Question

当我将数据从数据库导出到客户端时，我遇到了SQL问题

我的工作要求是：

select 
    datmvt as DATE_MVT,
    (select TRUNC(datmvt,-4) from dual ) as exercice,
    codpro as CODPRO,
    SUM(qtemvt)-SUM(C01) as QT
from myTable
where codpro = 'XXX'
and datmvt BETWEEN YYYYMMDD and YYYYMMDD
GROUP BY datmvt, codpro
HAVING abs(SUM(qtemvt)-SUM(C01)) > 0

给我这个结果：

DATMVT    exercice   codpro        QT
20190119  20190000  0828765332927  1
20190126  20190000  0828765332927  -1
20180117  20180000  0828765332927  -1
20180118  20180000  0828765332927  -1
20190122  20190000  0828765332927  1

在您询问之前，是的，日期以YYYYMMDD格式存储为整数值。
我的请求正在运行，但我想添加以下内容：

exercice  codpro         QT
20190000  0828765332927  -1
20180000  0828765332927  -2

因此，我试图按exercice订购，但Oracle回答：00904. 00000 - "%s: invalid identifier"

这是我无法使用但需要的请求：

select 
    (select TRUNC(datmvt,-4) from dual ) as exercice,
    codpro as CODPRO,
    SUM(qtemvt)-SUM(C01) as QT
from myTable
where codpro = 'XXX'
and datmvt BETWEEN YYYYMMDD and YYYYMMDD
GROUP BY exercice, codpro
HAVING abs(SUM(qtemvt)-SUM(C01)) > 0

Answer 1

您只能引用GROUP BY中数据集中存在的列，因此也需要在该位置重复该计算，例如：

select 
    TRUNC(datmvt,-4) as exercice,
    codpro as CODPRO,
    SUM(qtemvt)-SUM(C01) as QT
from myTable
where codpro = 'XXX'
and datmvt BETWEEN YYYYMMDD and YYYYMMDD
GROUP BY TRUNC(datmvt,-4), codpro
HAVING abs(SUM(qtemvt)-SUM(C01)) > 0

Working example on dbfiddle

Answer 2

我不知道您为什么在没有日期的情况下使用子查询(select TRUNC(datmvt,-4) from dual)作为日期。您可以从select中进行选择，但是此语法更简单，应该可以完成工作：

select trunc(datmvt,-4) exercice, codpro, sum(qtemvt)-sum(c01) qt
  from mytable
  where codpro = '0828765332927' and datmvt between 20150101 and 20191231
  group by trunc(datmvt, -4), codpro
  having sum(qtemvt) - sum(c01) <> 0

^demo

SQL Group By在计算列上-无效的标识符

2 个答案: