SQL Server：Min/Max/Percentiles

Question

我正在嘗試爲每個主工作職位獲得基本工資的計數，最小值，最大值和一些百分位數（第10，第25，第50，第75，第90）。

，我發現了以下錯誤：

Msg 8120, Level 16, State 1, Line 1
Column 'dbo.ps_employee.Base' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.

Answer 1

您必須在您的選擇列表中的所有非聚集項，在GROUP BY部分的底部。

Documentation: https://docs.microsoft.com/en-us/sql/t-sql/queries/select-group-by-transact-sql

Answer 2

我能夠使用以下查詢來獲得百分位數。我用一個單獨的計數/分鐘/最大/平均。

SELECT DISTINCT mj.title, PERCENTILE_DISC(.1) WITHIN GROUP (ORDER BY e.base) 
OVER (PARTITION BY mj.title) AS '10th', PERCENTILE_DISC(.25) WITHIN GROUP (ORDER 
BY e.base) OVER (PARTITION BY mj.title) AS '25th', PERCENTILE_DISC(.5) WITHIN 
GROUP (ORDER BY e.base) OVER (PARTITION BY mj.title) AS '50th', 
PERCENTILE_DISC(.75) WITHIN GROUP (ORDER BY e.base) OVER (PARTITION BY mj.title) 
AS '75th', PERCENTILE_DISC(.9) WITHIN GROUP (ORDER BY e.base) OVER (PARTITION BY 
mj.title) AS '90th' FROM dbo.ps_employee e FULL OUTER JOIN dbo.ps_jobs j on 
e.title = j.job FULL OUTER JOIN dbo.ps_masterjobs mj 
ON j.masterID = mj.ID;