美文网首页
根据日期求对应周次的同期上期,再根据求出的周次求对应周的开始时间

根据日期求对应周次的同期上期,再根据求出的周次求对应周的开始时间

作者: 圈半球 | 来源:发表于2018-12-27 10:27 被阅读0次

    select k.rq_min rq_bz_min,m.rq_max rq_sq_max,m.rq_min rq_sq_min,n.rq_max rq_tz_max,n.rq_min rq_tz_min from

    (select date (max(a.rqdm)) + 1 days rq_max,date (min(a.rqdm)) + 1 days rq_min

    from dm.td_rq a

    left join

    (select nfdm nf_bq,djz zc_bq, char(int(nfdm)-1) nf_tq,djz zc_tq,case when INT(djz)=1 then trim(char(int(nfdm)-1)) else nfdm end nf_sq,case when INT(djz)=1 then 53 else INT(djz)-1 end zc_sq

    from dm.td_rq

    where rqdm='2018-12-12') b

    on a.nfdm = b.nf_bq and a.djz=b.zc_bq

    where a.nfdm = b.nf_bq and a.djz=b.zc_bq) k

    inner join

    (select date (max(a.rqdm)) + 1 days rq_max,date (min(a.rqdm)) + 1 days rq_min

    from dm.td_rq a

    left join

    (select nfdm nf_bq,djz zc_bq, char(int(nfdm)-1) nf_tq,djz zc_tq,case when INT(djz)=1 then trim(char(int(nfdm)-1)) else nfdm end nf_sq,case when INT(djz)=1 then 53 else INT(djz)-1 end zc_sq

    from dm.td_rq

    where rqdm='2018-12-12') b

    on a.nfdm = b.nf_sq and a.djz=b.zc_sq

    where a.nfdm = b.nf_sq and a.djz=b.zc_sq) m

    on 1=1

    inner join

    (select date (max(a.rqdm)) + 1 days rq_max,date (min(a.rqdm)) + 1 days rq_min

    from dm.td_rq a

    left join

    (select nfdm nf_bq,djz zc_bq, char(int(nfdm)-1) nf_tq,djz zc_tq,case when INT(djz)=1 then trim(char(int(nfdm)-1)) else nfdm end nf_sq,case when INT(djz)=1 then 53 else INT(djz)-1 end zc_sq

    from dm.td_rq

    where rqdm='2018-12-12') b

    on a.nfdm = b.nf_tq and a.djz=b.zc_tq

    where a.nfdm = b.nf_tq and a.djz=b.zc_tq) n

    on 1=1

    相关文章

      网友评论

          本文标题:根据日期求对应周次的同期上期,再根据求出的周次求对应周的开始时间

          本文链接:https://www.haomeiwen.com/subject/betrlqtx.html