用于1值的SQL查询匹配2行

Question

我该如何编写一个查询来查找哪一个列中的每个值为1的2个特定值匹配的位置？我正在使用MySQL数据库。

例如：

| test_id | animal_type | classification_id | 
---------------------------------------------- 
    1   cat    456 
    2   dog    456 
    3   mongoose  456 
    4   cat    123 

我想编写发现2行碰巧具有“123”和“456”一个classification_id对于相同animal_type查询哪里，表中出现。

如果我写下以下它不起作用，因为我不知道如何在此查询中包含animal_type。

SELECT * 
FROM test_table 
WHERE classification_id = '123' AND classification_id = '456' 

我该如何编写一个查询来选择同时具有123和456作为特定animal_type的classification_id的所有行？

Answer 1

你可以使用：

SELECT animal_type 
FROM test_table 
WHERE classification_id IN (123, 456) 
GROUP BY animal_type 
HAVING COUNT(DISTINCT classification_id) = 2; 

如果你需要整行：

SELECT * 
FROM test_table 
WHERE animal_type IN (SELECT animal_type 
         FROM test_table 
         WHERE classification_id IN (123, 456) 
         GROUP BY animal_type 
         HAVING COUNT(DISTINCT classification_id) = 2) 

Answer 2

SELECT count(test_id) as animal_count,test_table.* 
FROM test_table 
WHERE classification_id = '123' AND classification_id = '456' 
GROUP BY animal_type HAVING animal_count > 2 

使用GROUP BY和HAVING。