XMLHttpRequest POST 到 PHP

2022-01-05 00:00:00 request xmlhttprequest php javascript html

我在管理 JS 上向我自己的服务器发出的简单 XMLHttpRequest 的答案时遇到了一些问题.我有一些令人不安的答案，这是我的代码:

I'm having some trouble managing the answers to a simple XMLHttpRequest made on JS to my own server. I have some troubling answers, here's my code:

JavaScript:

JavaScript:

function callPHP () {
    var xml = new XMLHttpRequest();
    xml.open("POST", "http://localhost/ajaxTest", false);
    xml.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xml.send("format=json");
    var resp = xml.responseText;
    console.log(resp);
}

和 PHP:

<?php
    if (isset($_POST["format"])){
        if($_POST["format"] == "json"){
            echo '
                {
                    "name": "Name",
                    "lastName": "LastName", 
                    "dob" : "dd/mm/yyyy",
                }
                ';
        }
    }else{
        print_r($_POST);
        echo "Error";
        http_response_code(400);
    }


?>

每当我执行我的 JS 时，我都会收到 PHP 代码的错误部分，即使根据请求发送了format=json"数据.但是，如果我将异步更改为 true，则会收到 400 Bad Request 的 GET 错误(在 chrome 的控制台中)，但没有执行 PHP 的回显.

And whenever I execute my JS I get the error part of the PHP code, even having sent the "format=json" data on request. But if I change asynchronous to true, I get a GET error (in chrome's console) of 400 Bad Request, but no echo of the PHP is executed.

我知道我必须检查 xmlhttprequest 状态和响应代码才能做异步，我是直接在控制台上测试的.

I know I have to check the xmlhttprequest status and response code to do asynchronous, I was testing it directly on the console.

我不知道我在这里做错了什么，应该发送POST数据，并且与异步与否无关，对吗?

I don't know what I'm doing wrong here, the POST data should be sent, and independently of asynchronous or not, right?

谢谢大家！

推荐答案

看起来你的 javascript 函数的结构不正确.您应该在发送请求之前设置一个侦听器.在您的代码中可能类似于:-

looks like the structure of your javascript function is incorrect. You should setup a listener before you send the request. In your code perhaps something like:-

var xml = new XMLHttpRequest();
xml.onreadystatechange = function() {
    if( xml.readyState==4 && xml.status==200 ){
        console.log( xml.responseText );
    }
};

xml.open("POST", "http://localhost/ajaxTest", false);
xml.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xml.send("format=json");

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