php - 在选择菜单中获取mysql数据库的值

Question

我在 PHP 上选择菜单时遇到问题。我试图让 mysql 数据库选择菜单。但是，它没有显示任何内容。

这是我的代码: 默认:

        mysql_select_db($database_conn, $conn);
        $query_Rsenroll = "SELECT * FROM `tbl_enroll` WHERE `tbl_enroll`.`courseid` ='".$_GET['courseid']."'";
        $Rsenroll = mysql_query($query_Rsenroll, $conn) or die(mysql_error());
        $row_Rsenroll = mysql_fetch_assoc($Rsenroll);
        $totalRows_Rsenroll = mysql_num_rows($Rsenroll);
   $courseid = $row_Rsenroll['courseid'];
   $er_staffid = "";

        break;
        }
    ?>

        <select name="courseid">
        <option value="" SELECTED>Selected Course ID</option>
          <?php 
    foreach( $Course as $course_id) {
        if ( $course_id == $courseid) {
            $selected = " SELECTED";
        } else {
            $selected = "";
        }
            ?>
         <option value="<?php echo $course_id; ?>"<?php echo $selected; ?>><?php echo $row_Rsenroll['courseid']; ?></option>
         <?php
    }
    ?>
        </select>

感谢您的帮助和建议。

Answer 1

假设 courseid 作为发送 URL 中的变量传递 (file.php?courseid=COURSEID)，我认为这应该满足您的要求:

这可能会稍微清理你的脚本(尽管我将其切换为 mysql_fetch_array，因为我比 mysql_fetch_assoc 更熟悉它。随意使用 assoc):

   <?php
   $cid = '6116';
   ?>


   <select name="courseidMenu">
   <option value="" SELECTED>Selected Course ID</option> 
   <?php

   $query = mysql_query("SELECT * FROM tbl_enroll WHERE courseid = '$cid'", $conn)or die(mysql_error());
   $total_rows = mysql_num_rows($query);
   while($row = mysql_fetch_array($query)){
      $courseId = $row['courseid'];
      ?>
      <option value="<?=$courseId?>" ><?=$courseId?></option>      
      <?

   }
   ?>
   </select>