select H01SQ , count(*) from usra01 where H01SQ like '01030201%' group by H01SQ;--如果你硬是要放在外面的话 select * from (select H01SQ , count(*) from usra01 group by H01SQ) where H01SQ like '01030201%' ;
select H01SQ , count(*) from usra01 where b0110 like '01030201%' group by H01SQ ;0.0 是这样么?
where条件确实应该是b0110我以为是H01SQ! sorry
select H01SQ , count(*) from usra01 where b0110 like '01030201%' group by H01SQ ;
SELECT `code`,COUNT(*) FROM productinfo WHERE `name` LIKE '%沙发' GROUP BY `code`
这么写:select H01SQ , count(*) from usra01 where H01SQ like '01030201%' group by H01SQ;
select H01SQ , count(*)from usra01 group by H01SQ having b0110 like '01030201%'
恩,where 肉 having 都可以啊
select H01SQ , count(*) from usra01 group by H01SQ,b0110 having b0110 like '01030201%'
select H01SQ , count(*) from usra01 where H01SQ like '01030201%' group by H01SQ;--如果你硬是要放在外面的话
select * from (select H01SQ , count(*) from usra01 group by H01SQ) where H01SQ like '01030201%' ;
sorry