[英]SQL Query using Partition By
I have following table name JobTitle 我有以下表名JobTitle
JobID LanaguageID
-----------------
1 1
1 2
1 3
2 1
2 2
3 4
4 5
5 2
I am selecting all records from table except duplicate JobID's for which count > 1. I am selecting only one record/first row from the duplicate JobID's. 我从表中选择所有记录,除了重复JobID的计数> 1。我只从重复的JobID中选择一个记录/第一行。 Now I am passing LanguageID as paramter to stored procedure and I want to select duplicate JobID for that languageID along with the other records Also.
现在我将LanguageID作为参数传递给存储过程,我想为该languageID选择重复的JobID以及其他记录。 If I have passed languageID as 1 then output should come as follows
如果我已将languageID作为1传递,则输出应如下所示
JobID LanaguageID
-----------------
1 1
2 1
3 4
4 5
5 2
I have tried using following query. 我尝试过使用以下查询。
with CTE_RN as
(
SELECT ROW_NUMBER() OVER(PARTITION BY JobTitle.JobID ORDER BY JobTitle.JobTitle) AS RN
FROM JobTitle
INNER JOIN JobTitle_Lang
ON JobTitle.JobTitleID = JobTitle_Lang.JobTitleID
)
But I am unable to use WHERE clause in the above query. 但我无法在上面的查询中使用WHERE子句。 Is any different approch should be followed.
是否应遵循任何不同的方法。 Or else how can i modify the query to get the desired output
或者我如何修改查询以获得所需的输出
with CTE_RN as
(
SELECT
JobID, LanaguageID,
ROW_NUMBER() OVER(PARTITION BY JobTitle.JobID ORDER BY JobTitle.JobTitle) AS RN
FROM JobTitle
INNER JOIN JobTitle_Lang ON JobTitle.JobTitleID = JobTitle_Lang.JobTitleID
)
select
from CTE_RN
where RN = 1 or LanguageID = @LanguageID
update 更新
simplified a bit (join removed), but you'll get the idea: 简化了一点(加入删除),但你会明白:
declare @LanguageID int = 2
;with cte_rn as
(
select
JobID, LanguageID,
row_number() over(
partition by JobTitle.JobID
order by
case when LanguageID = @LanguageID then 0 else 1 end,
LanguageID
) as rn
from JobTitle
)
select *
from cte_rn
where rn = 1
SELECT b.[JobID], b.[LanaguageID]
FROM
(SELECT a.[JobID], a.[LanaguageID],
ROW_NUMBER() OVER(PARTITION BY a.[JobID] ORDER BY a.[LanaguageID]) AS [row]
FROM [JobTitle] a) b
WHERE b.[row] = 1
Result 结果
| JOBID | LANAGUAGEID | --------|-------------| | 1 | 1 | | 2 | 1 | | 3 | 4 | | 4 | 5 | | 5 | 2 |
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