dounuo9921 2017-02-10 19:59
浏览 74
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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>
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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?
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