dounuo9921 2017-02-10 19:59
浏览 74
已采纳

当我将表单上的提交按到我的数据库时,说无法连接

I have been trying to create a form that updates to my database but... when I click "update" just says could not connect !!! WHY??? I know it hasn't updated because when I go on to my database nothing has changed... :-( .

<html>

   <head>
      <title></title>
   </head>

   <body>
      <?php

         if(isset($_POST['update'])) {


            $dbhost = 'localhost';
            $dbuser = '***';
            $dbpass = '******';

            $conn = mysqli_connect($dbhost, $dbuser, $dbpass);

            if(! $conn ) {
               die('Could not connect: ' . mysqli_error());
            }

            $Userid = $_POST['UserID'];
            $TableID = $_POST['tableID'];
            $Life_points = $_POST['Life_points'];
            $xp_points = $_POST['xp_points'];

            $sql = "UPDATE points"." SET TableID = $TableID". "WHERE UserID=  $UserID"."WHERE life_points= $Life_points"."WHERE  xp_points= $xp_points" ;

mysqli_select_db('womath');

            $retval = mysqli_query( $sql, $conn );

            if(! $retval ) {
               die('Could not update data: ' . mysqli_error());
            }
            echo "Updated data successfully
";

            mysqli_close($conn);
         }
         else {


            ?>
               <form method = "post" action = "<?php $_PHP_SELF ?>">
                  <table width = "400" border =" 0" cellspacing = "1" 
                     cellpadding = "2">

                     <tr>
                        <td width = "100">UserID</td>
                        <td><input name = "UserID" type = "int" 
                           id = "UserID"></td>
                     </tr>
                  <tr>
                     <td width = "100">TableID</td>
                     <td><input name = "TableID" type = "int" 
                        id = "TableID"></td>
                  </tr>
                  <tr>
                     <td width = "100">life_points</td>
                     <td><input name = "life_points" type = "int" 
                        id = "life_points"></td>
                  </tr>

                     <tr>
                        <td width = "100">xp_points</td>
                        <td><input name = "xp_points" type = "int" 
                           id = "xp_points"></td>
                     </tr>

                     <tr>
                        <td width = "100"> </td>
                        <td> </td>
                     </tr>

                     <tr>
                        <td width = "100"> </td>
                        <td>
                           <input name = "update" type = "submit" 
                              id = "update" value = "Update">
                        </td>
                     </tr>

                  </table>
               </form>
               <?php
         }
      ?>

   </body>
</html>
  • 写回答

2条回答 默认 最新

  • doumibi6899 2017-02-10 20:31
    关注

    There are few things you need to change in your code, such as:

    • If you're taking only integer input for all of your <input> fields, then make the type as number instead of int.

      <input name = "UserID" type = "number" id = "UserID">
      <input name = "TableID" type = "number" id = "TableID">
      // and so on ...
      
    • There are few syntax errors in your code. Also, your SQL query is wrong. So change your code in the following way,

      if(isset($_POST['update'])) {
          // your code
          $Userid = $_POST['UserID'];
          $TableID = $_POST['TableID'];
          $Life_points = $_POST['life_points'];
          $xp_points = $_POST['xp_points'];
      
          $sql = "UPDATE points SET TableID = " . $TableID . " WHERE UserID = ". $Userid . " AND life_points = " . $Life_points . " AND xp_points= " . $xp_points;
          // your code
      }else{
          // your code
      }
      
    • mysqli_query( $sql, $conn ) should be mysqli_query( $conn, $sql ). mysqli_query() takes the first argument as your connection handler and second argument as your query string.

    • Since you're doing UPDATE operation, use mysqli_affected_rows() function to check how many rows are affected from this UPDATE operation.

    • Last but not the least, don't mix mysql and mysqli APIs. Use either mysqli or pdo. Also, learn about prepared statement since it prevents your database from any kind of SQL injection attack.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(1条)

报告相同问题?

悬赏问题

  • ¥500 我要找大模型一体机产品手册和应用案例
  • ¥60 用r语言进行数据分析
  • ¥20 关于游戏c++语言代码问题
  • ¥15 如何制作永久二维码,最好是微信也可以扫开的。(相关搜索:管理系统)
  • ¥15 delphi indy cookie 有效期
  • ¥15 labelme打不开怎么办
  • ¥35 按照图片上的两个任务要求,用keil5写出运行代码,并在proteus上仿真成功,🙏
  • ¥15 免费的电脑视频剪辑类软件如何盈利
  • ¥30 MPI读入tif文件并将文件路径分配给各进程时遇到问题
  • ¥15 pycharm中导入模块出错