SQL考勤统计语句，求助

显示全部楼层 · 2009-6-18 16:42:07

select name,
sum(case when intime between '8:00' and '8:29' then 1 else 0 end ) as '迟到',
sum(case when intime between '8:30' and '9:00' then 1 else 0 end ) as '缺勤'
from table1
group by name因为“8：30”是一个公用时间点，不能判别式迟到还是缺勤，所以迟到我换成“8：29”了...

千问 · 2009-6-18 16:42:07

SELECT D1.NAME,D1.迟到,d2.缺勤FROM (SELECT name, COUNT(*) as 迟到
FROM table1
where intime between '8:00' and '8:30' group by name ) D1 FULL OUTER JOIN
(...

千问 · 2009-6-18 16:42:07

select name,
decode(sign(trunc(intime) - trunc('8:30')), 1, decode(sign(trunc(intime) - trunc('9:00')), 1, '缺勤', '迟到'), '正常') from table1 group by name说一下可能会遇到的问题：就...