PHP登录（ajax提交数据和后台校验）实例分享

1.前台ajax数据提交

<form id="login_form" action="" method="POST">
 <div class="login_frame" style="position:relative";>
 <div class="login_gl" style="margin-top:35px;">
 <span class="login_wz" >后台管理系统</span>
 </div>
 <div class="login_user">
 <input id="username" name="username" type="text" placeholder="请输入您的用户名" value="" style="width:100%;height:32px;border-style:none;font-size:16px;color:#959595;"/>
 </div>
 <div class="login_user">
 <input id="password" name="password" type="password" placeholder="请输入您的密码" value="" style="width:100%;height:32px;border-style:none;font-size:16px;color:#959595;"/>
 </div>
 <div id="login_btn" class="login_log">
 <span style="font-size:16px;">登录</span>
 </div>
 </div>
 </form>
</div>
<script type="text/javascript">
 $("#login_btn").click(function(){
 var username = $.trim($("#username").val());
 var password = $.trim($("#password").val());
 if(username == ""){
 alert("请输入用户名");
 return false;
 }else if(password == ""){
 alert("请输入密码");
 return false;
 }
 //ajax去服务器端校验
 var data= {username:username,password:password};
 $.ajax({
 type:"POST",
 url:"__CONTROLLER__/check_login",
 data:data,
 dataType:'json',
 success:function(msg){
 //alert(msg);
 if(msg==1){
 window.location.href = "{:U('Index/personal')}"; 
 }else{
 alert("登录失败，请重试!");
 }
 }
 });
}); 
</script>

2.后台校验：

* */
 public function check_login(){
 $password=I('param.password');
 $username=I('param.username');
 $data["name"]=$username;
 $user=M('systemuser');
 $list=$user->where($data)->find();
 $return=0;
 if($list!=""){
 if($list['password']==md5($password) && $list['status'] == 1){
 //登录时间和登录IP
 $public = new PublicController();
 $lastlogonip=$public->ip_address();
 
 $time=$time=date("Y-m-d H:i:s", time());
 $where=array('id'=>$list['id']);
 
 $user->where($where)->save(array('lastlogonip'=>$lastlogonip,'lastlogontime'=>$time));
 $this->login($list);
 $return=1;//登录成功
 }
 }else{
 $return=2;//登录失败
 }
 $this->ajaxReturn($return);
 }

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