如何将这些结果合并为一行

Question

我在下面有这个代码。我正在尝试根据日期合并行。

SELECT

TICKETS.TICKETID,
RECEIPTS.DATENEW,
TAXCATEGORIES.NAME = 'GCT' as GCT,
TAXCATEGORIES.NAME = 'Tax 25%' as Tax25,
TAXLINES.AMOUNT,

SUM(TAXLINES.AMOUNT) AS TOTAL,
SUM(CASE WHEN taxcategories.NAME = 'GCT' THEN taxlines.AMOUNT ELSE 0 END) AS GCTTOTAL,
SUM(CASE WHEN taxcategories.NAME = 'Tax 25%' THEN taxlines.AMOUNT ELSE 0 END) AS TAX25TOTAL

   FROM
     RECEIPTS,
     TAXLINES,
     TAXES,
     TAXCATEGORIES,
     TICKETS,
     PAYMENTS

  WHERE
    PAYMENTS.RECEIPT = RECEIPTS.ID
    AND RECEIPTS.ID = TAXLINES.RECEIPT
    AND RECEIPTS.ID = TICKETS.ID
    AND TAXLINES.TAXID = TAXES.ID
    AND TAXES.CATEGORY = TAXCATEGORIES.ID
    AND DATENEW >= '2016-07-14 00:00:00' AND DATENEW <= '2016-07-14 23:00:00'


GROUP BY gct, Tax25, CAST(RECEIPTS.DATENEW AS DATE)

查询结果附在下面的屏幕截图中：

现在我需要帮助将具有相同日期的行合并到一行中。我不知道我哪里出错了，我已经尝试了一系列连接但是我空白了。

Answer 1

具有相同日期的记录似乎实际上在GCT和Tax25列的值不同。如果您从GROUP BY子句中删除这些列，而是将它们汇总到SELECT列表中，那么您将留下重复日期的单个记录。

请注意，在下面的查询中，我已使用显式内连接替换了您的隐式连接（使用逗号分隔的FROM子句中的表），并使用ON子句中的连接条件。这是现在编写查询的标准方法，它使它们更容易阅读。如果INNER JOIN限制性太强，那么您可能打算使用LEFT JOIN代替。

SELECT TICKETS.TICKETID,
       RECEIPTS.DATENEW,
       MAX(TAXCATEGORIES.NAME) = 'GCT' as GCT,         -- one value per date
       MAX(TAXCATEGORIES.NAME) = 'Tax 25%' as Tax25,   -- one value per date
       TAXLINES.AMOUNT,
       SUM(TAXLINES.AMOUNT) AS TOTAL,
       SUM(CASE WHEN taxcategories.NAME = 'GCT' THEN taxlines.AMOUNT ELSE 0 END) AS GCTTOTAL,
       SUM(CASE WHEN taxcategories.NAME = 'Tax 25%' THEN taxlines.AMOUNT ELSE 0 END) AS TAX25TOTAL
FROM RECEIPTS
INNER JOIN TAXLINES
    ON RECEIPTS.ID = TAXLINES.RECEIPT
INNER JOIN TAXES
    ON TAXLINES.TAXID = TAXES.ID
INNER JOIN TAXCATEGORIES
    ON TAXES.CATEGORY = TAXCATEGORIES.ID
INNER JOIN TICKETS
    ON RECEIPTS.ID = TICKETS.ID
INNER JOIN PAYMENTS
    ON PAYMENTS.RECEIPT = RECEIPTS.ID
WHERE DATENEW >= '2016-07-14 00:00:00' AND
      DATENEW <= '2016-07-14 23:00:00'
GROUP BY CAST(RECEIPTS.DATENEW AS DATE)