Question

我在显示数据库中的数据时遇到了一些问题...

我有两个数据库表：

类别（cat_id，cat_name）
menu_items（menu_id，menu_name，cat_id，menu_description，menu_price）

我想将数据显示为：

分类1名称

cat_id = 1，menu_name，menu_description，menu_price
cat_id = 1，menu_name，menu_description，menu_price
cat_id = 1，...

分类2名称

cat_id = 2，menu_name，menu_description，menu_price
cat_id = 2，menu_name，menu_description，menu_price
cat_id = 2，...

分类3名称

cat_id = 3，menu_name，menu_description，menu_price
cat_id = 3，menu_name，menu_description，menu_price
cat_id = 3，....

...

我的代码：

function get_menu_items() {
       // query database for a list of categories
       $conn = db_connect();
       $query = "select cat_id, cat_name from categories";
       $result = @$conn->query($query);
       if (!$result) {
         return false;
       }
       $num_cats = @$result->num_rows;
       if ($num_cats == 0) {
          return false;
       }
       $result = db_result_to_array($result);
       return $result;
    }


function display_menu_items($menu_array) {
  if (!is_array($menu_array)) {
     echo "<p>No menu items currently available</p>";
     return;
  }
  foreach ($menu_array as $row)  {
    echo "<section id='".$row['cat_name']."'>";
    echo "<div class='group course'>";
    echo "<h3>".$row['cat_name']."</h3>";

    ***/* Problem */***

     echo "<ul>";
foreach... {
    echo "<li>cat_id = #, menu_name, menu_description, menu_price</li>";
    }
    echo "</ul>

    ***/* Problem */***

    echo "</div>";
    echo "</section>";
  }
}

有什么建议吗？

感谢。

Answer 1

catRecordset = select * from categories
foreach cat in catRecordset
{
    print( cat.name );  
    menuRecordset = select * from menu where menyu.cat_id = cat.cat_id
    foreach( menuItem in menuRecordSet )
    {
        print( menuItem.whatever );
    }
}

Answer 2

甚至更好地使用join并只运行一个查询

$Recordset = 'select * from categories cat LEFT JOIN menu ON menu.cat_id = cat.cat_id';

$category = '';  
foreach $record in $Recordset  
{
    if ($category != $record['cat_id']) {  
        print( cat.name );   
        $category = $record['cat_id'];  
    }   
    print( $record['menu_name'] ..... );  
}

显示多个MySQL表中的数据

2 个答案: