1. Trying to imitate a login program. This login form is on the home page. After successful login, it is required to use ajax on the home page to return the user name information (saved session value) without refreshing and automatically hide the form. Now it is true that user information can be returned through js, but as long as it is refreshed, the user information and form will be restored to their original state. How can it be maintained for a long time until the user clicks to log out or the session disappears?
2. I can see the returned user information through the xhr mode of F12. It can also be achieved through the append method and remove() of jquery. However, the problem is that the information does not exist and the form returns to its original state after refreshing.
2. Key code:
(1) IndexController.class.php:
<code>public function checkUser(){ //接收值 $userName=$_POST['username']; $userPass=$_POST['userpass']; //空值检测->function if(!trim($userName)){ return show(0,'用户名不能为空'); } if(!trim($userPass)){ return show(0,'密码不能为空'); } //对用户密码真实性进行检验->Model $res=D("Stuser")->getUser($userName); if(!$res['username']){ return show(0,'用户名不存在'); } //密码处理->function if($res['userpass']!=getMd5Pass($userPass)){ return show(0,'密码不正确'); } //echo $res['username']; //$_SESSION('username',$res); //设置session $_SESSION['username']=$res; //dump($username) ; //var_dump($username); return show(1,'登录成功',$this->getSessionNames()); } //判断session情况->index public function getSessionNames(){ if($_SESSION['username']['username']){ $username = $_SESSION['username']['username']; //$a=$this->ajaxReturn($username); //$this->assign('username',$username); return $username; } }请输入代码</code>
(2)function.php
<code>function show($status,$message,$data=array()){ $result=array( 'status' => $status, 'message' => $message, 'data' => $data, ); //JSON编码数据 exit(json_encode($result)); }请输入代码</code>
(3) login.js
<code>var login = { checkUser : function() { //获取登录页面中的用户名、密码 var userName=$('input[name="username"]').val(); var userPass=$('input[name="userpass"]').val(); if(!userName) { dialog.error("用户名不能为空"); } if(!userPass) { dialog.error("密码不能为空"); } var url="/stfjzd-12/index.php/Home/Index/checkUser"; var data={'username':userName,'userpass':userPass}; //执行异步请求 $.post(url,data,function(result){ if(result.status==0) { return dialog.error(result.message); } if(result.status==1) { if(data!=""){ //alert(data.username); $('#index_form2').remove(); $('#test').append(data.username); } return dialog.success(result.message,"/stfjzd-12/index.php/Home/Index/checkUser"); //alert(result.data['username']) ; } },'JSON'); } }请输入代码</code>
Reply content:
1. Trying to imitate a login program. This login form is on the home page. After successful login, it is required to use ajax on the home page to return the user name information (saved session value) without refreshing and automatically hide the form. Now it is true that user information can be returned through js, but as long as it is refreshed, the user information and form will be restored to their original state. How can it be maintained for a long time until the user clicks to log out or the session disappears?
2. I can see the returned user information through the xhr mode of F12. It can also be achieved through the append method and remove() of jquery. However, the problem is that the information does not exist and the form returns to its original state after refreshing.
2. Key code:
(1) IndexController.class.php:
<code>public function checkUser(){ //接收值 $userName=$_POST['username']; $userPass=$_POST['userpass']; //空值检测->function if(!trim($userName)){ return show(0,'用户名不能为空'); } if(!trim($userPass)){ return show(0,'密码不能为空'); } //对用户密码真实性进行检验->Model $res=D("Stuser")->getUser($userName); if(!$res['username']){ return show(0,'用户名不存在'); } //密码处理->function if($res['userpass']!=getMd5Pass($userPass)){ return show(0,'密码不正确'); } //echo $res['username']; //$_SESSION('username',$res); //设置session $_SESSION['username']=$res; //dump($username) ; //var_dump($username); return show(1,'登录成功',$this->getSessionNames()); } //判断session情况->index public function getSessionNames(){ if($_SESSION['username']['username']){ $username = $_SESSION['username']['username']; //$a=$this->ajaxReturn($username); //$this->assign('username',$username); return $username; } }请输入代码</code>
(2)function.php
<code>function show($status,$message,$data=array()){ $result=array( 'status' => $status, 'message' => $message, 'data' => $data, ); //JSON编码数据 exit(json_encode($result)); }请输入代码</code>
(3) login.js
<code>var login = { checkUser : function() { //获取登录页面中的用户名、密码 var userName=$('input[name="username"]').val(); var userPass=$('input[name="userpass"]').val(); if(!userName) { dialog.error("用户名不能为空"); } if(!userPass) { dialog.error("密码不能为空"); } var url="/stfjzd-12/index.php/Home/Index/checkUser"; var data={'username':userName,'userpass':userPass}; //执行异步请求 $.post(url,data,function(result){ if(result.status==0) { return dialog.error(result.message); } if(result.status==1) { if(data!=""){ //alert(data.username); $('#index_form2').remove(); $('#test').append(data.username); } return dialog.success(result.message,"/stfjzd-12/index.php/Home/Index/checkUser"); //alert(result.data['username']) ; } },'JSON'); } }请输入代码</code>
If you write it in TP and use {$Think.session.username} in the template, the refresh through js assignment will definitely be gone
Just create a hidden field on the page to put the information. The session information on the page will only be requested after you log in successfully. Refreshing the page will not trigger the request
<code>public function index(){ //你显示页面的函数 $user=$this->getSessionNames(); $this->assing('user',$user); ... }</code>
<code>//index.html <script> $(function(){ var username="{{$user}}"; if(username != ""){ $('#index_form2').remove(); $('#test').append(username); } }) </script></code>
<code> //login.js</code>
The poster’s question doesn’t seem to be difficult at all! Just make the form where you store user information dynamic, as follows:
<code>//这个是php的处理代码块 if($this->chen_user()) { //这个是验证是否登陆成功 $_SESSION['username'] = $username; //这里把用户信息存入session $_SESSION['sex] = $sex; } //这个是前端显示的 <?php if($_SESSION['username']) { ?> 如果session中有username的值就输出用户信息 没有就不输出 <td>姓名</td> <td><?php echo $_SESSION['username'];?></td> <?php } ?></code>
Do you think this is the case?

