如何使用PHP中的选择选项从数据库获取数据？

Question

我将使用select选项从数据库中获取数据，但没有任何显示任何输出......任何人都可以找到问题是什么问题，无论是我的代码还是我们采用了错误的技术？这是我的代码`

            <div class="heading" style="margin-bottom:10px; float:left; color:black;">Select Records</div>
            <div style="float:right; color:black;">
                <form >
                <select  name="session" >
                <option value='Null'><b>Select Session</b></option>             
                <?php
                $status_on = 'no';
                            $query1 = "select * from sessions where 
                          status='$status_on'";

                            $run = mysqli_query($con,$query1);
                            while($row=mysqli_fetch_array($run)){                   
                            $id = $row['session_id'];
                            $p_id = $row['program_id'];
                            $session_name  = $row['session_name'];
                            echo "      
                            <option value='$id'><b>$session_name</b></option>";

                        }
                            ?>          

                </select>       
                <input type="submit" value="go"  />
                <div style="color:red;"> 
                    <?php 
                        if(isset($_GET['go'])){

                        $session = $_GET['session'];

                            $sql = "select * from students where session_id like'%$session%'";
                          //$sql = "select * from students where session_id='$session%'";
                            // also tried

                            $run = mysqli_query($con,$sql);
                            while($row = mysqli_fetch_array($run)){

                                $s_name = $row['s_name'];



                    ?>
                        <table> 
                            <tr> 
                            <td><?php echo $s_name;?></td>
                            </tr>
                                    <?php }?>
                                    <?php }?>
                        </table>
                </div>
                </form>

            </div>

            <?php include("footer.php");?>

Answer 1

对输出表进行了修正：

            <?php 
                if(isset($_GET['go'])){

                $session = $_GET['session'];

                    $sql = "select * from students where session_id like'%$session%'";
                  //$sql = "select * from students where session_id='$session%'";
                    // also tried

                    $run = mysqli_query($con,$sql);
            ?>
                <table>
            <?
                    while($row = mysqli_fetch_array($run)){

                        $s_name = $row['s_name'];



            ?>
                    <tr> 
                    <td><?php echo $s_name;?></td>
                    </tr>
                            <?php }?>
                </table>
                            <?php }?>

Answer 2

实际上只是name="go"失踪了。

错误：

<input type="submit" value="go">

正确：

<input type="submit" value="go" name="go">

现在一切都好。