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PHP开发实例大全(提高卷) 中文完整pdf扫描版[244MB]
php ajax的实现原理
html处理页 index.html
<html>
<head>
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script>
function getdetails(){
var name = $('#name').val();
var rno = $('#rno').val();
$.ajax({
type: "POST",
url: "details.php",
data: {fname:name, id:rno}
}).done(function( result ) {
$("#msg").html( " Address of Roll no " +rno +" is "+result );
});
}
</script>
</head>
<body>
<table>
<tr>
<td>Your Name:</td>
<td>《input type="text" name="name" id="name" /><td>
</tr>
<tr>
<td>Roll Number:</td>
<td><input type="text" name="rno" id="rno" /><td>
</tr>
<tr>
<td></td>
<td><input type="button" name="submit" id="submit" value="submit" onClick = "getdetails()" /></td>
</tr>
</table>
<div id="msg"></div>
</body>
</html>
php处理页 details.php
$name = $_POST['fname'];
$rno = $_POST['id'];
$con = mysql_connect("localhost","root","");
$db= mysql_select_db("school", $con);
$sql = "SELECT address from students where name='".$name."' AND rno=".$rno;
$result = mysql_query($sql,$con);
$row=mysql_fetch_array($result);
echo $row['address'];
?>
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