SQL语句复习--group by 的用法

Posted 2021-03-08 dongyaotou

今天自己在写需求的时候，运行自己写过的sql，总是报不是group by 分组函数，自己都搞了两个小时了，实在没有办法，最后请教林哥，记得上次请教林哥的也是一个关于group by函数的使用方法，并且，人家已经叮嘱自己下来之后上网上多看看group by函数的使用方法，看来自己真的应该长点记性了，不要再不会用了，因为这可是工作呀！就得要有认真的态度。

select lb.contno,
        decode(cont.conttype, \'1\', cont.appntno, \'2\', cont.insuredno),
        decode(cont.conttype, \'1\', cont.appntname, \'2\', cont.insuredname),
        to_date(lb.create_time),
        to_char(lb.sum_total, \'9999999990.99\'),
        to_char(lb.sum_price, \'9999999990.99\'),
        to_char(lb.sum_tax, \'9999999990.99\'),
        max(lb.sid),
        lb.managecom
   from LIS_BUSI_TRANSACTIONS lb, lccont cont
  where \'1581260514000\' = \'1581260514000\'
    and cont.contno = lb.contno
    and lb.invoiceflag in (\'00\')
    and lb.successflag = \'1\'
    and not exists
  (select 1 from lcgrpcont c where c.grpcontno = lb.contno)
    and lb.sum_total > 0
    and (cont.currency = \'01\' or cont.currency is null)
    and not exists (select 1
           from ljagetendorse a, LIS_BUSI_TRANSACTIONS b
          where a.actugetno = b.sourceid
            and a.getflag = \'1\'
            and b.ruleid in (\'3\', \'6\', \'10\', \'12\')
            and a.actugetno = lb.sourceid)
    and exists (SELECT 1 FROM ljapay d where d.otherno = cont.contno)
/*    and lb.contno = \'1\'
    and to_date(lb.create_time) >= \'2019-02-10\'
    and to_date(lb.create_time) <= \'2020-02-14\'
    and lb.managecom like \'86%\'
    and lb.managecom like \'86%\'*/
    group by lb.contno,
        decode(cont.conttype, \'1\', cont.appntno, \'2\', cont.insuredno),
        decode(cont.conttype, \'1\', cont.appntname, \'2\', cont.insuredname),
        to_date(lb.create_time),
        to_char(lb.sum_total, \'9999999990.99\'),
        to_char(lb.sum_price, \'9999999990.99\'),
        to_char(lb.sum_tax, \'9999999990.99\'),
        lb.sid,
        lb.managecom

　　现在上面的函数是能够运行的，group by 函数就是要将查询结果中的每一列都要放到group by的后面。自己当时在写这个sql的时候，也不是怎么搞的，就是报错，可能是因为当时group by后面少放了字段的原因，看来遇到问题不要慌张，要冷静下来，仔细思考，结果一问林哥，人家也没有说啥，自己又尝试着改了一遍就好使了。

不过林哥说：你见哪个开发这样写sql了。

 

 下面这一段，确实，自己都看着恶心。改造一下呗！

select lb.contno,
        decode(cont.conttype, \'1\', cont.appntno, \'2\', cont.insuredno),
        decode(cont.conttype, \'1\', cont.appntname, \'2\', cont.insuredname),
        to_date(lb.create_time),
        to_char(lb.sum_total, \'9999999990.99\'),
        to_char(lb.sum_price, \'9999999990.99\'),
        to_char(lb.sum_tax, \'9999999990.99\'),
        max(lb.sid),
        lb.managecom
   from LIS_BUSI_TRANSACTIONS lb, lccont cont
  where \'1581260514000\' = \'1581260514000\'
    and cont.contno = lb.contno
    and lb.invoiceflag in (\'00\')
    and lb.successflag = \'1\'
    and not exists
  (select 1 from lcgrpcont c where c.grpcontno = lb.contno)
    and lb.sum_total > 0
    and (cont.currency = \'01\' or cont.currency is null)
    and not exists (select 1
           from ljagetendorse a, LIS_BUSI_TRANSACTIONS b
          where a.actugetno = b.sourceid
            and a.getflag = \'1\'
            and b.ruleid in (\'3\', \'6\', \'10\', \'12\')
            and a.actugetno = lb.sourceid)
    and exists (SELECT 1 FROM ljapay d where d.otherno = cont.contno)
/*    and lb.contno = \'1\'
    and to_date(lb.create_time) >= \'2019-02-10\'
    and to_date(lb.create_time) <= \'2020-02-14\'
    and lb.managecom like \'86%\'
    and lb.managecom like \'86%\'*/
    group by lb.contno,
        cont.conttype, cont.appntno, cont.insuredno, cont.appntname,  cont.insuredname,
        lb.create_time,
        lb.sum_total,
        lb.sum_price, 
        lb.sum_tax,
        lb.sid,
        lb.managecom

　　这样就好看多了，哈哈！

值得注意的是，像查询结果中使用到的decode函数，等等，一定要将函数中使用到的本表中的字段，一个不落的写在group by的后面才行。

假如我给要查询的结果列起别名，行不行呢？

select lb.contno a,
        decode(cont.conttype, \'1\', cont.appntno, \'2\', cont.insuredno) b,
        decode(cont.conttype, \'1\', cont.appntname, \'2\', cont.insuredname) c,
        to_date(lb.create_time) d,
        to_char(lb.sum_total, \'9999999990.99\') e,
        to_char(lb.sum_price, \'9999999990.99\') f,
        to_char(lb.sum_tax, \'9999999990.99\') g,
        max(lb.sid) h,
        lb.managecom j
   from LIS_BUSI_TRANSACTIONS lb, lccont cont
  where \'1581260514000\' = \'1581260514000\'
    and cont.contno = lb.contno
    and lb.invoiceflag in (\'00\')
    and lb.successflag = \'1\'
    and not exists
  (select 1 from lcgrpcont c where c.grpcontno = lb.contno)
    and lb.sum_total > 0
    and (cont.currency = \'01\' or cont.currency is null)
    and not exists (select 1
           from ljagetendorse a, LIS_BUSI_TRANSACTIONS b
          where a.actugetno = b.sourceid
            and a.getflag = \'1\'
            and b.ruleid in (\'3\', \'6\', \'10\', \'12\')
            and a.actugetno = lb.sourceid)
    and exists (SELECT 1 FROM ljapay d where d.otherno = cont.contno)
/*    and lb.contno = \'1\'
    and to_date(lb.create_time) >= \'2019-02-10\'
    and to_date(lb.create_time) <= \'2020-02-14\'
    and lb.managecom like \'86%\'
    and lb.managecom like \'86%\'*/
    group by a,b,c,d,e,f,g,h,j

　　答案是不行的，会报J标识符无效。

哦！对了，这个是oracle数据库的。