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Php 用户添加/登录表单问题_Php_Mysql - Fatal编程技术网

Php 用户添加/登录表单问题

Php 用户添加/登录表单问题,php,mysql,Php,Mysql,我制作了一个简单的用户添加/登录表单。但我不能使用数组提取数据或将数据添加到数据库中,需要简单的过程 <?php //User_Add_Page if(isset($_POST['userid'],$_POST['username'],$_POST['password'],$_POST['sex'],$_POST['desig'])) { include("connect.php"); $badchars='!@£$%^&*()+=

我制作了一个简单的用户添加/登录表单。但我不能使用数组提取数据或将数据添加到数据库中,需要简单的过程

<?php
 //User_Add_Page
    if(isset($_POST['userid'],$_POST['username'],$_POST['password'],$_POST['sex'],$_POST['desig']))
    {

       include("connect.php");
       $badchars='!@£$%^&*()+=-][\;/.,`~<>?:"|{} \'';
       $userid=$_POST['userid'];
       $username=$_POST['username'];
       $password=$_POST['password'];
       $sex=$_POST['sex'];
       $desig=$_POST['desig'];
       $errors=array();

               if(empty($userid)&& empty($username) && empty($password) && empty($sex) && empty($design))
               {
                        $errors[]= "All Fields Require";
               }
                else if($userid=="")
               {
                        $errors[]= "Enter User ID";
               }

               else if (strpbrk($userid, $badchars) !== false)
               {

                       $errors[]= 'INVALID User ID...';
               }
               else  if(strlen($userid)<5 && $username != "")
               {
                       $errors[]='User ID Must be Greater than 5 Characters';
               }
               else  if(is_numeric ($userid))
               {

                      $errors[]= 'User ID Must Conatin Both letters AND character';
               }

               else  if(strlen($userid)>20 && $username != "")
               {

                     $errors[]= 'User ID Must be Less Than 20 Characters';
               }

               else if($username=="")
               {

                       $errors[]= "Enter username";
               }
               else if(strlen($username)>30)
               {

                      $errors[]= "Enter username less than 30 char.";
               }

               else if(!ctype_alpha(str_replace(' ','',$username)))//search alphabetic char. from username
               {

                     $errors[]='Name Must be Character Only';
               }
               else if($password == "")
               {

                     $errors[]= 'Please Enter Password';
               }

               else  if(strlen($password)<5 && $password != "")
               {

                     $errors[]= 'Password Must Be Greater Than 5 Characters';
               }
               else  if($desig == "")
               {

                     $errors[]= 'Please Enter Designation';
        }


            else if(empty($errors))
            {
                       $write=("INSERT INTO adduser VALUES ('','$userid','$username','$password','$sex','$desig')");


                       $time = 1; //Time (in seconds) to wait.
                   $url = "add_user_process.php"; //Location to send to.
               header("Refresh: $time; url=$url");
            }
}



 ?>

插入后,必须指定mysql\u查询对象, like$result=mysql\u query$write

像这样,

 $write="INSERT INTO adduser VALUES ('','$userid','$username','$password','$sex','$desig')";
 $result = mysql_query($write);

你需要做的不是else ifempty$errors,而是else ifcount$errors count不起作用…解析错误:语法错误,在C:\xampp\htdocs\add_user2.php第79行出现意外..它给出了此错误检查括号可能是你离开了或添加了一个额外的括号。好的…它现在起作用了..谢谢。有更多疑问,,我不能显示表单错误,例如,如果我没有输入用户名,则应回显“输入用户名”?更多问题,,我不能显示表单错误,如果我没有输入用户名,则应回显“输入用户名”,,首先感谢您的帮助..但是您共享的链接,我无法理解…所以我做了一个更改,而不是使用数组我尝试使用普通全局变量在同一页上显示表单验证…但我没有回显任何内容…但没有错误,,我已经尝试了整整几天…我的意思是这样说,
 $write="INSERT INTO adduser VALUES ('','$userid','$username','$password','$sex','$desig')";
 $result = mysql_query($write);