我想创建一个“票”,计算每个ID的通过次数。当我们在任何ID上都有黄金通行证时,这意味着该通行证适用于所有预订的人。因此,对于此示例,我们要计数5。对于其他pass_code,我们只想计算通过次数,并排除那些为空的次数。我在下面有一个预期的输出。
说我有这个数据:
Passes
ID | GuestID | Pass_code
----------------------------
100 | 001 | Bronze
100 | 002 | Bronze
101 | 103 | Gold
101 | 104 | NULL
101 | 105 | NULL
101 | 106 | NULL
101 | 107 | NULL
102 | 208 | Silver
103 | 209 | Steel
103 | 210 | Steel
103 | 211 | NULL
Passengers
ID | Passengers
-----------------
100 | 2
101 | 5
102 | 1
103 | 3
我想算一下,然后在以下输出中创建票证:
ID 100 | 2 pass (bronze)
ID 101 | 5 pass (because it is gold, we count all passengers)
ID 102 | 1 pass (silver)
ID 103 | 2 pass (steel) (2 passes rather than than 3 as we just want to count only the passes for steel, bronze silver)
我想做这样的事情,但作为组合查询。
DECLARE @ID = 101; -- i will want to pass in IDs
-- for gold, we want to count all passengers when the gold pass is on
SELECT pp.Passengers
FROM passes
JOIN Passengers pp ON p.ID = pp.ID
WHERE p.pass_code IN'%gold%'
AND PP.id = @id
-- for bronze, silver and steel
SELECT
count(p.ID)
FROM Passes
WHERE p.ID = @id
AND P.pass_code IN ('Bronze', 'silver', 'steel') -- Dont want to check based on NUlls as this may chnage to something else.
)
任何帮助或建议,将不胜感激。
这对您有用吗?
with Passes as (
select 100 as id, 001 as guestid, 'Bronze' as passcode from dual union all
select 100 as id, 002 as guestid, 'Bronze' as passcode from dual union all
select 101 as id, 103 as guestid,'Gold' as passcode from dual union all
select 101 as id, 104 as guestid, NULL as passcode from dual union all
select 101 as id, 105 as guestid, NULL as passcode from dual union all
select 101 as id, 106 as guestid, NULL as passcode from dual union all
select 101 as id, 107 as guestid, NULL as passcode from dual union all
select 102 as id, 208 as guestid, 'Silver' as passcode from dual union all
select 103 as id, 209 as guestid, 'Steel' as passcode from dual union all
select 103 as id, 210 as guestid, 'Steel' as passcode from dual union all
select 103 as id, 211 as guestid, NULL as passcode from dual
)
SELECT
id,passcode,count(ID)
FROM Passes
where passcode is not null and passcode<>'Gold'
group by id,passcode
union all
SELECT
id,'Gold',count(ID)
FROM Passes
where id in
(
select id from Passes where passcode='Gold'
)
group by id
order by id
结果:
100 Bronze 2
101 Gold 5
102 Silver 1
103 Steel 2
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