Sql 使用union以不同的分组方式连接不同的查询

我有不同的查询（3），它们有不同的条件来获取不同的计数器

其中两个组的分组相同，但其中一个组没有

数线拣选

select 
carusrrsppck as 'users', 
COUNT(carusrrsppck) AS 'lines picking', 
cardepid as 'deposit'  
from order_products
where DATEADD(dd, 0, DATEDIFF(dd, 0, carfchcar)) 
between '20190220' and '20190220' 
and cardepid in ('D32', 'DA', 'DA1', 'DB', 'DC', 'DD', 'DE', 'DR')
and carusrrsppck<>''
and ordprdcnt<>0
group by carusrrsppck, cardepid
order by cardepid, COUNT(*) DESC

即使3个查询分组不同，也可以加入它们，或者我只能加入具有相同“分组依据”的查询

---

谢谢。

分拣的

计数行

和

拒绝计数行

查询使用相同的分组列，仅在条件上有所不同，因此可以简化为：

SELECT 
  carusrrsppck as 'users', 
  SUM(CASE WHEN carusrrsppck <> '' and ordprdcnt <> 0 THEN 1 ELSE 0 END) AS 'lines picking', 
  SUM(CASE WHEN carusrrsppck <> '' THEN 1 ELSE 0 END) AS 'lines denied',
  cardepid AS 'deposit'  
FROM order_products
WHERE 
  DATEADD(dd, 0, DATEDIFF(dd, 0, carfchcar)) BETWEEN '20190220' and '20190220' 
  AND 
  cardepid IN ('D32', 'DA', 'DA1', 'DB', 'DC', 'DD', 'DE', 'DR')
GROUP BY carusrrsppck, cardepid
ORDER BY cardepid, COUNT(*) DESC

选择
carusrrsppck作为“用户”，
求和（CARUSRRSPCK“”和ordprdcnt 0，然后1，否则0结束时的情况）为“行拾取”，
总和（CARUSRRSPCK为“”时为1，否则为0结束）为“拒绝的行”，
信用卡作为“存款”
从订购产品
哪里
“20190220”和“20190220”之间的日期添加（dd，0，DATEDIFF（dd，0，carfchcar））
及
cardepid IN（'D32'，'DA'，'DA1'，'DB'，'DC'，'DD'，'DE'，'DR'））
carusrrsppck集团，cardepid
按卡片订购，计数（*）说明

---

已确认的

计数行

查询是另一种情况，无法与上述情况关联。

我认为您只需要条件聚合：

select carusrrsppck as users, 
       cardepid as deposit  
       sum(case when carusrrsppck <> '' and ordprdcnt <> 0 then 1 else 0 end) AS lines_picking, 
       sum(case when carusrrsppck <> '' and ordprdcnt = 0 then 1 else 0 end) AS lines_denied, 
       count(*) as lines_confirmed
from order_products
where carfchar >= '20190220' and 
      carfchar < '20190221' and
      cardepid in ('D32', 'DA', 'DA1', 'DB', 'DC', 'DD', 'DE', 'DR')
group by carusrrsppck, cardepid
order by cardepid, COUNT(*) DESC;

选择carusrrsppck作为用户，
卡德皮德存款
求和（当carusrrsppck“”和ordprdcnt 0时，则为1，否则为0结束）作为行，
求和（当carusrrsppck“”和ordprdcnt=0时，则为1，否则为0结束）作为行被拒绝，
将（*）计算为已确认的行
从订购产品
其中carfchar>='20190220'和
carfchar<'20190221'和
cardepid in（'D32'，'DA'，'DA1'，'DB'，'DC'，'DD'，'DE'，'DR'））
carusrrsppck集团，cardepid
按卡片订购，计数（*）说明；

注：不进行转换时，日期算法更好。以上内容应与您拥有的内容相同

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另外，不要对列别名使用单引号。相反，选择不需要转义的列名。

只要它们具有相同的列数和类型，就可以合并它们。请注意，默认情况下，UNION语句只返回不同的结果（如果这很重要的话）。语法应该是“UNION UNION”。对于同一个carusrrspck，“lines picking”匹配“lines denied”的行@BoCoKeith-point将获得重复数据消除，这可能会偏离您的预期结果。最好添加一个指示结果集是什么的附加列（例如，

SELECT rsType='Lines Picking'，carusrrsppck为'usres'，COUNT（*）…

）@Forty3是的，我需要为每个计数添加一个附加列。在这种情况下，应该在子查询中进行分隔吗？

SELECT 
  carusrrsppck as 'users', 
  SUM(CASE WHEN carusrrsppck <> '' and ordprdcnt <> 0 THEN 1 ELSE 0 END) AS 'lines picking', 
  SUM(CASE WHEN carusrrsppck <> '' THEN 1 ELSE 0 END) AS 'lines denied',
  cardepid AS 'deposit'  
FROM order_products
WHERE 
  DATEADD(dd, 0, DATEDIFF(dd, 0, carfchcar)) BETWEEN '20190220' and '20190220' 
  AND 
  cardepid IN ('D32', 'DA', 'DA1', 'DB', 'DC', 'DD', 'DE', 'DR')
GROUP BY carusrrsppck, cardepid
ORDER BY cardepid, COUNT(*) DESC

select carusrrsppck as users, 
       cardepid as deposit  
       sum(case when carusrrsppck <> '' and ordprdcnt <> 0 then 1 else 0 end) AS lines_picking, 
       sum(case when carusrrsppck <> '' and ordprdcnt = 0 then 1 else 0 end) AS lines_denied, 
       count(*) as lines_confirmed
from order_products
where carfchar >= '20190220' and 
      carfchar < '20190221' and
      cardepid in ('D32', 'DA', 'DA1', 'DB', 'DC', 'DD', 'DE', 'DR')
group by carusrrsppck, cardepid
order by cardepid, COUNT(*) DESC;