dssqq82402 2015-09-17 14:32
浏览 85

MySql PHP选择下拉列表

I have the following code

echo '<table class="bookings">';
while($row = mysql_fetch_array($result)){

//set variables for events
    $id = $row['CourseDateID'];
    $begin = $row['CourseStartDate'];
    $end = $row['CourseEndDate'];
    $title = $row['CourseTitle'];
    $att = $row['Attendees'];
    $venue = $row['CourseLocation'];
    $formatted = date('Y-m-d', strtotime($begin));

    echo '<tr>';
    echo '<td>' . $formatted . '</td>';
    echo '<td>' . $title . '</td>';
    echo '<td>' . '<select>' . '<option>' . $att . '</option>' . '</select>' . '</td>';
    echo ' ' . '<td>' . '<a href="/cal.php?id=' . $id . '&begin=' . $begin . '&end=' . $end . '&title=' . $title . '&att=' . $att . '&venue=' . $venue . '">Add Bookings to Calendar</a>' . '</td>';
    echo '</tr>';
}
    echo '</table>';

I want the <select> to list the attendees as a dropdown but atm it creates a <select> box with all the attendee names in rather than a dropdown list.

Hope that makes sense.

  • 写回答

2条回答 默认 最新

  • dqj96395 2015-09-17 14:40
    关注

    Assuming the attendee value is a CSV list, e.g. John,Fred,Jane,Mary, you have to explode that into an array and loop on it:

    while($row = ...) {
        ... start table row+select
        $names = explode(',', $row['Attendees']);
        foreach($names as $name) {
            echo "<option>$name</option>"
        }
        ... end select + table row
    }
    
    评论

报告相同问题?

悬赏问题

  • ¥15 seatunnel-web使用SQL组件时候后台报错,无法找到表格
  • ¥15 fpga自动售货机数码管(相关搜索:数字时钟)
  • ¥15 用前端向数据库插入数据,通过debug发现数据能走到后端,但是放行之后就会提示错误
  • ¥30 3天&7天&&15天&销量如何统计同一行
  • ¥30 帮我写一段可以读取LD2450数据并计算距离的Arduino代码
  • ¥15 飞机曲面部件如机翼,壁板等具体的孔位模型
  • ¥15 vs2019中数据导出问题
  • ¥20 云服务Linux系统TCP-MSS值修改?
  • ¥20 关于#单片机#的问题:项目:使用模拟iic与ov2640通讯环境:F407问题:读取的ID号总是0xff,自己调了调发现在读从机数据时,SDA线上并未有信号变化(语言-c语言)
  • ¥20 怎么在stm32门禁成品上增加查询记录功能