Question

我正在针对Progress 10.2B07数据库编写SQL并收到以下错误＆＃34; Column＆＃39; OUTERINVOICEHEADER.MEMBERID＆＃39;无法找到或未指定查询（13865）。

以下是查询：

select concat(substring(OuterInvoiceHeader.sold_to_cust_nbr, 1, 6) + '-', OuterInvoiceHeader.sold_to_cust_seq) as MemberID, 
sum(OuterInvoiceHeader.net_weight) as TotalInvoicePounds, 
sum(OuterInvoiceHeader.net_weight / 2000) as TotalTons, 
sum(OuterInvoiceHeader.invoice_amt) as InvoiceAmount, 
sum(InvoiceSurcharges.Surcharge) as Surcharges, 
sum(OuterInvoiceHeader.invoice_amt - InvoiceSurcharges.Surcharge) as Total,
sum(Returns.qty_received) as PoundsReturned
from AXS.PUB.ivc_header OuterInvoiceHeader

inner join
(select m.invoice_nbr, sum(m.extension) Surcharge  from AXS.PUB.ivc_mchgs m
inner join
AXS.PUB.ivc_header h
on h.invoice_nbr = m.invoice_nbr
group by m.invoice_nbr) InvoiceSurcharges
on OuterInvoiceHeader.invoice_nbr = InvoiceSurcharges.invoice_nbr

left outer join
(select concat(substring(ReturnHeader.ship_to_nbr, 1, 6)+'-',InnerInvoiceHeader.sold_to_cust_seq) as ReturnMemberID, 
ReturnHeader.invoice_nbr as ReturnInvoiceNum, 
qty_received
from AXS.PUB.return_hdr ReturnHeader
inner join
AXS.PUB.ivc_header InnerInvoiceHeader
on ReturnHeader.invoice_nbr = InnerInvoiceHeader.invoice_nbr
inner join AXS.PUB.return_line ReturnLine
on ReturnHeader.claim_nbr = ReturnLine.claim_nbr
where ReturnInvoiceNum = '0001010914'
group by ReturnMemberID, ReturnInvoiceNum, qty_received) Returns
on OuterInvoiceHeader.MemberID = Returns.ReturnMemberID
--on OuterInvoiceHeader.invoice_nbr = Returns.ReturnInvoiceNum

where OuterInvoiceHeader.sold_to_cust_nbr =  '000837' and     OuterInvoiceHeader.invoice_date between '06/01/2016' and '06/30/2016' and  OuterInvoiceHeader.invoice_status = '5804' and OuterInvoiceHeader.invoice_type='5601'
group by MemberID

问题在于左连接;在OuterInvoiceHeader.invoice_nbr = Returns.ReturnInvoiceNum＆＃34;上注释掉的条款＆＃34;如果没有注释将会工作。＆＃34; on OuterInvoiceHeader.MemberID = Returns.ReturnMemberID＆＃34;子句给了我错误。

我不明白的是，这两个引用了顶级SELECT语句中的一列，唯一的区别是一个是串联而另一个不是。

我希望我在这里看不到森林的树木，答案很简单，所以如果有人有任何建议或问题，我会全神贯注。

Answer 1

试试这个：

我将对别名MemberID的引用替换为实际的连续列CONCAT(SUBSTRING(OuterInvoiceHeader.sold_to_cust_nbr, 1, 6)+'-', OuterInvoiceHeader.sold_to_cust_seq)

SELECT                     CONCAT(SUBSTRING(OuterInvoiceHeader.sold_to_cust_nbr, 1, 6)+'-', OuterInvoiceHeader.sold_to_cust_seq) AS MemberID
                    , SUM(OuterInvoiceHeader.net_weight) AS TotalInvoicePounds
                    , SUM(OuterInvoiceHeader.net_weight / 2000) AS TotalTons
                    , SUM(OuterInvoiceHeader.invoice_amt) AS InvoiceAmount
                    , SUM(InvoiceSurcharges.Surcharge) AS Surcharges
                    , SUM(OuterInvoiceHeader.invoice_amt - InvoiceSurcharges.Surcharge) AS Total
                    , SUM(Returns.qty_received) AS PoundsReturned
FROM                       AXS.PUB.ivc_header OuterInvoiceHeader
                      INNER JOIN
                              (SELECT m.invoice_nbr
                                   , SUM(m.extension) Surcharge
                               FROM   AXS.PUB.ivc_mchgs m
                                    INNER JOIN AXS.PUB.ivc_header h ON h.invoice_nbr = m.invoice_nbr
                               GROUP BY m.invoice_nbr) InvoiceSurcharges ON OuterInvoiceHeader.invoice_nbr = InvoiceSurcharges.invoice_nbr
                      LEFT OUTER JOIN
                                  (SELECT CONCAT(SUBSTRING(ReturnHeader.ship_to_nbr, 1, 6)+'-', InnerInvoiceHeader.sold_to_cust_seq) AS ReturnMemberID
                                       , ReturnHeader.invoice_nbr AS ReturnInvoiceNum
                                       , qty_received
                                   FROM   AXS.PUB.return_hdr ReturnHeader
                                        INNER JOIN AXS.PUB.ivc_header InnerInvoiceHeader ON ReturnHeader.invoice_nbr = InnerInvoiceHeader.invoice_nbr
                                        INNER JOIN AXS.PUB.return_line ReturnLine ON ReturnHeader.claim_nbr = ReturnLine.claim_nbr
                                   WHERE  ReturnInvoiceNum = '0001010914'
                                   GROUP BY ReturnMemberID
                                        , ReturnInvoiceNum
                                        , qty_received) Returns ON CONCAT(SUBSTRING(OuterInvoiceHeader.sold_to_cust_nbr, 1, 6)+'-', OuterInvoiceHeader.sold_to_cust_seq) = Returns.ReturnMemberID
--on OuterInvoiceHeader.invoice_nbr = Returns.ReturnInvoiceNum

WHERE OuterInvoiceHeader.sold_to_cust_nbr = '000837'
     AND OuterInvoiceHeader.invoice_date BETWEEN '06/01/2016' AND '06/30/2016'
     AND OuterInvoiceHeader.invoice_status = '5804'
     AND OuterInvoiceHeader.invoice_type = '5601'
GROUP BY CONCAT(SUBSTRING(OuterInvoiceHeader.sold_to_cust_nbr, 1, 6)+'-', OuterInvoiceHeader.sold_to_cust_seq);

基本上，您需要记住执行SQL语句的顺序：

FROM clause
WHERE子句
GROUP BY子句
HAVING条款
SELECT条款
ORDER BY子句

Answer 2

这是一个计算列别名，因此也就是错误。您应该考虑使用整个表达式而不是

on concat(substring(OuterInvoiceHeader.sold_to_cust_nbr, 1, 6) + '-', OuterInvoiceHeader.sold_to_cust_seq) = Returns.ReturnMemberID

而不是on OuterInvoiceHeader.MemberID = Returns.ReturnMemberID。同样，更改使用相同别名的任何其他位置。您可以而且应该仅在外部查询中使用该别名，而不是在同一查询中。

无法找到Progress SQL列或未为查询指定Progress SQL列

2 个答案: