如何使用ajax和php更新mysql数据库而不刷新页面
这里有一个好例子 http://www.w3schools.com/PHP/php_ajax_database.asp，它显示了一个SELECT声明，但它应该是直接的，并且可以轻松地将脚本更新为您需要的内容。
示例页面中的 HTML：
<html>
<head>
<script src="selectuser.js"></script>
</head>
<body>
<form>
Select a User:
<select name="users" onchange="showUser(this.value)">
<option value="1">Peter Griffin</option>
<option value="2">Lois Griffin</option>
<option value="3">Glenn Quagmire</option>
<option value="4">Joseph Swanson</option>
</select>
</form>
<p>
<div id="txtHint"><b>User info will be listed here.</b></div>
</p>
</body>
</html>
JavaScript：
var xmlHttp;
function showUser(str)
{
xmlHttp=GetXmlHttpObject();
if (xmlHttp==null)
{
alert ("Browser does not support HTTP Request");
return;
}
var url="getuser.php";
url=url+"?q="+str;
url=url+"&sid="+Math.random();
xmlHttp.onreadystatechange=stateChanged;
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
}
function stateChanged()
{
if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete")
{
document.getElementById("txtHint").innerHTML=xmlHttp.responseText;
}
}
function GetXmlHttpObject()
{
var xmlHttp=null;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
//Internet Explorer
try
{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
}
return xmlHttp;
}
Ajax 调用 PHP：
<?php
$q = $_GET["q"];
$con = mysql_connect('localhost', 'peter', 'abc123');
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("ajax_demo", $con);
$sql = "SELECT * FROM user WHERE id = '" . $q . "'";
$result = mysql_query($sql);
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Hometown</th>
<th>Job</th>
</tr>";
while ($row = mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "<td>" . $row['Age'] . "</td>";
echo "<td>" . $row['Hometown'] . "</td>";
echo "<td>" . $row['Job'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>
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