Question

我正在花时间搞清楚如何总结这两个查询的总和：

select t.processed as title, count(t.processed) as ckos
from circ_longterm_history clh, title t
where t.bib# = clh.bib#
and clh.cko_location = 'dic'
group by t.processed
order by ckos DESC

select t.processed as title, count(t.processed) as ckos
from circ_history ch, title t, item i
where i.item# = ch.item#
and t.bib# = i.bib#
and ch.cko_location = 'dic'
group by t.processed
order by ckos DESC

基本上我想要一个结果集，其中一列为t.processed，另一列为第一个计数加上第二个计数的总和。

有什么想法吗？

Answer 1

SELECT A.title, A.ckos +B.ckos as ckos from
(select t.processed as title, count(t.processed) as ckos
from circ_longterm_history clh, title t
where t.bib# = clh.bib#
and clh.cko_location = 'dic'
group by t.processed
) as A inner join

(select t.processed as title, count(t.processed) as ckos
from circ_history ch, title t, item i
where i.item# = ch.item#
and t.bib# = i.bib#
and ch.cko_location = 'dic'
group by t.processed
) as B on A.title =B.title order by ckos DESC

Answer 2

; WITH CTE AS (
    SELECT T.PROCESSED AS TITLE, T.PROCESSED AS CKOS
    FROM dbo.CIRC_LONGTERM_HISTORY CLH
    INNER JOIN dbo.TITLE T
    WHERE T.[BIB#] = CLH.[BIB#]
      AND CLH.CKO_LOCATION = 'DIC'

    UNION ALL

    SELECT T.PROCESSED AS TITLE, T.PROCESSED AS CKOS
    FROM dbo.CIRC_HISTORY CH
    INNER JOIN dbo.ITEM I
      ON I.[ITEM#] = CH.[ITEM#]
    INNER JOIN dbo.TITLE T
      ON T.[BIB#] = I.[BIB#]
    WHERE CH.CKO_LOCATION = 'DIC'
)
SELECT TITLE, COUNT(*) AS CKOS
FROM CTE
GROUP BY TITLE
ORDER BY CKOS DESC

Answer 3

我不知道这是否对您有所帮助

如果你想从2个查询总和，我通常会喜欢这个

示例有一个这样的表：

id     name    
1     jenny     
2     jack

 
select count(id) as id, name 
from table 
group by name 
union all 
select count(id) as id, name 
from table 
group by name

如果显示如下：

id    name

1     jenny

1     jenny

2     jack

2     jack

所以如果你想加总id 就这样做：

select sum(id), name 
from (select count(name) as id, name from table group by name 
      UNION ALL 
      select count(name) as title, name from table group by name)aa 
group by name

所以看起来像这样：

sum   name

2     jenny

2     jack

但是如果你使用UNION代替UNION ALL 结果将是这样的：

sum   name

1     jenny

1     jack

Answer 4

我相信以下内容应该有效（尽管我没有样本数据来测试它......）：

SELECT t.processed as title, 
       COALESCE(SUM(clh.count), 0) + COALESCE(SUM(ch.count), 0) as ckos
FROM Title t
LEFT JOIN (SELECT bib#, COUNT(*) as count
           FROM Circ_Longterm_History
           WHERE cko_location = 'dic'
           GROUP BY bib#) clh
       ON clh.bib# = t.bib#
LEFT JOIN (SELECT i.bib#, COUNT(*) as count
           FROM Item i
           JOIN Circ_History ch
             ON ch.item# = i.item#
           WHERE ch.cko_location = 'dic'
           GROUP BY i.bib#) ch
       ON ch.bib# = t.bib#
GROUP BY t.processed
ORDER BY ckos DESC

总计数（与分组相乘）

4 个答案: