Question

我在Mysql中有一个像这样调用的函数：

  DB::raw("count_adults(rooms.id) as adults"),

count_adults是这个：

BEGIN

 DECLARE adults INT;

    SELECT  
     count(*) INTO adults 
    FROM 
     clients 
    WHERE 
     clients.room_id = r_id 
        and (age >= 18 
        or age = 0);
            RETURN adults;
    END

但是我想如果这个查询没有返回结果（0）那么做另一个select或function，如下所示：

   SELECT count(*) INTO adults  FROM client_room, clients    
WHERE client_id = clients.id and client_room.room_id = r_id and (age >18 or age = 0)

我可以创建if条件吗？在PHP代码？还是在查询？像这样：

IF EXISTS (SELECT 1 FROM clients WHERE clients.room_id = r_id) 


BEGIN

DECLARE adults INT;

    SELECT  
        count(*) INTO adults 
    FROM 
        clients 
    WHERE 
        clients.room_id = r_id 
        and (age >= 18 
        or age = 0);

   RETURN adults;
END
ELSE 

BEGIN
DECLARE adults INT;
SELECT count(*) INTO adults 
FROM 
client_room, clients
WHERE 
client_id = clients.id and client_room.room_id = r_id 
and (age >18 
or age = 0);

END

我试过这个似乎有效：

BEGIN

 DECLARE adults INT;

    SELECT COUNT(*) INTO adults
    FROM clients
    WHERE clients.room_id = r_id
     and (age >= 18 or age = 0);

    RETURN IF( adults>0, adults, (SELECT count(*) FROM client_room, clients
         WHERE client_id = clients.id
         and client_room.room_id = r_id)
             );

   END

Answer 1

您可以在查询中使用Array，例如：

CASE
  WHEN function1() > 0
    THEN function1()
  WHEN function2() > 0
    THEN function2()
  ELSE
    0
END

如果function1()大于0，则将评估返回的值。如果不是function2()返回的值大于0，则执行此操作。否则就是0。

可以使用列，子查询（即，如果它返回标量），文字等，而不是function1()，function2()等。当然，它可以改变到或多或少的条件。

用if条件选择mysql

1 个答案: