我可以保证 onreadystatechange 最终总是会在 readState == 4 的情况下被调用吗？

发布于 2024-12-02 17:18:13 字数 592 浏览 0 评论 0原文

我在嵌入式设备上使用 XMLHttpRequest，该设备为 API 提供非标准扩展，以允许在请求完成后手动清理资源。

我是否可以假设，对于所有情况（成功或其他情况，例如 404、DNS 查找失败等），对 send() 方法的调用最终将导致我的 onreadstatechange 处理程序被调用，并且 readstate == 4 ？

或者，换句话说，假设此实现的 XHR 在所有其他方面的行为与标准浏览器的行为类似，那么以下代码是否总是会导致调用 destroy() 方法？

var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function () {
    if (xhr.readyState == 4) {
        callback(xhr.responseText);
        if (xhr.destroy) { // must call this to prevent memory leak
            xhr.destroy();
        }
    }       
};
xhr.open(method, url, true);
xhr.send(null);

原文

I'm using XMLHttpRequest on an embedded device that provides a non-standard extension to the API to allow manual cleanup of resources after the request has finished.

Can I assume that, for all cases (successful or otherwise, eg 404, DNS lookup failed, etc), a call to the send() method will eventually result in my onreadstatechange handler being called with readyState == 4?

Or, to put it another way, assuming that this implementation's XHR behaves similarly to that of the standard browsers in all other ways, will the following code always result in the destroy() method being called?

var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function () {
    if (xhr.readyState == 4) {
        callback(xhr.responseText);
        if (xhr.destroy) { // must call this to prevent memory leak
            xhr.destroy();
        }
    }       
};
xhr.open(method, url, true);
xhr.send(null);

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