Question

+-------------+--------------+----------+-------+
| ticketRefNo | nameOnTicket | boughtBy | event |
+-------------+--------------+----------+-------+
|          38 | J XXXXXXXXX  |        2 |    13 |
|          39 | C YYYYYYY    |        1 |    13 |
|          40 | M ZZZZZZZZZZ |        3 |    14 |
|          41 | C AAAAAAA    |        3 |    15 |
|          42 | D BBBBBB     |        3 |    16 |
|          43 | A CCCCC      |        3 |    17 |
+-------------+--------------+----------+-------+

+-------------+------------------+--------------+---------------------+--------+
| ticketRefNo | cardNo           | cardHolder   | exp                 | issuer |
+-------------+------------------+--------------+---------------------+--------+
|          38 | 4444111133332222 | J McKenny    | 2016-01-01 00:00:00 | BOS    |
|          39 | 4434111133332222 | C Dempsey    | 2016-04-01 00:00:00 | BOS    |
|          40 | 4244111133332222 | M Gunn-Davis | 2018-02-01 00:00:00 | RBS    |
+-------------+------------------+--------------+---------------------+--------+

+-------------+-------------+----------+
| ticketRefNo | boxOfficeID | paidWith |
+-------------+-------------+----------+
|          41 |           1 | card     |
|          42 |           2 | cash     |
|          43 |           3 | chequ    |
+-------------+-------------+----------+

我有一个包含上述数据的数据库。它代表了购票系统。我希望能够看到以活动名称和boxOfficeID或借记卡发行者购买的门票列表。

我试过运行以下代码，但无济于事。

SELECT t.ticketRefNo AS 'Reference', t.event AS 'Event', 
IF(COUNT(SELECT * FROM Online WHERE t.ticketRefNo=o.ticketRefNo;) >= 1,
o.issuer, InPerson.boxOfficeID) AS 'Card Issuer or Box Office'
FROM Ticket AS t, InPerson, Online AS o 
WHERE t.ticketRefNo=o.ticketRefNo;

提前干杯！

Answer 1

一些注意事项：分号字符的语法无效;如果您需要分隔子查询，请将其包装在parens中。转义列别名就像你要转义任何其他标识符：使用反引号，而不是单引号。单引号用于字符串文字。

假设issuer表中的Online为非空，并且假设ticketRefNo和Online表中的InPerson都是唯一的，则可以做这样的事情：

SELECT t.ticketRefNo    AS `Reference`
     , t.event          AS `Event`
     , IF(o.ticketRefNo IS NOT NULL,o.issuer,i.boxOfficeId) 
       AS `Card Issuer or Box Office`
  FROM Ticket t
  LEFT
  JOIN InPerson i 
    ON i.ticketRefNo = t.ticketRefNo
  LEFT
  JOIN Online o
    ON o.ticketRefNo  = t.ticketRefNo

使用外部联接操作查找InPerson和Online表中的匹配行，并使用条件测试查看是否从Online表中获得了匹配的行。如果找不到匹配的行，将返回NULL。

Answer 2

将一列连接到两个不同的表并且两个表中的每一个都有一些值，这不是一个好主意。

但无论如何这里仍然存在：

( SELECT ... FROM Ticket t JOIN InPerson x USING(ticketRefNo) ... )
UNION ALL
( SELECT ... FROM Ticket t JOIN Online x USING(ticketRefNo) ... )
ORDER BY ...

ALL假设InPerson和Online永远不会有任何重叠的ticketRefNos。最后的ORDER BY是为了你想要排序的东西，虽然我在你尝试的SELECT中看不到它。两个SELECT必须具有相同的列数。

MySQL在IF中计数

2 个答案: