Question

我想从一个表调整一个简单的MySQL查询到另一个表。第一个查询按预期工作，但第二个修改后的查询没有显示预期结果，但我无法弄清楚结构上的任何差异。

这个问题与我之前认为已经解决的问题有关：

SQL IF ELSE / CASE clause in WHERE condition

简要解释了查询应该做的事情：

我想在SQL SELECT中编写一个WHERE条件，该条件应该显示每个ID最多一个结果。但它应该只在LANG是FR或EN时显示结果。如果没有FR可用于ID，则应优先选择FR并且EN应仅显示为替代。所以结果看起来像这样。

这里是工作查询。这是SQL小提琴LINK

SELECT * FROM `table1`
WHERE ID = 4
AND lang = 'FR'
OR (lang = 'EN' AND ID = 4 NOT IN (SELECT ID FROM table1 WHERE lang = 'FR'))

这是结构相同的查询。这是SQL小提琴LINK

SELECT * FROM `epf_application_detail`
WHERE application_id = 281656475
AND language_code = 'PL'
OR (language_code = 'EN' AND application_id = 281656475 NOT IN (SELECT application_id FROM `epf_application_detail` WHERE language_code = 'PL'))

如果第二个小提琴能正常工作，它应该显示

+--------------+----------------+---------------+-----------+
|  export_date | application_id | language_code |    title  |
+--------------+----------------+---------------+-----------+
|1377594004198 | 281656475      | 'EN'          | 'PAC-MAN' |
+--------------+----------------+---------------+-----------+

我绝对不知道什么是差别，所以任何帮助都很有意义

编辑：最适合我的解决方案（因为我需要将它与union all结合起来，每个application_id都需要一个结果）

SELECT *
FROM `epf_application_detail`
WHERE application_id = 281656475 AND
      (language_code = 'PL' OR
       (language_code = 'EN' AND
        application_id NOT IN (SELECT application_id
                               FROM `epf_application_detail`
                               WHERE language_code = 'PL' and application_id is not null
                              )
       )
      )

Answer 1

您的查询失败，因为您正在进行比较和 not in：

SELECT *
FROM `epf_application_detail`
WHERE (application_id = 281656475 AND language_code = 'PL') OR
      (language_code = 'EN' AND
       application_id = 281656475 NOT IN (SELECT application_id
----------------------------------^
                                          FROM `epf_application_detail` WHERE language_code = 'PL'))

MySQL将进行第一次比较，并将布尔结果转换为not in的整数。根据第一个查询的结构，您需要：

SELECT *
FROM `epf_application_detail`
WHERE (application_id = 281656475 AND language_code = 'PL') OR
      (language_code = 'EN' AND
       application_id NOT IN (SELECT application_id
                              FROM `epf_application_detail`
                              WHERE language_code = 'PL' and application_id is not null
                             )
      )

我还添加了application_id is not null，因为NULL会导致NOT IN失败。

编辑：

根据您的评论，这应该捕捉您的逻辑：

SELECT *
FROM `epf_application_detail`
WHERE application_id = 281656475 AND
      (language_code = 'PL' OR
       (language_code = 'EN' AND
        application_id NOT IN (SELECT application_id
                               FROM `epf_application_detail`
                               WHERE language_code = 'PL' and application_id is not null
                              )
       )
      )

如果您只是从详细信息表中查找一行，则以下内容要简单得多：

SELECT *
FROM `epf_application_detail`
WHERE application_id = 281656475 AND
      language_code in ('PL', 'EN')
ORDER BY language_code = 'PL' desc
LIMIT 1;

这使用特定于MySQL的语法。您的问题没有标记为MySQL，但确实使用了MySQL语法。

Answer 2

你可以替换

 not IN

与

 and NOT exists


SELECT * FROM `epf_application_detail`
WHERE application_id = 281656475
AND language_code = 'PL'
OR (language_code = 'EN' AND
    application_id = 281656475 and NOT exists (SELECT application_id FROM `epf_application_detail` WHERE language_code = 'PL'))

MySQL - 在WHERE子句中的NOT IN查询，在第一个表上使用相同的结构但在第二个表上没有

2 个答案: