[英]Ajax Post to PHP and Get Return Data
我正在开发一个滑块来设置预算,并希望实现在slider1.php中设置的值。 但是,当我尝试使用下面的代码时,遇到错误“注意:未定义的索引:C:\\ xampp \\ htdocs \\ 1204763e \\ slider1 \\ slider1.php在第4行上的slideStatus谢谢,说的是PHP文件”
在slide.php中,我插入了以下代码集:
<html>
<head>
<script language="JavaScript" type="text/javascript">
function ajax_post(val){
var hr = new XMLHttpRequest();
var url = "slider1.php";
var ss = document.getElementById('sliderStatus').innerHTML = val;
var vars = "sliderStatus="+ss;
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;
}
}
// Send the data to PHP now... and wait for response to update the status div
hr.send(vars); // Actually execute the request
document.getElementById("status").innerHTML = "processing...";
}
</script>
</head>
<body>
<h2>Ajax Post to PHP and Get Return Data</h2>
<input type="range" name="slide" min="0" max="100" value="50" step="2" onChange="ajax_post(this.value)" />
<br /><br />
<span id="sliderStatus">50</span>
<br/><br/>
<div id="status"></div>
</body>
</html>
In slider1.php, I inserted this set of code:
<?php
echo 'Thank you '. $_GET['slideStatus'] . ', says the PHP file';
?>
您似乎尝试访问“ slideStatus”,但要发布“ sliderStatus”。
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