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Php 从ajax加载数据_Php_Ajax - Fatal编程技术网

Php 从ajax加载数据

Php 从ajax加载数据,php,ajax,Php,Ajax,我有下面的代码,我想用它来生成一个页面,在这个页面中,我从下拉菜单中选择一个选项,然后应该加载所选选项的html,但现在它不加载 <form> <fieldset> <legend>Select transaction type:</legend> <select name="stakeholders" onchange="fetchStakeholder(this.value)">

我有下面的代码,我想用它来生成一个页面,在这个页面中,我从下拉菜单中选择一个选项,然后应该加载所选选项的html,但现在它不加载

<form>
    <fieldset>
        <legend>Select transaction type:</legend>
        <select name="stakeholders" onchange="fetchStakeholder(this.value)">
            <option value="">Select a Stakeholder:</option>
            <option value="CHALS">CHALS</option>
            <option value="juggernaut">juggernaut</option>
            <option value="STFU CHALS">STFU CHALS</option>
        </select>

        <script type="text/javascript">
            function fetchStakeholder(name) {
                if(name.length ==0){
                    return;
                }
                var request = new XMLHttpRequest();
                request.onreadystatechange=function(){
                    if(request.readyState==4 && request.status == 200){
                        document.getElementById("stakeholder").innerHTML=
                                request.responseText;
                    }

                }

                request.open("POST","stakeholder.php",true);
                request.setRequestHeader("Content-type","application/x-www-form-urlencoded");
                request.send("stakeholder=" + name);
            }


        </script>

    </fieldset>
</form>

选择交易类型:
选择一个利益相关者:
查尔斯
无法控制的强大机构
STFU CHALS
职能部门(名称){
如果(name.length==0){
返回;
}
var request=new XMLHttpRequest();
request.onreadystatechange=函数(){
if(request.readyState==4&&request.status==200){
document.getElementById(“涉众”).innerHTML=
request.responseText;
}
}
open(“POST”,“涉众.php”,true);
setRequestHeader(“内容类型”、“应用程序/x-www-form-urlencoded”);
请求。发送(“干系人=”+名称);
}
然后在PHP文件中:

<?php
    if($_SERVER["REQUEST_METHOD"] == "POST" ){
        $stakeholder = $_POST["stakeholder"];
        if($stakeholder === "CHALS"){
            include("CHALS.html");
        }elseif ($stakeholder === "juggernaut"){
            include("CHALS.html");
        }elseif ($stakeholder === "STFU CHALS"){
            include("STFU.html");
        }    
    }
?>

您什么也得不到,因为您没有id为
涉众的元素。此元素可以是任何
div
span
输入
选择
。我在这里使用的是
div

<form>
    <fieldset>
        <legend>Select transaction type:</legend>
        <select name="stakeholders" onchange="fetchStakeholder(this.value)">
            <option value="">Select a Stakeholder:</option>
            <option value="CHALS">CHALS</option>
            <option value="juggernaut">juggernaut</option>
            <option value="STFU CHALS">STFU CHALS</option>
        </select>
        <div id="stakeholder"></div>

        <script type="text/javascript">
            function fetchStakeholder(name) {
                if(name.length ==0){
                    return;
                }
                var request = new XMLHttpRequest();
                request.onreadystatechange=function(){
                    if(request.readyState==4 && request.status == 200){
                        document.getElementById("stakeholder").html=
                                request.responseText;
                    }

                }

                request.open("POST","stakeholder.php",true);
                request.setRequestHeader("Content-type","application/x-www-form-urlencoded");
                request.send("stakeholder=" + name);
            }


        </script>

    </fieldset>
</form>

选择交易类型:
选择一个利益相关者:
查尔斯
无法控制的强大机构
STFU CHALS
职能部门(名称){
如果(name.length==0){
返回;
}
var request=new XMLHttpRequest();
request.onreadystatechange=函数(){
if(request.readyState==4&&request.status==200){
document.getElementById(“涉众”).html=
request.responseText;
}
}
open(“POST”,“涉众.php”,true);
setRequestHeader(“内容类型”、“应用程序/x-www-form-urlencoded”);
请求。发送(“干系人=”+名称);
}

什么是
文档.getElementById(“干系人”)。
?我在你的codeNothing lol上看不到这个元素。这是我们给出的一个例子,应该是什么,你有一个例子吗?那么你的浏览器控制台对它发送的请求和接收的响应有什么说明?你的错误是什么?实际上我得到了它,只需要在页面中添加