php - 从不同的表中获取并添加下拉菜单

标签 php mysql

我已经有了一个启动连接,我想添加另一个下拉菜单,该菜单将从不同的表中获取数据,并在我尝试工作时将其分配给所有用户,但它只显示一个用户,而数据库中有很多用户。当我删除下拉菜单时,它会显示数据库中的所有用户。

 <div class="table-responsive">
         <?php
         include 'config.php';
         $sql = "SELECT * FROM tbl_department";
         $result = $conn->query($sql);


         if ($result->num_rows > 0) {?>

      <table>
          <tr>
              <th>NO</th>
              <th>Department</th>
              <th>Status</th>
              <th>Action</th>
          </tr>

      <tbody>
          <?php
          $no = 1;
         while($row = $result->fetch_assoc()) {
             $session = $row['session'];
             if ($session == "AM") {
             $st = 'Morning';
             }else{
             $st = 'Afternoon';                                          
             }?>


             <tr>
      <td><?php echo $no; ?></td>
      <td><?php echo $row['department'] ?></td>
      <td><?php echo $row['status'] ?></td>
      <td><select class="form-control" id="school" required>
      <option value="" selected disabled>-Select School-</option>
      <?php
          include '../database/config.php';
          $sql = "SELECT * FROM tbl_school WHERE status = 'Active' ORDER BY name";
          $result = $conn->query($sql);

          if ($result->num_rows > 0) {

          while($row = $result->fetch_assoc()) {
          print '<option value="'.$row['school_id'].'">'.$row['name'].'</option>';
          }
         } else {

          }
           ?>

  </td>
               <?php
                  $no++;
                 }}?>

          </tr>


     </tbody>
     </table>

最佳答案

是否包含“../database/config.php”;并包含“config.php”;来自不同的数据库源,如果是,您应该在两个不同的资源连接之间分开,或者正如@Sean所说,您应该更改 tbl_school 行和 tbl_department 行的变量名称

我创建了 $select_options 作为示例,用于保留 tbl_school 中的所有行,防止您查询每个 tbl_department 行:

<div class="table-responsive">
    <?php


    include '../database/config.php';
    $sql = "SELECT * FROM tbl_school WHERE status = 'Active' ORDER BY name";
    $result = $conn->query($sql);
    $select_options = "";
    if ($result->num_rows > 0) {

      while($row = $result->fetch_assoc()) {
        $select_options .= '<option value="'.$row['school_id'].'">'.$row['name'].'</option>';
      }
    }
    ?>
    <?php
    include 'config.php';
    $sql = "SELECT * FROM tbl_department";
    $result = $conn->query($sql);

    ?>
    <table>
        <thead>
            <tr>
              <th>NO</th>
              <th>Department</th>
              <th>Status</th>
              <th>Action</th>
            </tr>
        </thead>
    <?php
    if ($result->num_rows > 0) { 
        ?>
        <tbody>
          <?php
            $no = 1;
            while($row = $result->fetch_assoc()) {
                 $session = $row['session'];
                 if ($session == "AM") {
                 $st = 'Morning';
                 }else{
                 $st = 'Afternoon';                                          
                 }
            ?>
            <tr>
                <td><?php echo $no; ?></td>
                <td><?php echo $row['department'] ?></td>
                <td><?php echo $row['status'] ?></td>
                <td>
                    <select class="form-control" id="school" required>
                        <option value="" selected disabled>-Select School-</option>
                        <?php
                        echo $select_options;
                        ?>
                    </select>
                </td>
            </tr>
            <?php
                $no++;
            }
        ?>
        </tbody>
        <?php
        }
        ?>
    </table>
</div>

关于php - 从不同的表中获取并添加下拉菜单,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53491628/

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