1
我回來了;)這次我有一個相當繁重的任務(我認爲)。總結爲每個客戶提供的公司(ID)
這裏就是我的了:
|customerID ||company |compdel |Street |Code |Date 1 |Date 2 |
+-------------------------------+--------------------------------------+
|1 ||Example1 |DELExam1|ABC Rd.1|4025 |01.01.2015 |01.08.2015 |
|1 ||Example1 |DELExam1|ABC Rd.1|4025 |13.04.2015 |01.12.2015 |
|1 ||Example1 |DELExam2|DEL St.1|0212 |13.03.2015 |09.07.2015 |
|1 ||Example1 |DELExam3|REF Wy.1|9875 |26.05.2015 |16.09.2015 |
|2 ||Example2 |DELExam4|REG St.1|6754 |21.02.2015 |16.05.2015 |
|2 ||Example2 |DELExam5|HIO Wy.1|9999 |01.03.2015 |06.08.2015 |
|2 ||Example2 |DELExam5|HIO Wy.1|9999 |01.01.2015 |06.02.2015 |
我想告訴每一個的customerID在與時間1的最早日期和日期2.最新的日期一行總結每交付公司(compdel)爲了使它更容易理解,我想這樣的結果:
|customerID ||company |compdel |Street |Code |Date 1 |Date 2 |
+-------------------------------+--------------------------------------+
|1 ||Example1 |DELExam1|ABC Rd.1|4025 |01.01.2015 |01.12.2015 |
|1 ||Example1 |DELExam2|DEL St.1|0212 |13.03.2015 |09.07.2015 |
|1 ||Example1 |DELExam3|REF Wy.1|9875 |26.05.2015 |16.09.2015 |
|2 ||Example2 |DELExam4|REG St.1|6754 |21.02.2015 |16.05.2015 |
|2 ||Example2 |DELExam5|HIO Wy.1|9999 |01.01.2015 |06.08.2015 |
我這個select語句嘗試它已經,但它不需額外的工作:我知道,這隻能是答案的一部分....
SELECT *
FROM
(SELECT
customerID, company, compdel, Street, Code, Date 1, Date 2,
ROW_NUMBER() OVER(PARTITION BY compdel ORDER BY customerID) rn
FROM
table 1) as Y
WHERE
rn = 1