基于字符串和null计数
以下是我的示例数据：
id FirstName LastName HouseNo MyCount
1 A C 1-1 2
2 B C 1-1 2
4 D A 3
5 F A 3
6 J A 3
7 Q X 1-2 3
8 D X 1-2 3
9 D X 1-2 3
10 A C 1-3 3
11 B C 1-3 3
12 C C 1-3 3
14 F K 2
15 J K 2
16 Q X 1-5 1
根据以上数据，我想计算具有相同HouseNo和LastName的记录数。
为此，我正在使用
SELECT COUNT(ID) AS _COUNT FROM MYTABLE GROUP BY LASTNAME, HOUSENO
但上述声明有一个问题。在数据中，一些记录没有HouseNo。在上面的示例中，ID 4,5,6和14,15没有HouseNo。所以，上面的语句返回5但它应该分别返回3和2。
主要目标
- 根据
LastName和HouseNo计算
- 记录那些没有
HouseNo的记录（它们会串联出来）。 - 即将到来的计数应在
MyCount中更新
- 示例数据
MyCount和CountId为空，应填写。 -
MyCount将基于HouseNo和LastName，请参阅ID 1至3，其姓氏为khan，房屋号为1-1，因此MyCount为ID 1到3将为3，CountId将为1. - 在示例数据中，有许多记录没有
HouseNo，因此在这种情况下，系列中的同名姓氏将被计算在内。请参阅ID 7，其计数为1.另见ID 18和19，其计数为2。 -
CountId是一个ID号的序列号。请参阅ID 1至3，由于同一房屋号和姓氏相同，因此为1。
我如何计算这个数字？
编辑赏金：
示例数据
id FirstName LastName HouseNo MyCount CountId
1 Imran Khan 1-1
2 Waseem Khan 1-1
3 Rihan Khan 1-1
4 Moiz Shaikh 1-2
5 Zbair Shaikh 1-2
6 Sultan Shaikh 1-2
7 Zaid Khan
10 Parvez Patel 1-3
11 Ahmed Patel 1-3
12 Rahat Syed 1-4
13 Talha Khan
14 Zia Khan
15 Arshad Patel 1-3
16 Samad Patel 1-3
17 Raees Syed 1-4
18 Azmat Khan
19 Imran Khan
预期结果：
id FirstName LastName HouseNo MyCount CountId
1 Imran Khan 1-1 3 1
2 Waseem Khan 1-1 3 1
3 Rihan Khan 1-1 3 1
4 Moiz Shaikh 1-2 3 2
5 Zbair Shaikh 1-2 3 2
6 Sultan Shaikh 1-2 3 2
7 Zaid Khan 1 3
10 Parvez Patel 1-3 2 4
11 Ahmed Patel 1-3 2 4
12 Rahat Syed 1-4 1 5
13 Talha Khan 2 6
14 Zia Khan 2 6
15 Arshad Patel 1-3 2 7
16 Samad Patel 1-3 2 7
17 Raees Syed 1-4 1 8
18 Azmat Khan 2 9
19 Imran Khan 2 9
8 个答案:
答案 0 :(得分：4)
看起来主要的混淆是由问题开头的SQL语句引起的，只是GROUP BY LASTNAME, HOUSENO。
如果您想要一个简单的分组，您的查询将是正确的。但是，然后您向我们展示了具有预期结果的更详细的示例数据，并且很明显您不仅需要分组（不关心数据中行的顺序），而是希望基于它们对行进行分组序列。
这是一个名为gaps-and-islands的经典问题。在SQL Server 2008中，可以使用少量调用ROW_NUMBER函数来完成。
示例数据
DECLARE @T TABLE
(id int PRIMARY KEY
,FirstName nvarchar(50)
,LastName nvarchar(50)
,HouseNo nvarchar(50)
,MyCount int
,CountId int);
INSERT INTO @T (id, FirstName, LastName, HouseNo) VALUES
(1 , 'Imran ', 'Khan ', '1-1'),
(2 , 'Waseem', 'Khan ', '1-1'),
(3 , 'Rihan ', 'Khan ', '1-1'),
(4 , 'Moiz ', 'Shaikh', '1-2'),
(5 , 'Zbair ', 'Shaikh', '1-2'),
(6 , 'Sultan', 'Shaikh', '1-2'),
(7 , 'Zaid ', 'Khan ', NULL),
(10, 'Parvez', 'Patel ', '1-3'),
(11, 'Ahmed ', 'Patel ', '1-3'),
(12, 'Rahat ', 'Syed ', '1-4'),
(13, 'Talha ', 'Khan ', NULL),
(14, 'Zia ', 'Khan ', NULL),
(15, 'Arshad', 'Patel ', '1-3'),
(16, 'Samad ', 'Patel ', '1-3'),
(17, 'Raees ', 'Syed ', '1-4'),
(18, 'Azmat ', 'Khan ', NULL),
(19, 'Imran ', 'Khan ', NULL);
SELECT查询
WITH
CTE_RN
AS
(
SELECT
id
,FirstName
,LastName
,HouseNo
,MyCount
,CountId
,ROW_NUMBER() OVER (PARTITION BY LastName, HouseNo ORDER BY ID) AS rn1
,ROW_NUMBER() OVER (ORDER BY ID) AS rn2
FROM @T AS T
)
,CTE_GRoups
AS
(
SELECT
id
