Question

我想我应该知道这一点，特别是在阅读了很多关于“条件必须进入ON子句而不是WHERE子句”的问题和答案之后。但是，我还是输了。

我有三个表，我通常使用LEFT（OUTER）连接加入它们。连接表看起来像这样（retty标准）：


task_id task_questions_taskId   taskQuestions_questionId    question_id
1         1                     5                           5
1         1                     8                           8
2         2                     8                           8

SELECT `t`.`id` AS `task_id` , 
       `task_questions`.`taskId` AS `task_questions_taskId` ,
       `task_questions`.`questionId` AS `task_questions_questionId` , 
       questions.id AS question_id
FROM `task` `t`
LEFT OUTER JOIN `task_questions` `task_questions` 
    ON ( `task_questions`.`taskId` = `t`.`id` )
LEFT OUTER JOIN `question` `questions` 
    ON ( `task_questions`.`questionId` = `questions`.`id` )

这是获取所有记录的标准查询。（它来自Yii;我实际上想要使用Active Record，但是甚至无法获得普通的SQL）。

现在我想得到那些有问题2和8（例如）的任务因此，如果任务没有这两个question.ids，我不希望它在结果集中。在这种情况下，任务也可以有其他question_ids。虽然如果它只返回那些具有那些2（或任何其他集合）的那些查询将会如何看起来会很有趣。使用WHERE question.id = 2，很容易得到所有有一个问题的任务，但是WHERE子句中的AND会导致空结果。

Answer 1

WHERE子句一次只能将条件应用于一行。但是你对不同id的问题出现在不同的行上。怎么解决这个？使用自联接将两行连接到一行。

以下是一个例子：

SELECT t.`id` AS `task_id`, ...
FROM `task` AS t
INNER JOIN `task_questions` AS tq2 ON ( tq2.`taskId` = t.`id` )
INNER JOIN `questions` AS q2 ON ( tq2.`questionId` = q2.`id` )
INNER JOIN `task_questions` AS tq8 ON ( tq8.`taskId` = t.`id` )
INNER JOIN `questions` AS q8 ON ( tq8.`questionId` = q8.`id` )
WHERE q2.`id` = 2 AND q8.`id` = 8

另一个解决方案是找到有问题2 OR 8的任务，然后使用GROUP BY和HAVING按照其中两个具有正确值的组进行过滤。

SELECT t.`id` AS `task_id`, ...
FROM `task` AS t
INNER JOIN `task_questions` AS tq ON ( tq.`taskId` = t.`id` )
INNER JOIN `questions` AS q ON ( tq.`questionId` = q.`id` )
WHERE tq.`questionId` IN (2, 8)
GROUP BY t.`id`
HAVING COUNT(DISTINCT q.`id`) = 2

Answer 2

即使没有使用和，你也可以这样做 ...其中question.id IN（2,8）

Answer 3

使用IN：

SELECT `t`.`id` AS `task_id` , 
       `task_questions`.`taskId` AS `task_questions_taskId` ,
       `task_questions`.`questionId` AS `task_questions_questionId` , 
       questions.id AS question_id
FROM `task` `t`
LEFT OUTER JOIN `task_questions` `task_questions` 
    ON ( `task_questions`.`taskId` = `t`.`id`)
LEFT OUTER JOIN `question` `questions` 
    ON ( `task_questions`.`questionId` = `questions`.`id` )
WHERE  `task_questions`.`questionId` IN (2, 8)

Answer 4

这应该这样做

SELECT `t`.`id` AS `task_id` , 
       `task_questions`.`taskId` AS `task_questions_taskId` ,
       `task_questions`.`questionId` AS `task_questions_questionId` , 
       questions.id AS question_id
FROM `task` `t`
LEFT OUTER JOIN `task_questions` `task_questions` 
    ON ( `task_questions`.`taskId` = `t`.`id` )
LEFT OUTER JOIN `question` `questions` 
    ON ( `task_questions`.`questionId` = `questions`.`id` )
WHERE  questions.id in (2,8)

Answer 5

你不是在寻找AND，而是在寻找OR，或者是IN：

WHERE `questions`.`id` IN (2,8) -- grab everything in the parens.

或

WHERE `questions`.`id` = 2 OR -- grab each item individually
      `questions`.`id` = 8

如果您使用AND，那意味着ID必须同时为8和2。糟糕的交易。

加入：三个表和一个或条件

5 个答案: