我只想在一个表中显示我的数据库，表中的每一行都有一个复选框。用户可以选择选中一个或多个框，然后点击提交按钮，在另一个页面中查看所选数据。下面是我尝试的代码。
索引.php

<!--Adds column titles -->
      <div style="overflow-x:auto;">
         <table id="test_data">
           <tr>
             <th> Select </th>
             <th> Test ID </th>
             <th> Path </th>
             <th> Video 1 Path </th>
             <th> Video 2 Path</th>
             <th> Video 3 Path</th>
             <th> Video 4 Path</th>
           </tr>

           <!--Prints out data from test_data table-->
           <?php
             $sql = "SELECT * FROM test_data";
             $result= mysqli_query($conn, $sql);
             $queryResults = mysqli_num_rows($result);

               if ($queryResults > 0) {
                 echo "<form action= 'search.php' method='get'>";
                 while ($row = mysqli_fetch_assoc($result)) {
                   echo "<tr>
                           <td><input type='checkbox' name='checkbox_id' value='" . $test_id . "'> </td>
                           <td> ".$row['test_id']." </td>
                           <td> ".$row['path']." </td>
                           <td> ".$row['video1_path']." </td>
                           <td> ".$row['video2_path']." </td>
                           <td> ".$row['video3_path']." </td>
                           <td> ".$row['video4_path']." </td>
                         </tr>";

                 }
                echo "<input type= 'submit' value='submit' >";
               echo "</form>";
               }
           ?>

         </table>
       </div>

搜索.php

<div style="overflow-x:auto;">
          <table id="test_data">
            <tr>
              <th> Test ID </th>
              <th> Path </th>
              <th> Video 1 Path </th>
              <th> Video 2 Path</th>
              <th> Video 3 Path</th>
              <th> Video 4 Path</th>
            </tr>

            <?php
              if (isset($_POST['checkbox_id'])) {
                $checkbox_id = $_POST['checkbox_id'];
                $sql= "SELECT * FROM test_data WHERE test_id IN $checkbox_id";
                $result = mysqli_query($conn, $sql);
                $queryResults = mysqli_num_rows($result);

                if ($queryResults > 0) {
                  while ($row = mysqli_fetch_assoc($result)){
                    $field1name = $row["test_id"];
                    $field2name = $row["path"];
                    $field3name = $row["video1_path"];
                    $field4name = $row["video2_path"];
                    $field5name = $row["video3_path"];
                    $field6name = $row["video4_path"];

                    echo "<tr>
                            <td> ".$field1name." </td>
                            <td> ".$field2name." </td>
                            <td> ".$field3name." </td>
                            <td> ".$field4name." </td>
                            <td> ".$field5name." </td>
                            <td> ".$field6name." </td>
                          </tr>";
                  }
                }else {
                  echo "There are no results matching your search";
                }
              }

            ?>
          </table>
        </div>

会创建复选框，但单击“提交”按钮时，不会显示任何数据。有什么帮助吗？