SELECT tb.Type, Count(tb.Type) AS Type之Count, Sum(tb.Price) AS Price之Sum
FROM tb
GROUP BY tb.Type;
union
SELECT 'zzzz',Count(tb.Type) AS Type之Count, Sum(tb.Price) AS Price之Sum
FROM tb
FROM tb
GROUP BY tb.Type;
union
SELECT 'zzzz',Count(tb.Type) AS Type之Count, Sum(tb.Price) AS Price之Sum
FROM tb
解决方案 »
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我自己的笨办法如下:
表格式
送检科室,类别,应收金额类别只有('门诊'和'住院')两种
语句如下:SELECT tzs.科别, tzs.人次 AS 总人次, (IIF(ISNULL(tzs.收费),0,tzs.收费)) AS 总收费, (IIF(ISNULL(tmz.人次),0,tmz.人次)) AS 门诊人次, tmz.收费 AS 门诊收费, (总人次-iif(isnull(门诊人次),0,门诊人次)) AS 住院人次, (总收费-iif(isnull(门诊收费),0,门诊收费)) AS 住院收费
FROM [select max(送检科室) as 科别 ,sum(应收金额) as 收费,count(*) as 人次 from data1 group by 送检科室]. AS tzs LEFT JOIN [select max(送检科室) as 科别 ,sum(应收金额) as 收费,count(*) as 人次 from data1 where 类别='门诊' group by 送检科室]. AS tmz ON tzs.科别=tmz.科别
不知道有没有更精简更高效的?
FROM tb
where type=m
union
SELECT Count(tb.Group) AS MyCount, Sum(tb.Price) AS MySum
FROM tb
where type=n
union
SELECT Count(tb.Group) AS MyCount, Sum(tb.Price) AS MySum
FROM tb
不过只是你的type只有两种才这么做。
FROM tb
where type=m
union
SELECT Count(*) AS MyCount, Sum(tb.Price) AS MySum
FROM tb
where type=n
union
SELECT Count(*) AS MyCount, Sum(tb.Price) AS MySum
FROM tb