Question

背景：我需要将此查询修改为仅输出不平衡为零的发票号（可能是+/-）。我还需要输出以包括发票编号上的所有项目，这些项目不会平衡为零（无分组）。

所以，如果发票余额，则将其从输出中抑制。

查询：

SELECT     invoices.account, invoices.invoice_no, transact.item, transact.date_time, transact.operator, transact.salespoint, transact.extension
FROM         transact INNER JOIN
                  invoices ON transact.invoice_no = invoices.invoice_no
WHERE     (invoices.account = '*GUESTS*') AND (transact.extension <> 0))
ORDER BY invoices.invoice_no DESC

输出：

account invoice_no  item    date_time   operator    salespoint   extension 
Test    1   **TRANS**   1/0/00 12:25 AM SUNNY   RTL2     $(2.69)
Test    1   BT_DIET     1/0/00 12:25 AM SUNNY   RTL2     $2.69 
Test    2   **TRANS**   1/0/00 12:08 AM SUNNY   RTL2     $(14.55)
Test    2   **TRANS**   1/0/00 12:08 AM SUNNY   RTL2     $(1.00)
Test    2   QUICHE      1/0/00 12:08 AM SUNNY   RTL2     $7.01 
Test    2   FRUITSALAD  1/0/00 12:08 AM SUNNY   RTL2     $7.54 
Test    2   **TIPS**    1/0/00 12:08 AM SUNNY   RTL2     $1.00 
Test    3   **TRANS**   1/0/00 12:07 AM SUNNY   RTL2     $(40.67)
Test    3   BURRITO     1/0/00 12:07 AM SUNNY   RTL2     $16.17 
Test    3   ENGMUFFSAU  1/0/00 12:07 AM SUNNY   RTL2     $7.54 
Test    3   DANISH      1/0/00 12:07 AM SUNNY   RTL2     $4.30 
Test    3   SUMPLYJUIC  1/0/00 12:07 AM SUNNY   RTL2     $6.47 
Test    3   SUMPLYJUIC  1/0/00 12:07 AM SUNNY   RTL2     $3.23 
Test    3   COFFEE_CUP  1/0/00 12:07 AM SUNNY   RTL2     $2.96 
Test    4   QUICHE      1/0/00 12:01 AM SUNNY   RTL2     $7.01 
Test    4   DANISH      1/0/00 12:07 AM SUNNY   RTL2     $4.30

期望的输出：

account invoice_no  item    date_time   operator    salespoint   extension 
Test    4   QUICHE      1/0/00 12:01 AM SUNNY   RTL2     $7.01 
Test    4   DANISH      1/0/00 12:07 AM SUNNY   RTL2     $4.30

此致 DH

Answer 1

我会使用窗口/分析函数，如果这是一个选项。然后你可以通过两个步骤来解决这个问题：

添加df[cbind(seq(df$var), df$var)] [1] "3050" "2062" "1036" "4001" "3075" "4083" "1085" "3061"
将整个查询包装在另一个查询中以过滤sum(extension) over (partition by invoice_no) as invoice_balance

Answer 2

;WITH list AS (
    SELECT     invoices.account, invoices.invoice_no, transact.item, transact.date_time, transact.operator, transact.salespoint, transact.extension
    , SUM(transact.salespoint) OVER (PARTITION BY transact.item ORDER BY transact.item) AS Total
    FROM         transact INNER JOIN
              invoices ON transact.invoice_no = invoices.invoice_no
    WHERE     (invoices.account = '*GUESTS*') AND (transact.extension <> 0)
)
SELECT *
FROM list
WHERE Total > 0
ORDER BY invoice_no DESC

Answer 3

问题是扩展名不是数字。您需要编写一个存储函数，该函数可以正确地将任何值从扩展转换为数值。您还需要group by，因为0的数量是一个聚合术语。所以，你需要这样的东西：

SELECT     account, invoice_no, item, date_time, operator, salespoint, extension
FROM         transact INNER JOIN
                  invoices ON transact.invoice_no = invoices.invoice_no
group by invoices.account, invoices.invoice_no, transact.item, transact_date_time, transact.operator, transact.salespoint, myfunction(transact.extension) as extension
having     (account = '*GUESTS*') AND (sum(extension) <> 0))
ORDER BY invoice_no DESC

使用条件求和过滤查询而不进行分组

3 个答案: