15

我有一张桌子：

custID    orderID    orderComponent
=====================================
1          123        pizza
1          123        wings
1          234        breadsticks
1          239        salad
2          456        pizza
2          890        salad

我有一份价值清单——比萨饼、鸡翅、面包棒和沙拉。如果客户至少有一条记录包含这些记录，我需要一种方法来获取真/假值。这可能与 mysql 查询，还是我只需要select distinct(orderComponent)为每个用户做一个并使用 php 来检查结果？

4

4 回答 4

14

如果您只是想查看客户是否订购了所有商品，那么您可以使用：

select t1.custid,
  case when t2.total is not null 
    then 'true'
    else 'false'
  end OrderedAll
from yourtable t1
left join
(
  select custid, count(distinct orderComponent) Total
  from yourtable
  where orderComponent in ('pizza', 'wings', 'breadsticks', 'salad')
  group by custid
  having count(distinct orderComponent) = 4
) t2
  on t1.custid = t2.custid

请参阅带有演示的 SQL Fiddle

如果您想扩展它，查看是否custid已在一个订单中订购了所有商品，那么您可以使用：

select t1.custid,
  t1.orderid,
  case when t2.total is not null 
    then 'true'
    else 'false'
  end OrderedAll
from yourtable t1
left join
(
  select custid, orderid, count(distinct orderComponent) Total
  from yourtable
  where orderComponent in ('pizza', 'wings', 'breadsticks', 'salad')
  group by custid, orderID
  having count(distinct orderComponent) = 4
) t2
  on t1.custid = t2.custid
  and t1.orderId = t2.orderid

请参阅带有演示的 SQL Fiddle

如果您只需要 custid 和真/假值，则可以添加distinct到查询中。

select distinct t1.custid,
  case when t2.total is not null 
    then 'true'
    else 'false'
  end OrderedAll
from yourtable t1
left join
(
  select custid, count(distinct orderComponent) Total
  from yourtable
  where orderComponent in ('pizza', 'wings', 'breadsticks', 'salad')
  group by custid
  having count(distinct orderComponent) = 4
) t2
  on t1.custid = t2.custid

请参阅带有演示的 SQL Fiddle

或者通过 custid 和 orderid：

select distinct 
  t1.custid,
  t1.orderid,
  case when t2.total is not null 
    then 'true'
    else 'false'
  end OrderedAll
from yourtable t1
left join
(
  select custid, orderid, count(distinct orderComponent) Total
  from yourtable
  where orderComponent in ('pizza', 'wings', 'breadsticks', 'salad')
  group by custid, orderID
  having count(distinct orderComponent) = 4
) t2
  on t1.custid = t2.custid
  and t1.orderId = t2.orderid

请参阅带有演示的 SQL Fiddle

于 2013-01-15T17:00:22.297 回答
5 
select case when 
count(distinct orderComponent) = 4 
then 'true' 
else 'false' 
end as bool
from tbl
where custID=1
于 2013-01-15T16:51:12.810 回答
2

这是一种方法。此方法不需要内联视图（派生表），如果您想包含多个条件的标志，则可能会很有效：

编辑：

这将返回custID所有四个项目都有一行：

SELECT t.custID
     , MAX(IF(t.orderComponent='breadsticks',1,0))
       + MAX(IF(t.orderComponent='pizza',1,0))
       + MAX(IF(t.orderComponent='salad',1,0))
       + MAX(IF(t.orderComponent='wings',1,0)) AS has_all_four
  FROM mytable t
 GROUP BY t.custID
HAVING has_all_four = 4

原始答案：

（这会检查包含所有四个项目的客户“订单”，而不仅仅是“客户 ID”。）

SELECT t.custID
     , t.orderID
     , MAX(IF(t.orderComponent='breadsticks',1,0))
       + MAX(IF(t.orderComponent='pizza',1,0))
       + MAX(IF(t.orderComponent='salad',1,0))
       + MAX(IF(t.orderComponent='wings',1,0)) AS has_all_four
  -- , MAX(IF(t.orderComponent='breadsticks',1,0)) AS has_breadsticks
  -- , MAX(IF(t.orderComponent='pizza',1,0)) AS has_pizza
  -- , MAX(IF(t.orderComponent='salad',1,0)) AS has_salad
  -- , MAX(IF(t.orderComponent='wings',1,0)) AS has_wings
  FROM mytable t
 GROUP BY t.custID, t.orderID
HAVING has_all_four = 4

这将获得包含所有四个项目的“订单”。如果您只想返回 custID 的值，则将上面的查询用作内联视图（将其包装在另一个查询中）

SELECT s.custID
  FROM (
         SELECT t.custID
              , t.orderID
              , MAX(IF(t.orderComponent='breadsticks',1,0))
                + MAX(IF(t.orderComponent='pizza',1,0))
                + MAX(IF(t.orderComponent='salad',1,0))
                + MAX(IF(t.orderComponent='wings',1,0)) AS has_all_four
             FROM mytable t
            GROUP BY t.custID, t.orderID
           HAVING has_all_four = 4
       ) s
 GROUP BY s.custID
于 2013-01-15T17:13:19.663 回答
0

@EmmyS：你可以双向进行。如果您想使用 MySql 进行检查，请使用：

SELECT @rowcount:=COUNT(*) FROM orderComponent Where (Your Conditions);
IF (@rowcount > 0) THEN
    'True'
ELSE
    'False'
END IF
于 2013-01-15T16:52:34.930 回答