2

自从我上次做 Oracle SQL 以来已经有一段时间了，希望有人能告诉我为什么我得到一个 933：

   SELECT TRIM(A.ACCOUNTNUMBER) AS INDBDebnmbr
, TRIM(A.VOUCHER) AS INinvoicenmbr
, A.DATE_ AS INinvoiceDate
, A.DUEDATE AS INinvoiceDueDate
, A.TXT AS INDescription
, A.EXCHANGECODE AS INCurrencyCode
, subq.AMOUNTMST AS INOriginalamount
, subq.SETTLEAMOUNTMST AS INpaidAmount
, subq.OPENAMOUNT AS INOpenAmount
FROM (
  SELECT DEBTRANS.VOUCHER AS VOUCHER, SUM(DEBTRANS.AMOUNTMST) AS AMOUNTMST
  , SUM(DEBTRANS.SETTLEAMOUNTMST) AS SETTLEAMOUNTMST
  , SUM(DEBTRANS.AMOUNTMST - DEBTRANS.SETTLEAMOUNTMST) AS OPENAMOUNT
  FROM XAL_SUPERVISOR.DEBTRANS DEBTRANS 
  WHERE DEBTRANS.OPEN = 1 AND
  DEBTRANS.TRANSTYPE <> 9 AND
  (DEBTRANS.AMOUNTMST - DEBTRANS.SETTLEAMOUNTMST) <> 0 AND
  DEBTRANS.DATASET = 'FIK'
  GROUP BY DEBTRANS.VOUCHER) subq INNER JOIN DEBTRANS A ON A.VOUCHER = subq.VOUCHER

在 SQL Plus 中提前致谢，

迈克尔

4

3 回答 3

2

您对使用 8i 的评论解释了这一点。ANSI '92 Join 语法直到 9i 才在 Oracle 中实现。

您将需要修改您的查询：

       SELECT TRIM(A.ACCOUNTNUMBER) AS INDBDebnmbr
    , TRIM(A.VOUCHER) AS INinvoicenmbr
    , A.DATE_ AS INinvoiceDate
    , A.DUEDATE AS INinvoiceDueDate
    , A.TXT AS INDescription
    , A.EXCHANGECODE AS INCurrencyCode
    , subq.AMOUNTMST AS INOriginalamount
    , subq.SETTLEAMOUNTMST AS INpaidAmount
    , subq.OPENAMOUNT AS INOpenAmount
    FROM (
      SELECT DEBTRANS.VOUCHER AS VOUCHER, SUM(DEBTRANS.AMOUNTMST) AS AMOUNTMST
      , SUM(DEBTRANS.SETTLEAMOUNTMST) AS SETTLEAMOUNTMST
      , SUM(DEBTRANS.AMOUNTMST - DEBTRANS.SETTLEAMOUNTMST) AS OPENAMOUNT
      FROM XAL_SUPERVISOR.DEBTRANS DEBTRANS 
      WHERE DEBTRANS.OPEN = 1 AND
      DEBTRANS.TRANSTYPE <> 9 AND
      (DEBTRANS.AMOUNTMST - DEBTRANS.SETTLEAMOUNTMST) <> 0 AND
      DEBTRANS.DATASET = 'FIK'
      GROUP BY DEBTRANS.VOUCHER) subq,
      DEBTRANS A
 WHERE A.VOUCHER = subq.VOUCHER;
于 2011-06-15T14:39:29.753 回答
1

您要加入的 DEBTRANS 是什么，它是 XAL_SUPERVISOR.DEBTRANS 的另一个实例吗？如果是这样，请不要在子查询中使用 DEBTRANS 作为别名，这会造成混淆。将其更改为其他内容并重试，例如

 SELECT TRIM(A.ACCOUNTNUMBER) AS INDBDebnmbr
, TRIM(A.VOUCHER) AS INinvoicenmbr
, A.DATE_ AS INinvoiceDate
, A.DUEDATE AS INinvoiceDueDate
, A.TXT AS INDescription
, A.EXCHANGECODE AS INCurrencyCode
, subq.AMOUNTMST AS INOriginalamount
, subq.SETTLEAMOUNTMST AS INpaidAmount
, subq.OPENAMOUNT AS INOpenAmount
FROM (
  SELECT dt.VOUCHER AS VOUCHER
  , SUM(dt.AMOUNTMST) AS AMOUNTMST
  , SUM(dt.SETTLEAMOUNTMST) AS SETTLEAMOUNTMST
  , SUM(dt.AMOUNTMST - dt.SETTLEAMOUNTMST) AS OPENAMOUNT
  FROM XAL_SUPERVISOR.DEBTRANS dt
  WHERE dt.OPEN = 1 AND
  dt.TRANSTYPE <> 9 AND
  (dt.AMOUNTMST - dt.SETTLEAMOUNTMST) <> 0 AND
  dt.DATASET = 'FIK'
  GROUP BY dt.VOUCHER) subq INNER JOIN DEBTRANS A ON A.VOUCHER = subq.VOUCHER
于 2011-06-15T13:09:40.747 回答
0

而不是最后一行

A.VOUCHER 上的内部连接 ​​DEBTRANS A = subq.VOUCHER

采用

A.VOUCHER 上的内部连接 ​​DEBTRANS A = subq.INinvoicenmbr

于 2011-06-15T13:25:27.240 回答