,FirstName
,LastName
,HouseNo
,MyCount
,CountId
,rn1
,rn2
,rn2-rn1 AS GroupNumber
,COUNT(ID) OVER (PARTITION BY LastName, HouseNo, rn2-rn1) AS NewMyCount
,MIN(ID) OVER (PARTITION BY LastName, HouseNo, rn2-rn1) AS GroupMinID
FROM CTE_RN
)
SELECT
id
,FirstName
,LastName
,HouseNo
,rn1
,rn2
,GroupNumber
,NewMyCount
,GroupMinID
,DENSE_RANK() OVER (ORDER BY GroupMinID) AS NewCountId
FROM CTE_GRoups
ORDER BY ID;
<强>结果
+----+-----------+----------+---------+-----+-----+-------------+------------+------------+------------+
| id | FirstName | LastName | HouseNo | rn1 | rn2 | GroupNumber | NewMyCount | GroupMinID | NewCountId |
+----+-----------+----------+---------+-----+-----+-------------+------------+------------+------------+
| 1 | Imran | Khan | 1-1 | 1 | 1 | 0 | 3 | 1 | 1 |
| 2 | Waseem | Khan | 1-1 | 2 | 2 | 0 | 3 | 1 | 1 |
| 3 | Rihan | Khan | 1-1 | 3 | 3 | 0 | 3 | 1 | 1 |
| 4 | Moiz | Shaikh | 1-2 | 1 | 4 | 3 | 3 | 4 | 2 |
| 5 | Zbair | Shaikh | 1-2 | 2 | 5 | 3 | 3 | 4 | 2 |
| 6 | Sultan | Shaikh | 1-2 | 3 | 6 | 3 | 3 | 4 | 2 |
| 7 | Zaid | Khan | NULL | 1 | 7 | 6 | 1 | 7 | 3 |
| 10 | Parvez | Patel | 1-3 | 1 | 8 | 7 | 2 | 10 | 4 |
| 11 | Ahmed | Patel | 1-3 | 2 | 9 | 7 | 2 | 10 | 4 |
| 12 | Rahat | Syed | 1-4 | 1 | 10 | 9 | 1 | 12 | 5 |
| 13 | Talha | Khan | NULL | 2 | 11 | 9 | 2 | 13 | 6 |
| 14 | Zia | Khan | NULL | 3 | 12 | 9 | 2 | 13 | 6 |
| 15 | Arshad | Patel | 1-3 | 3 | 13 | 10 | 2 | 15 | 7 |
| 16 | Samad | Patel | 1-3 | 4 | 14 | 10 | 2 | 15 | 7 |
| 17 | Raees | Syed | 1-4 | 2 | 15 | 13 | 1 | 17 | 8 |
| 18 | Azmat | Khan | NULL | 4 | 16 | 12 | 2 | 18 | 9 |
| 19 | Imran | Khan | NULL | 5 | 17 | 12 | 2 | 18 | 9 |
+----+-----------+----------+---------+-----+-----+-------------+------------+------------+------------+
这里我在结果中包含了所有中间步骤，因此您可以看到它是如何工作的。主要部分是两组ROW_NUMBER s。 rn1序列将针对每个LastName, HouseNo重新启动。它由LastName, HouseNo分区。 rn2是一个简单的增加序列，没有间隙。我们需要它，因为原始ID定义了顺序，但可能有差距。
然后我们减去这两个序列，差异给我们GroupNumber。
计算组中元素的数量很简单COUNT，这会给我们NewMyCount。
使用无间隙的序列号枚举组分两步完成。首先MIN为组提供标识符，然后DENSE_RANK生成一系列NewCountId无间隙。
如果您想要使用计算出的NewMyCount和NewCountId实际更新原始表格，可以轻松将上面的SELECT查询转换为UPDATE查询：
更新查询
WITH
CTE_RN
AS
(
SELECT
id
,FirstName
,LastName
,HouseNo
,MyCount
,CountId
,ROW_NUMBER() OVER (PARTITION BY LastName, HouseNo ORDER BY ID) AS rn1
,ROW_NUMBER() OVER (ORDER BY ID) AS rn2
FROM @T AS T
)
,CTE_GRoups
AS
(
SELECT
id
,FirstName
,LastName
,HouseNo
,MyCount
,CountId
,rn1
,rn2
,rn2-rn1 AS GroupNumber
,COUNT(ID) OVER (PARTITION BY LastName, HouseNo, rn2-rn1) AS NewMyCount
,MIN(ID) OVER (PARTITION BY LastName, HouseNo, rn2-rn1) AS GroupMinID
FROM CTE_RN
)
,CTE_Update
AS
(
SELECT
id
,FirstName
,LastName
,HouseNo
,MyCount
,CountId
,rn1
,rn2
,GroupNumber
,NewMyCount
,GroupMinID
,DENSE_RANK() OVER (ORDER BY GroupMinID) AS NewCountId
