php - 查询多个表输出重复的结果

Question

我有一个ajax搜索引擎。如果没有打开任何过滤器，我想查询每个表中的标题。在此示例中，只有两个表。音乐页面和餐厅页面。

public function Any_Category_Name($search_title) {

                $query = mysql_query("SELECT music_page.title AS music_title, 
                                             restaurant_page.title AS restaurant_title
                                      FROM music_page,
                                           restaurant_page
                                      WHERE music_page.title 
                                            LIKE '%$search_title%' OR
                                              restaurant_page.title
                                              LIKE '%$search_title%'");
                while($fetch = mysql_fetch_assoc($query)) {
                    $music_title = $fetch['music_title'];
                    $restaurant_title = $fetch['restaurant_title'];
                    echo '<b>Restaurant:</b> '.$restaurant_title.'<br />';
                    echo '<b>Music Spot:</b> '.$music_title.'<br />';
                    }

                }

这是输入字母/搜索时的输出:

Restaurant: Giordanoes
Music Spot: jazz showcase
Restaurant: Giordanoes
Music Spot: The Green Mill
Restaurant: Giordanoes
Music Spot: ayooooo
Restaurant: Giordanoes
Music Spot: ayoooooooo
Restaurant: Giordanoes
Music Spot: new one

Giordanoes 是restaurant_page 表中的唯一行。它是为 music_page 中的每一行输出的。我该如何解决这个问题？

Answer 1

替代

while($fetch = mysql_fetch_assoc($query)) {
    // Create the listing array
    $listing[$fetch['restaurant_title']] = array($fetch['music_title'] => 'music_title');
}

// Check if we have Restaurants
if(is_array($listing)) {
    // Loop through the new listings array
    foreach($listing as $restaurant_title => $music_titles) {
        echo '<b>Restaurant:</b> '.$restaurant_title.'<br />';

        // Check if we have Music
        if(is_array($music_titles)) {
            // Loop through the Music
            foreach($music_titles as $music_title => $placeholder) {
                echo '<b>Music Spot:</b> '.$music_title.'<br />';
            }
        } else {
            echo '<b>Music Spot:</b> No Music found<br />'; 
        }            
    }
} else {
    echo '<b>Restaurant:</b> No Restaurants found<br />';
}