原帖由 dingjun123 于 2011-1-12 15:48 发表
SQL> select sfzh,hblb,count(*)
2from (
3select sfzh,
4
hbrq,
5
hblb,
6
lag(hblb) over(partition by sfzh order by hbrq) lagval
7from SQL_TEST_TEMP
8)
9where hblb=lagval
10 group by sfzh,hblb ;
SFZH HBLB COUNT(*)
---------- ---------- ----------
1 1
3
1 2
1
这样简单点吧
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