Question

file_in.pdf

}

PHP：

> xmlhttp.onreadystatechange = function() {
    if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
       var jsondata = xmlhttp.responseText;
        console.log(xmlhttp.responseText);
        document.getElementById("jsondata").value = jsondata;


        console.log(innerHTML = jsondata.word);
        document.getElementById("approved").innerHTML = "Thank you! We will get back to you.";
    }
};
xmlhttp.open("POST", url,false);
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded; charset=UTF-8");
xmlhttp.setRequestHeader("Access-Control-Allow-Origin", "*");
xmlhttp.setRequestHeader("Access-Control-Allow-Methods", "GET, POST, OPTIONS");
xmlhttp.setRequestHeader("Access-Control-Allow-Headers", "Content-Type");
xmlhttp.send("a="+a+ "&b="+b);

Answer 1

首先，您要在请求中发送CORS 响应标题 - 这不是必需的，不做任何事情，除非由于非标准标题而保证预检 - 所以，删除：

use Cake\Auth\DefaultPasswordHasher;


        protected function _setPassword($password)
        {
            if (strlen($password) > 0) {
                return (new DefaultPasswordHasher)->hash($password);
            }
        }

其次，您将xmlhttp.setRequestHeader("Access-Control-Allow-Origin", "*"); xmlhttp.setRequestHeader("Access-Control-Allow-Methods", "GET, POST, OPTIONS"); xmlhttp.setRequestHeader("Access-Control-Allow-Headers", "Content-Type");与同步（已弃用）请求结合使用 - 在下面的代码中，使请求异步，并使用更现代的onreadystatechange

希望有些评论能够让你了解其余的变化

onload

和PHP

var xmlhttp = new XMLHttpRequest();
xmlhttp.onload = function() {
    if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
        var data = JSON.parse(xmlhttp.responseText); // we get JSON, so parse it to an Object
        console.log(xmlhttp.responseText);
        document.getElementById("jsondata").value = data;
        //console.log(innerHTML = jsondata.word); what is this supposed to do? removed because it looks wrong
    }
};
xmlhttp.open("POST", url, true); // use ASYNCHRONOUS request, because it's the 21st century
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded; charset=UTF-8");
xmlhttp.send("a="+a+ "&b="+b);

Answer 2

您的XMLhttprequest.send未发送空白请求。它实际上发送了一个＆amp;湾我通过将一些随机值设置为和来测试您的代码，并且$req变量实际上填充了a和b的值。

＆＃13;

<script>
    var xmlhttp = new XMLHttpRequest();
    var url = 'http://localhost/test/test.php';
    var a= 5;
    var b=6;
  xmlhttp.onreadystatechange = function() {
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
            var jsondata = xmlhttp.responseText;
            console.log(xmlhttp.responseText);
            document.getElementById("jsondata").value = jsondata;


            console.log(innerHTML = jsondata.word);
            document.getElementById("approved").innerHTML = "Thank you! We will get back to you.";
        }
    };
    xmlhttp.open("POST", url,false);
    xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded; charset=UTF-8");
    xmlhttp.setRequestHeader("Access-Control-Allow-Origin", "*");
    xmlhttp.setRequestHeader("Access-Control-Allow-Methods", "GET, POST, OPTIONS");
    xmlhttp.setRequestHeader("Access-Control-Allow-Headers", "Content-Type");
    xmlhttp.send("a="+a+ "&b="+b);

</script>

＆＃13;

当我跑步时，我得到a=5&b=6。我在你的代码中看到的一个问题是你没有回应结果。这样做：

$req = file_get_contents("php://input");
echo $req;

根据Jaromanda X的建议，我想指出您没有将JSON数据发送到PHP脚本，因此尝试解码它会产生错误。

XMLhttprequest.send在PHP中给出了空白数组

2 个答案: