SELECT pae.peID FROM PermissionAccessEntities pae 
INNER JOIN PermissionAccessEntityGroups paeg1 ON  pae.peID=paeg1.peID
WHERE pae.petID = 4 AND paeg1.peID NOT IN (SELECT DISTINCT peID FROM 
PermissionAccessEntityGroups WHERE gID NOT IN (2,3));

您可以通过首先查找与（2,3）匹配的条目，然后查看其中是否有与其他值之一匹配的条目来获得所需的结果。任何没有的( paeg2.peID IS NULL )完全匹配：

SELECT DISTINCT(pae.peID)
FROM PermissionAccessEntities pae 
JOIN PermissionAccessEntityGroups paeg1 ON paeg1.peID = pae.peID AND paeg1.gID IN (2,3)
LEFT JOIN PermissionAccessEntityGroups paeg2 ON paeg2.peID = pae.peID AND paeg2.gID NOT IN (2,3)
WHERE pae.petID = 4 AND paeg2.peID IS NULL

在dbfiddle上演示

Where 这种情况对你没有帮助。因为，您需要考虑所有行的计数以进行进一步筛选。你需要使用 Group By 使用条件筛选 Having 条款。
我们还可以简化查询，只使用一个 PermissionAccessEntityGroups table。 Count() 函数用于计算组中的行数。我们也可以使用 Sum( conditional statements ) 计算符合条件的行数。条件语句将为匹配行返回1，为不匹配行返回0。
请尝试以下查询：

SELECT pae.peID 
FROM PermissionAccessEntities AS pae 
LEFT JOIN PermissionAccessEntityGroups AS paeg 
  ON pae.peID = paeg.peID 
WHERE pae.petID = 4
GROUP BY pae.peID
HAVING COUNT(*) = SUM(paeg.gID IN (2,3))

db小提琴演示

现在，记住这一点 Count(NULL) = 0 ; 然而 COUNT(0) = COUNT(1) = 1 . 所以，另一种使用 Count() 功能仅为：

SELECT pae.peID 
FROM PermissionAccessEntities AS pae 
LEFT JOIN PermissionAccessEntityGroups AS paeg 
  ON pae.peID = paeg.peID 
WHERE pae.petID = 4
GROUP BY pae.peID
HAVING COUNT(*) = COUNT(CASE WHEN paeg.gID IN (2,3) THEN 1 END)

小提琴演示2