Question

嗨我试图在ajax函数中访问一个php变量，但显然它不工作...我使用了onClick事件来激活ajax函数，我将我的本地php变量作为参数参数传递...

<?php
$name = $_GET['name'];

?>

<html>
<head>
<script language="JavaScript" type="text/javascript">
function ajax_post(x){
var nm = x;
var hr = new XMLHttpRequest();

var url = "my_parse_file.php";
var fn = document.getElementById("first_name").value;


 var vars = "todo="+fn+"&name="+nm;
hr.open("POST", url, true);

hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
hr.onreadystatechange = function() {
    if(hr.readyState == 4 && hr.status == 200) {
        var return_data = hr.responseText;
        document.getElementById("status").innerHTML = return_data;
    }
}
hr.send(vars); // Actually execute the request
document.getElementById("status").innerHTML = "processing...";
}
</script>
</head>
<body>
<?php

$display =' Name of list:;   
 echo <label for="name"></label>
<input type="text" name="name" id="name">
 </p>
<p>Name of item:
<input id="first_name" name="first_name" type="text" /> 
 <br /><br />

 <input name="myBtn" type="submit" value="Submit Data" onClick="javascript:ajax_post('   . $name . ');">
</p>
<p>Your list has been succesfully created.</p>
<form name="form1" method="post" action="">
  <input type="submit" name="AddItem" id="AddItem" value="Add Items">
</form>
<p><br />
 <br />
</p>
<div id="status"></div>

</body>
</html>';
?>
<?php
 echo $display;
 ?>

Answer 1

将var回显到隐藏的跨度中，或者在需要时从那里输入它。

Answer 2

看起来您正在通过POST发送请求，然后尝试通过GET访问它。尝试将其更改为：

<?php
    $name = $_POST['name'];
?>

php变量为ajax函数

2 个答案: