<html>
<head>
</head>
<body>
<formaction="7.PHP">
地址：<inputtype="text"name="addr"value="localhost">
用户：<inputtype="text"name="name"value="root">
密码：<inputtype="text"name="pass"value="root">
<inputtype="button"value="提交"onclick="a();">
</form>
<scriptsrc="js\jquery-1.11.2.min.js"></script>
<script>
functiona(){
varcont=$("input").serialize();
$.ajax({
url:"7.PHP",type:'post',data:cont,success:function(data){
varres=eval("("+data+")");//转为Object对象
//用JQ来取值
//$.each(res,function(i,v){
//$.each(v,function(m,n){
//document.write(m+"-----"+n+"<br>");
//})
//})
//用for来取值
for(variinres){
for(varninres[i]){
document.write(n+"-----"+res[i][n]+"<br>");
}
}
//document.write("Host:"+res[0]["Host"]+"<br>");
//document.write("User:"+res[0]["User"]+"<br>");
//document.write("Pass:"+res[0]["Password"]+"<br>");
}
})
}
</script>
</body>
</html>
PHP页面
<?PHP
$name=$_POST["name"];
$pass=$_POST["pass"];
$addr=$_POST["addr"];
$conn=MysqL_connect($addr,$name,$pass);
if(!$conn){
die('Couldnotconnect:'.MysqL_error());
}
MysqL_select_db("MysqL");
$str="select*fromuser;";
$result=MysqL_query($str);
for($i=0;$arr[$i]=MysqL_fetch_assoc($result);$i++);
array_pop($arr);
$arr=json_encode($arr);
echo$arr;
MysqL_close($conn);
?>