FROM CTE_GRoups
)
UPDATE CTE_Update
SET
MyCount = NewMyCount
,CountId = NewCountId
;
<强>结果
SELECT *
FROM @T
ORDER BY ID;
+----+-----------+----------+---------+---------+---------+
| id | FirstName | LastName | HouseNo | MyCount | CountId |
+----+-----------+----------+---------+---------+---------+
| 1 | Imran | Khan | 1-1 | 3 | 1 |
| 2 | Waseem | Khan | 1-1 | 3 | 1 |
| 3 | Rihan | Khan | 1-1 | 3 | 1 |
| 4 | Moiz | Shaikh | 1-2 | 3 | 2 |
| 5 | Zbair | Shaikh | 1-2 | 3 | 2 |
| 6 | Sultan | Shaikh | 1-2 | 3 | 2 |
| 7 | Zaid | Khan | NULL | 1 | 3 |
| 10 | Parvez | Patel | 1-3 | 2 | 4 |
| 11 | Ahmed | Patel | 1-3 | 2 | 4 |
| 12 | Rahat | Syed | 1-4 | 1 | 5 |
| 13 | Talha | Khan | NULL | 2 | 6 |
| 14 | Zia | Khan | NULL | 2 | 6 |
| 15 | Arshad | Patel | 1-3 | 2 | 7 |
| 16 | Samad | Patel | 1-3 | 2 | 7 |
| 17 | Raees | Syed | 1-4 | 1 | 8 |
| 18 | Azmat | Khan | NULL | 2 | 9 |
| 19 | Imran | Khan | NULL | 2 | 9 |
+----+-----------+----------+---------+---------+---------+
答案 1 :(得分：3)
使用CTE然后更新您的表格如下：
;WITH T AS
(
SELECT
*,
ROW_NUMBER() OVER (ORDER BY ID) AS SrNo,
ROW_NUMBER() OVER (PARTITION BY LastName,HouseNo ORDER BY HouseNo) AS PartNo
FROM MYTABLE
),
X as
(
SELECT
T.LastName,
T.HouseNo,
(MAX(T.ID)-MIN(T.ID))+1 AS NoOfCount,
ROW_NUMBER() OVER(Order BY MAX(ID)) AS RowNo,
MAX(ID) AS ID
FROM T
GROUP BY T.LastName,T.HouseNo, (T.SrNo - T.PartNo)
)
Update MYTABLE
SET
MyCount=X.NoOfCount,
CountId=X.RowNo
FROM X
WHERE MYTABLE.LastName=X.LastName
AND MYTABLE.HouseNo=X.HouseNo
AND MYTABLE.ID<=X.ID
SELECT * FROM MYTABLE
<强>输出：
答案 2 :(得分：2)
这应该这样做
declare @temp table (id int, firstname varchar(5), lastname varchar(5), houseno varchar(5), mycount int)
insert into @temp values(1, 'A', 'C', '1-1', 2)
insert into @temp values(2, 'B', 'C', '1-1', 2)
insert into @temp values(4, 'D', 'A', null, 3)
insert into @temp values(5, 'F', 'A', null, 3)
insert into @temp values(6, 'J', 'A', null, 3)
insert into @temp values(7, 'Q', 'X', '1-2', 3)
insert into @temp values(8, 'D', 'X', '1-2', 3)
insert into @temp values(9, 'D', 'X', '1-2', 3)
insert into @temp values(10, 'A', 'C', '1-3', 3)
insert into @temp values(11, 'B', 'C', '1-3', 3)
insert into @temp values(12, 'C', 'C', '1-3', 3)
insert into @temp values(14, 'F', 'K', null, 2)
insert into @temp values(15, 'J', 'K', null, 2)
insert into @temp values(16, 'Q', 'X', '1-5', 1)
select count(ID) as _count
from @temp
group by isnull(lastname, ''), isnull(houseno, '')
返回
_count
3
2
2
3
3
1
你可以用这个来吐出更多细节：
select distinct
t.lastname,
isnull(t.houseno, '') as houseno,
(select count(ID) from @temp t2 where t2.lastname = t.lastname and t2.houseno = t.houseno) as _count_filled,
