I am trying to code a function which creates a dropdown of school names selected from a database. It is doing fine creating a dropdown but it is not putting anything in the dropdown. Here is the code:

我正在尝试编写一个函数,该函数创建从数据库中选择的学校名称下拉列表。它正在创建一个下拉列表,但它没有在下拉列表中放置任何内容。这是代码:

function schoolDD($name, $selected){
   $select = '';
   if( $selected != null )
   {
      $select = $selected;
   }

     $qry = "select *   
             from   school
             order by name, id
             where display = 'Y'";

     $schools = _execQry($qry);


   $html = '<select name="'.$name.'" >';

   foreach( $schools as $s ){
      $html .= '<option value="'. $s['id'] .'"';
      if( $select == $s['name'] ){
         $html .= 'selected="selected"';
      }
      $html .= '>'. $s['name'] . '</option>';

   }
   $html .= '</select>';
   return $html;
}

3 个解决方案

#1


Problem solved. It was because in the query I had order before where. It should have been:

问题解决了。这是因为在我之前订购的查询中。应该是:

$qry = "select *   
             from   school
             where display = 'Y'
             order by name, id";

Not:

 $qry = "select *   
             from   school
             order by name, id
             where display = 'Y'";

更多相关文章

  1. MySQL常用的函数
  2. MySQL查询返回用户的收件人列表
  3. Atitit 数据库 标准库 &#160;sdk 函数库 编程语言 mysql oracle
  4. C标准库中的函数定义在哪里?
  5. 如何使用Angularjs显示函数的返回值
  6. JavaScript 字符串函数 之查找字符方法(一)
  7. 如何将对象作为参数传播给函数?
  8. 来自jQuery的Javascript中的等效函数
  9. 如何在javascript函数中将URL编码为参数?

随机推荐

  1. 新安装Android(安卓)Studio遇到问题集合+
  2. android studio2.0解决办法 Plugin is to
  3. android 之JSON
  4. 在android上监听网络状态的变更
  5. Android 判断网络是否可用以及网络类型(WI
  6. android (22)
  7. Android:Galaxy Nexus升级到4.1.2,并root(设
  8. This Android SDK requires Android Deve
  9. Android(1)进程通信基础知识
  10. Android学习之线性布局管理器