gue*_*est 3 null join sql-server group-by
我有 4 个子查询，每个子查询都按“组名”分组。尝试将每个子查询作为一个列，全部按“组名”分组。这是查询：
select
coalesce(co.group_name, requests.group_Name, incidents.group_Name, problems.group_Name) as 'SD Groups'
, isnull(co.co, '') as 'CO'
, isnull(incidents.incidents, '' ) as 'Inc'
, isnull(problems.problems, '') as 'Prob'
, isnull(requests.requests, '') as 'Rqst'
from
(
select
groups.last_name AS Group_Name
,count(chg_ref_num) AS 'CO'
from chg
left join ca_contact groups
on chg.group_id = groups.contact_uuid
left join ca_contact assignee
on chg.assignee = assignee.contact_uuid
left join ca_company cc
on assignee.company_uuid = cc.company_uuid
where
groups.last_name in ('8197 Qlikview Support', '8202 OBIEE-BIP'
, '8205 BI SAS', '8206 BI CCBI', '8208 BI ePlan Reporting and Support'
, '8211 BI Data Quality', '8212 BI EPM Report Architect/Developer'
, '8213 BI EPM Framework Architect/Developer', '8214 BI EPM User Experience'
, '8215 BI EPM OLAP Architect/Developer', '8219 BI Data Warehouse')
and status = 'CL'
and convert(varchar, dateadd(hh,-8,dateadd(ss,chg.close_date, '1970')), 101) >= DATEADD(month, datediff(month, 0, getdate()), 0)
and parent is NULL
and cc.company_name = 'XYZ'
group by groups.last_name
) as CO
full outer join
(
select
groups.last_name AS Group_Name
,count(ref_num) AS Requests
from call_req cr
left join ca_contact groups
on cr.group_id = groups.contact_uuid
left join ca_contact assignee
on cr.assignee = assignee.contact_uuid
left join ca_company cc
on assignee.company_uuid = cc.company_uuid
where
groups.last_name in ('8197 Qlikview Support', '8202 OBIEE-BIP'
, '8205 BI SAS', '8206 BI CCBI', '8208 BI ePlan Reporting and Support'
, '8211 BI Data Quality', '8212 BI EPM Report Architect/Developer'
, '8213 BI EPM Framework Architect/Developer', '8214 BI EPM User Experience'
, '8215 BI EPM OLAP Architect/Developer', '8219 BI Data Warehouse')
and cr.status in ('CL')
and convert(varchar, dateadd(hh,-8,dateadd(ss,cr.close_date, '1970')), 101) >= DATEADD(month, datediff(month, 0, getdate()), 0)
and cr.parent is NULL
and cr.type = 'R'
and cc.company_name = 'XYZ'
group by groups.last_name, cr.type
) as Requests
on co.group_name = requests.group_name
full outer join
(
select
groups.last_name AS Group_Name
,count(ref_num) AS Problems
from call_req cr
left join ca_contact groups
on cr.group_id = groups.contact_uuid
left join ca_contact assignee
on cr.assignee = assignee.contact_uuid
left join ca_company cc
on assignee.company_uuid = cc.company_uuid
where
groups.last_name in ('8197 Qlikview Support', '8202 OBIEE-BIP'
, '8205 BI SAS', '8206 BI CCBI', '8208 BI ePlan Reporting and Support'
, '8211 BI Data Quality', '8212 BI EPM Report Architect/Developer'
, '8213 BI EPM Framework Architect/Developer', '8214 BI EPM User Experience'
, '8215 BI EPM OLAP Architect/Developer', '8219 BI Data Warehouse')
and cr.status in ('CL')
and convert(varchar, dateadd(hh,-8,dateadd(ss,cr.close_date, '1970')), 101) >= DATEADD(month, datediff(month, 0, getdate()), 0)
and cr.parent is NULL
and cr.type = 'P'
and cc.company_name = 'XYZ'
group by groups.last_name, cr.type
) as Problems
on requests.group_name = problems.group_name
full outer join
(
select
groups.last_name AS Group_Name
,count(ref_num) AS Incidents
from call_req cr
left join ca_contact groups
on cr.group_id = groups.contact_uuid
left join ca_contact assignee
on cr.assignee = assignee.contact_uuid
left join ca_company cc
on assignee.company_uuid = cc.company_uuid
where
groups.last_name in ('8197 Qlikview Support', '8202 OBIEE-BIP'
, '8205 BI SAS', '8206 BI CCBI', '8208 BI ePlan Reporting and Support'
