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<html>
<head>
<script language="JavaScript" type="text/javascript">
function ajax_post(){
    // Create our XMLHttpRequest object
    var hr = new XMLHttpRequest();
    // Create some variables we need to send to our PHP file
    var url = "index2.php";
    var fn = document.getElementById("first_name").value;
    var ln = document.getElementById("last_name").value;
    var vars = "firstname="+fn+"&lastname="+ln;
    hr.open("POST", url, true);
    // Set content type header information for sending url encoded variables in the request
    hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    // Access the onreadystatechange event for the XMLHttpRequest object
    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>
Your First Name: <input id="first_name" name="first_name" type="text" /> 
<br /><br />
Your Last Name: <input id="last_name" name="last_name" type="text" />
<br /><br />
<input name="myBtn" type="submit" value="Submit Data" onClick="javascript:ajax_post();">
<br /><br />
<div id="status"></div>
</body>
</html>
<?php 
echo 'Thank you '. $_POST['firstname'] . ' ' . $_POST['lastname'] . ', says the PHP file';
?>

上面的代码工作并提交数据并获取结果,以及再次呈现表单的结果。图片 --> http://oi47.tinypic.com/1bt02.jpg

如何解决?提前致谢。

4

1 回答 1

1

不要在您发送给 Ajax 请求的响应中包含该表单。

于 2013-01-19T23:40:10.380 回答