无法显示数据库中的 table

Question

无法显示数据库中的 table。每个程序都有一个 table，其中保存了一些分数。另外，还有一个程序table，其中保存了program id和program name。当我选择一个程序和一年时，出现以下错误：

No data found

Notice: Undefined variable: result in C:\XAMPP\htdocs\statistics\ac_directorPrograms.php on line 61

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, null given in C:\XAMPP\htdocs\statistics\ac_directorPrograms.php on line 61

操作代码中的 result 出了点问题，我猜 prog_name 和 program 形成了表单。

表格：

   <?php 
      include 'connect.php'; 
   ?>
    <form name="myform" action="ac_directorPrograms.php" method="POST">

      <b>Programs:<b/>
        <select name="program"> 
        <option value="Choose">Please select..</option>
        <?php
                $sql = mysql_query("SELECT prog_name FROM program");
                while ($row = mysql_fetch_array($sql)) {
                echo "<option value='" . $row['prog_name'] . "'>" .    $row['prog_name'] ."</option>";
                }
        ?>  
         </select><br/><br/>

<b>Year:<b/>
<select name="year"> 
<option value="Choose">Please select..</option>
<option value="2005">2005</option> 
<option value="2006">2006</option>
<option value="2007">2007</option></select><br/><br/>


<br/>
<input type="submit" value="submit" name="Submit">
<input type="reset" name="reset" value="Clear">

</form>
</div>

操作代码

      <?php 

        include 'connect.php';

        $year = $_POST['year'];
        $program = $_POST['program']; 
        $years     = array(
            2005,
            2006,
            2007
       );
       $programs = array(
          'bsc computer science',
          'bsc psychology',
          'ba finance',
          'ba marketing',
          'ba management'
       );

       if (in_array($program, $programs) && in_array($year, $years)) {

          $sql = "SELECT a1,a2,a3,l1,l2,l3,l4,l5,l6,l7,lavg,r1,r2,u1,u2,u3 FROM $program WHERE year=$year";

            $result = mysql_query($sql);
      }

      else {
            echo "No data found";
     }
     ?>
        <html>
         <head>
         <link rel="stylesheet" type="text/css" href="../../statistics/style.css">
         </head>
          <body>

           <div id="container">
            <table id="table" width="900" border="1" cellspacing="1">

             <tbody>
               <tr>
                <td>A1 </td>
                <td>A2 </td>
                <td>A3 </td>
                <td>L1 </td>
                <td>L2 </td>
                <td>L3 </td>
                <td>L4 </td>
                <td>L5 </td>
                <td>L6 </td>
                <td>L7 </td>
                <td>LAVG </td>
                 <td>R1 </td>
                <td>R2 </td>
                <td>U1 </td>
                <td>U2 </td>
               <td>U3 </td>
         </tr>
       </tbody>
       <?php
              while($program=mysql_fetch_assoc($result)){
                   echo "<tr>";
                   echo "<td>".$program['a1']."</td>";
                   echo "<td>".$program['a2']."</td>";
                   echo "<td>".$pprogram['a3']."</td>";
                   echo "<td>".$pprogram['l1']."</td>";
                   echo "<td>".$program['l2']."</td>";
                   echo "<td>".$program['l3']."</td>";
                   echo "<td>".$program['l4']."</td>";
                   echo "<td>".$program['l5']."</td>";
                   echo "<td>".$program['l6']."</td>";
                   echo "<td>".$program['l7']."</td>";
                   echo "<td>".$program['lavg']."</td>";
                   echo "<td>".$program['r1']."</td>";
                   echo "<td>".$program['r2']."</td>";
                   echo "<td>".$program['u1']."</td>";
                   echo "<td>".$program['u2']."</td>";
                   echo "<td>".$program['u3']."</td>";
                  echo "</tr>";    
               }
            ?>
      </table>
    </div>
   </body>
   </html>

Answer 1

这是变量作用域的问题。您已经在 if 条件内定义了 $result 并在 if 之外使用它。您需要在 if 条件之外定义 $result。

Answer 2

在操作代码中 include connect.php 之后，添加此... $con=mysqli_connect("localhost","my_user","my_password","my_db"); 并且不要忘记将 my_user、my_password、my_db 替换为您的用户名、密码和数据库名称。

也用这个 $result=mysqli_query($con,$sql);

替换此代码 $result = mysql_query($sql);

无法显示数据库中的 table

Cannot display the table from database

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php

mysql

html-table