, '8211 BI Data Quality', '8212 BI EPM Report Architect/Developer'
, '8213 BI EPM Framework Architect/Developer', '8214 BI EPM User Experience'
, '8215 BI EPM OLAP Architect/Developer', '8219 BI Data Warehouse')
and cr.status in ('CL')
and convert(varchar, dateadd(hh,-8,dateadd(ss,cr.close_date, '1970')), 101) >= DATEADD(month, datediff(month, 0, getdate()), 0)
and cr.parent is NULL
and cr.type = 'I'
and cc.company_name = 'XYZ'
group by groups.last_name, cr.type
) as Incidents
on requests.group_name = incidents.group_name
order by
'SD Groups' asc
结果如下：
Group_Name CO Inc Prob Rqst
8197 Qlikview Support 0 1 0 7
8202 OBIEE-BIP 0 4 0 11
8205 BI SAS 0 11 1 11
8206 BI CCBI 10 17 0 43
8208 BI ePlan Reporting and Support 0 0 0 4
8211 BI Data Quality 0 0 0 12
8212 BI EPM Report Architect/Developer 0 3 1 5
8214 BI EPM User Experience 0 2 0 0
8214 BI EPM User Experience 0 0 1 0
8215 BI EPM OLAP Architect/Developer 0 15 0 2
8219 BI Data Warehouse 16 71 4 13
请注意 Group 8214 的两行。每个组应表示一次，Group 8214 的所需结果应为：
Group_Name CO Inc Prob Rqst
8214 BI EPM User Experience 0 2 1 0
代码中是否存在明显缺陷？我最好的估计是查看 coalesce 函数 - 我尝试过但失败了。在我的尝试中，有一段不友好的文章：
isnull(isnull(isnull(co.group_name, requests.group_Name), incidents.group_Name), problems.group_Name)
但同样的结果。
代码中是否存在明显缺陷？我最好的估计是查看合并函数 - ...
是的。一个问题是第二个（和第三个）FULL JOIN使用（派生）表的group_name列。结果，第 2 次和第 3 次完全联接分别取消了第 1 次和第 2 次完全联接。您的查询只有基本结构，从细节中剥离：
select
coalesce(co.group_name, requests.group_Name, incidents.group_Name,
problems.group_Name) as 'SD Groups'
---
from
( --- ) as CO
full outer join
( --- ) as Requests
on co.group_name = requests.group_name
full outer join
( --- ) as Problems
on requests.group_name = problems.group_name
full outer join
( --- ) as Incidents
on requests.group_name = incidents.group_name
第一次完全加入后：
ON co.group_name = requests.group_name
表 (groups或requests) 中的行可能与另一个不匹配，并且空列填充了NULL值。
然后第二次完全连接发生：
ON requests.group_name = problems.group_name
但ON只有具有非 null 的行才能满足此条件requests.group_name。这基本上将这 2 个FULL连接转换为一个复杂的混乱，这几乎（但不完全）是 3 个表的完整连接。某些行不会匹配，特别地从表中的行co和problems具有相同group_name（但没有这样的行中的表中存在requests）将不被匹配，但将在不同的行中结束。
第三次完全加入：
ON requests.group_name = incidents.group_name
会让事情变得更加复杂。因此，您的查询相当于（requests左连接到其他 3 个表）和 3 个反连接（3 个表中的每一个都带有requests）的并集：
select
---
from
( --- ) as Requests
left outer join
( --- ) as CO
on co.group_name = requests.group_name
left outer join
( --- ) as Problems
on requests.group_name = problems.group_name
left outer join
( --- ) as Incidents
on requests.group_name = incidents.group_name
union all
select
---
from
( --- ) as CO
left outer join
( --- ) as Requests
on co.group_name = requests.group_name
where
requests.group_name is null
union all
select
---
from
( --- ) as Problems
left outer join
( --- ) as Requests
on problems.group_name = requests.group_name
where
requests.group_name is null
union all
select
---
from
( --- ) as Incidents
left outer join
( --- ) as Requests
on incidents.group_name = requests.group_name
where
requests.group_name is null
这就是同样group_name出现在两行的原因。您可以看到只有在CO没有具有此类值的行但其他表具有时才会发生这种情况。
如果你真的想有 3 个FULL连接，ON条件应该用ISNULL或重写COALESCE：
select
coalesce(co.group_name, requests.group_Name, incidents.group_Name,
problems.group_Name) as [SD Groups]
---
from
( ---
) as CO
full outer join
( ---
) as Requests
on co.group_name = requests.group_name
full outer join
( ---
) as Problems
on coalesce(co.group_name, requests.group_name)
= problems.group_name
full outer join
( ---
) as Incidents
on coalesce(co.group_name, requests.group_name, problems_group_name)
= incidents.group_name
order by
--- ;
其他小问题是'single quotes'别名的使用。您应该使用标准"double quotes"或 SQL Server 的[square brackets]. 或者最好不要引用您的标识符，并保持它们没有空格和其他奇怪的字符。单引号字符串应该只用于字符串文字，而不是标识符。