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HomeBackend DevelopmentPHP Tutorial请问:PHP js 实现复杂的按钮功能

请教:PHP js 实现复杂的按钮功能
请教各位大虾:我要制作的网页里有一个按钮,点击这个按钮要实现以下几个功能:

先判断页面上给出的姓名是否已经存在于SQL数据库中了。如果存在,就给个提示框,然后跳出。如果不存在,要将页面上的内容(如姓名、性别、Email、电话等)存入SQL数据库,再提示已经写入数据库,然后跳转到另一个页面。

我定义的按钮如下(a.php文件):



然后打算在a.js文件的function data_send()里实现按钮的功能,目前给出提示框和跳转网页我知道怎么实现,可是如何判断是否应该写入数据库,以及如何在js里写入数据库,我就不得而知了。

我是初来乍到的菜鸟,还望各位前辈多多指教,先谢谢大家帮忙啦~!


------解决方案--------------------
ajax检测和入库,然后再跳转
------解决方案--------------------
利用ajax请求php,由php来检查姓名是否存在。
------解决方案--------------------
我最近也在研究,分享下:
文件1、(conn.php) 代码如下:
$conn = mysql_connect('localhost','数据库用户名','数据库密码');
mysql_select_db('数据库名',$conn);
?>

文件2、(a.html)代码如下:


      
           
           
           
      
      
           
      
用户名:  


<script><br />var xmlHttp<br /><br />function check(str){<br /> <br />xmlHttp=GetXmlHttpObject()<br />if (xmlHttp==null)<br /> {<br /> alert ("Browser does not support HTTP Request")<br /> return<br /> } <br />var url="a.php"<br />url=url+"?n="+str<br />url=url+"&sid="+Math.random()<br />xmlHttp.onreadystatechange=stateChanged <br />xmlHttp.open("GET",url,true)<br />xmlHttp.send(null)<br />} <br /><br />function stateChanged() <br />{ <br />if (xmlHttp.readyState==4 <br><font color='#FF8000'>------解决方案--------------------<br> xmlHttp.readyState=="complete")<br /> { <br /> document.getElementById("name").innerHTML=xmlHttp.responseText <br /> } <br />}<br /><br /><br />function GetXmlHttpObject()<br />{<br />var xmlHttp=null;<br /><br />try<br /> {<br /> // Firefox, Opera 8.0+, Safari<br /> xmlHttp=new XMLHttpRequest();<br /> }<br />catch (e)<br /> {<br /> // Internet Explorer<br /> try<br /> {<br /> xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");<br /> }<br /> catch (e)<br /> {<br /> xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");<br /> }<br /> }<br />return xmlHttp;<br />}<br /></script>

文件3、(a.php) 代码如下:

include "conn.php";
$sql = 'select * from user';//user为你的数据库名,自己修改
$res = mysql_query($sql);
while($row=mysql_fetch_assoc($res)){
$u_name[]= $row['name'];//$row['name']里面的name为你数据库字段名

}
$name = $_GET['n'];
if(in_array($name,$u_name)){
echo "此用户名已被注册请输入新的用户名!";
}else{
echo "可以使用";
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