php把负数转为正整数的方法:1、使用abs()函数将负数转为正数,使用intval()函数对正数取整,转为正整数,语法“intval(abs($number))”;2、利用“~”位运算符将负数取反加一,语法“~$number + 1”。

实现方法:1、使用“sleep(延迟秒数)”语句,可延迟执行函数若干秒;2、使用“time_nanosleep(延迟秒数,延迟纳秒数)”语句,可延迟执行函数若干秒和纳秒;3、使用“time_sleep_until(time()+7)”语句。

php除以100保留两位小数的方法:1、利用“/”运算符进行除法运算,语法“数值 / 100”;2、使用“number_format(除法结果, 2)”或“sprintf("%.2f",除法结果)”语句进行四舍五入的处理值,并保留两位小数。

判断方法:1、使用“strtotime("年-月-日")”语句将给定的年月日转换为时间戳格式;2、用“date("z",时间戳)+1”语句计算指定时间戳是一年的第几天。date()返回的天数是从0开始计算的,因此真实天数需要在此基础上加1。

方法:1、用“str_replace(" ","其他字符",$str)”语句,可将nbsp符替换为其他字符;2、用“preg_replace("/(\s|\ \;||\xc2\xa0)/","其他字符",$str)”语句。

php判断有没有小数点的方法:1、使用“strpos(数字字符串,'.')”语法,如果返回小数点在字符串中第一次出现的位置,则有小数点;2、使用“strrpos(数字字符串,'.')”语句,如果返回小数点在字符串中最后一次出现的位置,则有。

在PHP中,可以利用implode()函数的第一个参数来设置没有分隔符,该函数的第一个参数用于规定数组元素之间放置的内容,默认是空字符串,也可将第一个参数设置为空,语法为“implode(数组)”或者“implode("",数组)”。

php字符串有下标。在PHP中,下标不仅可以应用于数组和对象,还可应用于字符串,利用字符串的下标和中括号“[]”可以访问指定索引位置的字符,并对该字符进行读写,语法“字符串名[下标值]”;字符串的下标值(索引值)只能是整数类型,起始值为0。


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