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ora 丢失右括号解决方案

热度:22   发布时间:2016-04-24 05:25:20.0
ora 丢失右括号
select * from ( select distinct * from ( select ctm.id_ ctmId , ctm.name_ ctmName , ctr.id_ ctrId , ctr.name_ ctrName , ctr.email_ ctrEmail , ctr.mphone_ ctrMphone , ctr.brithday_ ctrBrithday from contacter_ ctr inner join customer_ ctm on ctr.fk_customer_ = ctm.id_ inner join user_ u on ctm.fk_user_ = u.id_ inner join ctm_ctg_lk_ ctmCtg on ctmCtg.fk_ctm_ = ctm.id_ inner join ctm_ctg_ cc on ctmCtg.fk_ctm_ctg_ = cc.id_ where ctr.brithday_ like '%-04-12' and u.id_ in '5dfaf039298290ee0129d51355e5029e' regexp_like( ctr.mphone_ , '^((13[0-9])|(15[^4,\D])|(18[0,5-9]))\d{8}$' ) ) order by ctmId asc )where rownum <= ?


高手来帮忙看看这是怎么回事


------解决方案--------------------
SQL code
select *  from (select distinct *          from (select ctm.id_       ctmId,                       ctm.name_     ctmName,                       ctr.id_       ctrId,                       ctr.name_     ctrName,                       ctr.email_    ctrEmail,                       ctr.mphone_   ctrMphone,                       ctr.brithday_ ctrBrithday                  from contacter_ ctr                 inner join customer_ ctm                    on ctr.fk_customer_ = ctm.id_                 inner join user_ u                    on ctm.fk_user_ = u.id_                 inner join ctm_ctg_lk_ ctmCtg                    on ctmCtg.fk_ctm_ = ctm.id_                 inner join ctm_ctg_ cc                    on ctmCtg.fk_ctm_ctg_ = cc.id_                 where ctr.brithday_ like '%-04-12'                   and u.id_ in ('5dfaf039298290ee0129d51355e5029e',                 regexp_like(ctr.mphone_,                                   '^((13[0-9])|(15[^4,\D])|(18[0,5-9]))\d{8}$')))         order by ctmId asc) where rownum <=?
------解决方案--------------------
SQL code
select *  from (select distinct *          from (select ctm.id_       ctmId,                       ctm.name_     ctmName,                       ctr.id_       ctrId,                       ctr.name_     ctrName,                       ctr.email_    ctrEmail,                       ctr.mphone_   ctrMphone,                       ctr.brithday_ ctrBrithday                  from contacter_ ctr                 inner join customer_ ctm                    on ctr.fk_customer_ = ctm.id_                 inner join user_ u                    on ctm.fk_user_ = u.id_                 inner join ctm_ctg_lk_ ctmCtg                    on ctmCtg.fk_ctm_ = ctm.id_                 inner join ctm_ctg_ cc                    on ctmCtg.fk_ctm_ctg_ = cc.id_                 where ctr.brithday_ like '%-04-12'                   and u.id_ in ('5dfaf039298290ee0129d51355e5029e',                 regexp_like(ctr.mphone_,                                   '^((13[0-9])|(15[^4,\D])|(18[0,5-9]))\d{8}$')))         order by ctmId asc) where rownum <=?
------解决方案--------------------
SQL code
select *  from (select distinct *          from (select ctm.id_       ctmId,                       ctm.name_     ctmName,                       ctr.id_       ctrId,                       ctr.name_     ctrName,                       ctr.email_    ctrEmail,                       ctr.mphone_   ctrMphone,                       ctr.brithday_ ctrBrithday                  from contacter_ ctr                 inner join customer_ ctm on ctr.fk_customer_ = ctm.id_                 inner join user_ u on ctm.fk_user_ = u.id_                 inner join ctm_ctg_lk_ ctmCtg on ctmCtg.fk_ctm_ = ctm.id_                 inner join ctm_ctg_ cc on ctmCtg.fk_ctm_ctg_ = cc.id_                 where ctr.brithday_ like '%-04-12'                   and u.id_ = '5dfaf039298290ee0129d51355e5029e' --这个直接用=号                   and regexp_like(ctr.mphone_, --前面加个and关系符                                   '^((13[0-9])|(15[^4,\D])|(18[0,5-9]))\d{8}$'))         order by ctmId asc) where rownum <= ?
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