![Ich muss das U und A zählen, das jeder Layer-ID entspricht. Zum Beispiel HYD_NET_LN U 5 A 10 HYD_VAL_PT U 8 A 25 Wie schreibt man][1]
学习是最好的投资!
简单点,不知道这样可不可以。SELECT layerid,ChangeMold,COUNT(ChangeMold) FROM table_name GROUP BY layerid,ChangeMold
select count(*),layid,changeMold from tbl group by layid,changeMold
SELECT layerid, sum(case when changeMold='U' then 1 else 0 end) changeMold_U, sum(case when changeMold='A' then 1 else 0 end) changeMold_A FROM table_name GROUP BY layerid;
如果 要 统计这种大文件 的话 。还是都查出来 然后再处理数组方便点(sql比较弱 )
select layerid,changeMold,count(*) as num from TABLE where layerid in (select layerid from TABLE group by layerid) group by changeMold; 这条sql的性能很不好
简单点,不知道这样可不可以。
SELECT layerid,ChangeMold,COUNT(ChangeMold) FROM table_name
GROUP BY layerid,ChangeMold
select count(*),layid,changeMold from tbl group by layid,changeMold
如果 要 统计这种大文件 的话 。还是都查出来 然后再处理数组方便点(sql比较弱 )
select layerid,changeMold,count(*) as num from TABLE where layerid in (select layerid from TABLE group by layerid) group by changeMold; 这条sql的性能很不好