readyState
为 4 并不意味着服务器没有用表示错误(如 404):
if (xhr.readyState === 4) {
console.log(xhr.status); // Could be any HTTP status code
} else if (xhr.status === 400) {
console.log(xhr.readyState); // Could be any readyState besides 4
} else if (xhr.status != 400 && xhr.status != 200) {
console.log(xhr.readyState); // Could be any readyState besides 4...
console.log(xhr.status); // ...and any HTTP status code besides a Bad Request (400) and an OK (200)
}
所以让我们稍微处理一下那部分,而其余代码保持不变(尽管包装在一个函数中):
function sendRequest(data) {
const xhr = new XMLHttpRequest();
xhr.open('POST','/endpoint.php'); // All requests are asynchronous by default,// so we can drop the third parameter.
xhr.setRequestHeader('Content-Type','application/json');
// Since we've just created a client and initialized
// a request,we'll receive notifications for states
// 2-4 only (instead of 0-4).
xhr.onreadystatechange = function () {
console.log(xhr.readyState); // Let's watch the readyState changing
// We're interested in the final result of our request only (state 4),// so let's jump all other states.
if (xhr.readyState !== 4) {
return;
}
const status = xhr.status; // HTTP status code
const type = status.toString().charAt(0); // Get class of HTTP status code (4xx,5xx,...)
if ([4,5].includes(type)) {
console.log('An error occured',status,xhr.responseText);
return;
}
if (status == 200) {
console.log('OK',xhr.responseText);
return;
}
// Server answered with a status code of 1xx,3xx,or > 200.
console.log('Unexpected response',xhr.responseText);
}
xhr.send(data);
}
现在,您应该能够成功发送表单数据(并将其作为 JSON 发送)并评估 HTTP 响应状态代码。不过,您可能需要考虑使用 fetch()
,而不是使用 XMLHttpRequest
。
其他:
- https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest/readyState
- https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest#events
- https://developer.mozilla.org/en-US/docs/Web/API/FormData/Using_FormData_Objects
- How to convert FormData (HTML5 object) to JSON
- Send POST data using XMLHttpRequest