行转列问题,解决立马结贴

[happy664618843]

number  createdate                         username         caseno
1	   2018-05-29 18:57:54.150	戴婷	             201708220028
2	  2018-07-04 14:28:40.167	戴婷	             201708220028
3	 2019-01-02 11:34:53.533	戴婷 	    201708220028
4	 2019-01-16 18:17:01.313	戴婷	            201708220028

想要结果:

username   caseno                     createdate1                             createdate2                           createdate3                         createdate4
戴婷	           201708220028          2018-05-29 18:57:54.150      2018-07-04 14:28:40.167       2019-01-02 11:34:53.533     2019-01-16 18:17:01.313

[听雨停了]

SELECT username,
		caseno,
       MAX(CASE number WHEN '1' THEN createdate ELSE 0 END) createdate1,
       MAX(CASE number WHEN '2' THEN createdate ELSE 0 END) createdate2,
       MAX(CASE number WHEN '3' THEN createdate ELSE 0 END) createdate3,
       MAX(CASE number WHEN '4' THEN createdate ELSE 0 END) createdate4
FROM   tab
GROUP BY
username,
		caseno

[insus]

参考：



详细实现过程：
https://www.cnblogs.com/insus/p/10944163.html