今天下午郁悶，遇到了下面這個問題，應為以前對case用得少，所以開始沒有想到這上面來,還從什麼鬼游標等其它方面著首去想，真TMD的花了好久時間，後來看到CASE的用法後，呵呵，就簡單了
表名：TableName
字段名 日期
feild date
abcde 20005-01-01
abc 20005-01-01
abcd 20005-01-01
abcde 20005-01-01
ab 20005-01-02
abc 20005-01-02
abcd 20005-01-02
abcde 20005-01-02
abcd 20005-01-03
abc 20005-01-03
現在查各日期內的字符串的長度為>=1,>=2,>=3,>=4,>=5的各有多少個
如上結果應為：
1 2 3 4 5 日期
4 4 4 3 2 20005-01-01
4 4 3 2 1 20005-01-02
2 2 2 1 0 20005-01-03
select len(field) as lenfield,date into #countTable form TableName order by date
select sum(case when lenfield>=1 then 1 else 0) as fid1,
sum(case when lenfield>=2 then 1 else 0) as fid2,
sum(case when lenfield>=3 then 1 then 0) as fid3,
sum(case when lenfield>=4 then 1 then 0) as fid4,
sum(case when lenfield>=5 then 1 then 0) as fid5
into #returnTable
from #countTable
group by date
drop table #countTable
drop table #returnTable