Question

是否可以从一个表中选择值，它们在一个列表中不存在，但在另一个列表中存在...或者它们是否相反？

E.g。

SELECT COUNT(g.`property`) as `number`, g.`property` 
  FROM `foo` g 
 WHERE `theID` IS IN (SELECT `theID` 
                        FROM `tableofIDS` 
                       WHERE `theID` = '54252') 
          AND NOT IN (SELECT `theID` 
                        FROM `anotherTableofIDS` 
                       WHERE `theID` = '54252')

Answer 1

SELECT COUNT(g.`property`) as `number`, g.`property`
FROM `foo` g
WHERE `theID` IN (SELECT `theID` FROM `tableofIDS` WHERE `theID` = '54252')
  AND `theID` NOT IN (SELECT `theID` FROM `anotherTableofIDS` WHERE `theID` = '54252')
GROUP BY g.`property`

另外，您可以使用效果更好的联接：

SELECT COUNT(g.`property`) as `number`, g.`property`
FROM `foo` g JOIN (
    SELECT `theID`
    FROM `tableofIDS`
    WHERE `theID` = '54252'
   ) id1 ON g.theID = id1.theID
  LEFT JOIN (
    SELECT `theID`
    FROM `anotherTableofIDS`
    WHERE `theID` = '54252'
  ) id2 ON g.theID = id2.theID
WHERE id2.theID IS NULL
GROUP BY g.`property`

Answer 2

没有想到，但发现了2个语法错误。尝试

SELECT COUNT(g.`property`) as `number`, g.`property` FROM `foo` g WHERE `theID` IN (SELECT `theID` FROM `tableofIDS` WHERE `theID` = '54252') AND `theID` NOT EXISTS (SELECT `theID` FROM `anotherTableofIDS` WHERE `theID` = '54252')

修改：将NOT IN更改为NOT EXISTS

Answer 3

我想也许你正在寻找的只是这个：

    SELECT COUNT(g.property) as `number`,
           g.property
      FROM foo g
      JOIN tableofIDs i USING (theID)
RIGHT JOIN anothertableofIDs x USING (theID)
     WHERE g.theID = '54252'
       AND x.theID IS NULL
  GROUP BY g.property

mySQL：如何选择FROM表WHERE IN LIST而不是另一个列表

3 个答案: