ajax php 实现无刷新用户检查是否存在 (2011-12-04 15:02)
当输入用户名后，失去焦点时 就可以提示用户名是不是存在……
改连库语句就可以实现！
test.html
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
<title>无标题文档</title>
</head>
<script type="text/javascript">
function ajaxFunction()
{
var xmlHttp;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer
try
{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
try
{
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e)
{
alert("您的浏览器不支持AJAX！");
return false;
}
}
}
xmlHttp.onreadystatechange=function()
{
if(xmlHttp.readyState==4)
{
document.myForm.exsit.value=xmlHttp.responseText;
}
}
var name=document.myForm.username.value;
xmlHttp.open("get","test.php?name="+name,true);
xmlHttp.send(null);
}
</script>
<form name="myForm">
用户: <input type="text" name="username" onblur="ajaxFunction(name);" />
是否存在: <input type="text" name="exsit" />
</form>
</body>
</html>
test.php
<?php
$name=$_GET["name"];
$conn=mysql_connect("localhost","root","123456");
mysql_select_db(ddg,$conn);
mysql_query("set names UTF8");
$sql="select * from ddg_member where member_name ='$name'";
$rs=mysql_query($sql);
$result=mysql_num_rows($rs);
if($result>0)
{
echo "no";
}
else
{
echo "yes!";
}
?>