(select count(ID) from @temp t2 where t2.lastname = t.lastname and isnull(t2.houseno, '') = isnull(t.houseno, '') and t2.houseno is null) as _count_empty
from @temp t
它会返回：
lastname houseno _count_filled _count_empty
A 0 3
C 1-1 2 0
C 1-3 3 0
K 0 2
X 1-2 3 0
X 1-5 1 0
答案 3 :(得分：2)
首先创建一个视图来计算每个部分的计数和排名。
CREATE VIEW cnt AS
SELECT
T.LastName,
T.HouseNo, MIN(t.id) AS START , MAX(T.id) AS finish ,
(MAX(T.ID)-MIN(T.ID))+1 AS NoOfCount,
ROW_NUMBER() OVER(Order BY MAX(T.ID)) AS RowNo,
MAX(T.ID) AS ID
FROM (
SELECT
*,
ROW_NUMBER() OVER (ORDER BY ID) AS SrNo,
ROW_NUMBER() OVER (PARTITION BY LastName,HouseNo ORDER BY HouseNo) AS PartNo
FROM myTable
) T
GROUP BY T.LastName,T.HouseNo, (T.SrNo - T.PartNo)
然后将它用于您的目的：
SELECT a.*,
b.NoOfCount,
b.RowNo
FROM myTable AS a
INNER JOIN cnt AS b
ON a.id BETWEEN b.start AND b.finish
以下是结果：
答案 4 :(得分：1)
SELECT COUNT(ID) AS _COUNT
FROM MYTABLE
GROUP BY ISNULL(LASTNAME, ''), ISNULL(HOUSENO, '');
答案 5 :(得分：1)
我相信，您的第三个主要目标是在相应的行上 更新 MYCOUNT列和结果。一般来说，您要找的是 相关子查询 。
UPDATE MYTABLE T1
SET T1.MYCOUNT =
( SELECT COUNT (*)
FROM MYTABLE T2
WHERE T1.LASTNAME = B2.LASTNAME
AND NVL (T2.HOUSENO, 0) = NVL (T1.HOUSENO, 0)
GROUP BY T2.LASTNAME, T2.HOUSENO);
*注意：这是为 Oracle SQL
实现的答案 6 :(得分：1)
我同意@Vladimir Baranov的分析，所以我在此不再重复。 我只是想使查询更简单，如下所示（在SQL Server 2012中测试）
--drop table #temp
create table #temp (id int, firstname varchar(15), lastname varchar(15), houseno varchar(5));
go
insert into #temp (id, firstname, lastname, houseno)
values
(1 , 'Imran' ,'Khan' ,'1-1')
,(2 , 'Waseem' ,'Khan' ,'1-1')
,(3 , 'Rihan' ,'Khan' ,'1-1')
,(4 , 'Moiz' ,'Shaikh' ,'1-2')
,(5 , 'Zbair' ,'Shaikh' ,'1-2')
,(6 , 'Sultan' ,'Shaikh' ,'1-2')
,(7 , 'Zaid' ,'Khan' , null)
,(10 , 'Parvez' ,'Patel' ,'1-3')
,(11 , 'Ahmed' ,'Patel' ,'1-3')
,(12 , 'Rahat' ,'Syed' ,'1-4')
,(13 , 'Talha' ,'Khan' ,null )
,(14 , 'Zia' ,'Khan' ,null )
,(15 , 'Arshad' ,'Patel' ,'1-3')
,(16 , 'Samad' ,'Patel' ,'1-3')
,(17 , 'Raees' ,'Syed' ,'1-4')
,(18 , 'Azmat' ,'Khan' , null)
,(19 , 'Imran' ,'Khan' , null)
-- query
; with c as (
select id, firstname, lastname, houseno=isnull(houseno, '')
, new_id=row_number() over (partition by lastname, isnull(houseno, '') order by id)
, grp = id -row_number() over (partition by lastname, isnull(houseno, '') order by id)
FROM #temp
)
, d as (
select id, firstname, lastname, houseno, T.cnt, c.grp
, row_id=id-row_number() over ( partition by grp, houseno order by c.grp)
from c
cross apply (select cnt=count(*) from c as c2 where c.grp = c2.grp and c.lastname=c2.lastname and c.houseno=c2.houseno) T(cnt)
)
select id, FirstName, LastName, Houseno, MyCount=cnt, CountId= DENSE_RANK() over (order by row_id)
from d
结果如下：
答案 7 :(得分：0)
试试这个：
SELECT COUNT（ID）作为来自MYTABLE GROUP的_COUNT来自LASTNAME + ISNULL（HOUSENO，＆＃39;＆＃39;）
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