Php 使用if条件选择mysql

我在Mysql中有一个函数，叫做：

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

   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)

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

伯爵大人是这样的：

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，如果此查询未返回任何结果（0），我希望执行另一个select或函数，如下所示：

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

   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)

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

我可以创建一个if条件吗？在php代码中？还是在查询中？大概是这样的：

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

   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)

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

我已经试过了，它似乎很管用：

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

您可以在查询中使用，例如：

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

如果大于0，则将计算为从

function1（）

返回的值。如果不是，但

function2（）

返回的值大于0，则取该值。否则是0

---

可以使用列、子查询（即，如果返回标量）、文本等，而不是

function1（）

，

function2（）

等。当然，它可以更改为或多或少的条件。

这将类似于：当count\u adulst1（）>0时开始CASE，然后当count\u adulst1（）>0时开始CASE，然后count\u adulst1（）否则0结束，在php中，我将调用这个有CASE的函数，比如：检查成员？@Mary:就我理解你的要求而言，

CASE

在我看来是正确的。我不知道你调用“this function”是什么意思，因为从技术上讲，查询不是PHP函数。。。将

大小写

放在查询中需要计数的地方。比方说，如果您的查询是

从表中选择count_成人（）将“count_成人（）”替换为您的
案例
。并在PHP中使用结果查询，就像使用原始查询一样。希望有帮助。