php - SQL 按案例排序错误

Question

我创建了一个 SQL 语句，该语句应返回一个项目列表，该列表按与所提供的三个词的相关性排序。 (TAG_1、TAG_2 等)。这是代码:

$sql = mysql_query("SELECT * FROM SEARCH_POSTS WHERE(
    POST_TAG_1 LIKE '%{$_POST['TAG_1']}%'
    OR POST_TAG_1 LIKE '%{$_POST['TAG_2']}%'
    OR POST_TAG_1 LIKE '%{$_POST['TAG_3']}%'
    OR POST_TAG_2 LIKE '%{$_POST['TAG_1']}%'
    OR POST_TAG_2 LIKE '%{$_POST['TAG_2']}%'
    OR POST_TAG_2 LIKE '%{$_POST['TAG_3']}%'
    OR POST_TAG_3 LIKE '%{$_POST['TAG_1']}%'
    OR POST_TAG_3 LIKE '%{$_POST['TAG_2']}%'
    OR POST_TAG_3 LIKE '%{$_POST['TAG_3']}%'
) ORDER BY(
    CASE 
    WHEN  POST_TAG_1 LIKE '%{$_POST['TAG_1']}%' THEN 1
    ELSE 0
    END CASE + 
    CASE 
    WHEN  POST_TAG_1 LIKE '%{$_POST['TAG_2']}%' THEN 1
    ELSE 0 
    END CASE +    
    CASE 
    WHEN  POST_TAG_1 LIKE '%{$_POST['TAG_3']}%' THEN 1
    ELSE 0
    END CASE +
    CASE 
    WHEN  POST_TAG_2 LIKE '%{$_POST['TAG_1']}%' THEN 1
    ELSE 0
    END CASE +     
    CASE 
    WHEN  POST_TAG_2 LIKE '%{$_POST['TAG_2']}%' THEN 1
    ELSE 0 
    END CASE +    
    CASE 
    WHEN  POST_TAG_2 LIKE '%{$_POST['TAG_3']}%' THEN 1
    ELSE 0
    END CASE +
    CASE 
    WHEN  POST_TAG_3 LIKE '%{$_POST['TAG_1']}%' THEN 1
    ELSE 0
    END CASE +     
    CASE 
    WHEN  POST_TAG_3 LIKE '%{$_POST['TAG_2']}%' THEN 1
    ELSE 0 
    END CASE +    
    CASE 
    WHEN  POST_TAG_3 LIKE '%{$_POST['TAG_3']}%' THEN 1
    ELSE 0
    END CASE) DESC");  

WHERE 子句工作正常，但 ORDER BY(...) 似乎给我一个错误。我究竟做错了什么？一切看起来都井井有条。谢谢。

Answer 1

END CASE 应该只是 END。 END CASE用于结束存储过程中的CASE语句，不用于结束CASE表达式。