Php 合并或相交多个MySQL命令

我得搜索两张桌子

我的第二张桌子是：

+---------+---------------+-------+
| user_id | is_first_time | score |
+---------+---------------+-------+
|   44    |       1       |   20  |
+---------+---------------+-------+
|   44    |       0       |   20  |
+---------+---------------+-------+

我的第一个表只包含用户信息

如果我使用以下MySQL命令，我将得到一个结果：

SELECT * , users_score.user_id users_score_user_id 
FROM users JOIN users_score 
ON users_score.user_id = users.id 
WHERE (users_score.score='20' AND users_score.is_first_time='1')

如果我使用以下MySQL命令，我也会得到一个结果，我只需将is_first_time值更改为0：

SELECT * , users_score.user_id users_score_user_id 
FROM users JOIN users_score 
ON users_score.user_id = users.id 
WHERE (users_score.score='20' AND users_score.is_first_time='0')

但如果我将这两个命令与AND结合使用，则不会得到任何结果：

SELECT * , users_score.user_id users_score_user_id 
FROM users JOIN users_score 
ON users_score.user_id = users.id 
WHERE (users_score.score='20' AND users_score.is_first_time='1')
AND (users_score.score='20' AND users_score.is_first_time='0')

如何合并或相交两个命令

谢谢您的帮助。

您需要在此处使用或调节

WHERE (users_score.score='20' AND users_score.is_first_time='1') 
  OR  (users_score.score='20' AND users_score.is_first_time='0')

但在条件允许的情况下，它可以以更好的方式编写

SELECT *,
       users_score.user_id users_score_user_id
FROM   users
       JOIN users_score
         ON users_score.user_id = users.id
WHERE  users_score.score = '20'
       AND users_score.is_first_time IN ( '1', '0' ) 

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如果希望用户同时满足这两个条件，则有一种方法在中使用或存在：

与的问题是，单行不能同时为0和1

如果您只需要用户id，我建议group by并拥有：

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注意，我删除了单引号。如果列是数字，则不要使用单引号进行比较。当然，如果它们是字符串，请使用单引号。

谢谢您的回复。或者将获得is_first_time=1的行、is_first_time=0的行以及is_first_time=1和0的行。因此，它将获得三个用户id，而不是一个用户id。我认为戈登·林诺夫的答案更正确。无论如何，谢谢你。

SELECT u.* 
FROM users u
WHERE u.id IN (SELECT us.user_id FROM users_score us WHERE us.score = '20' AND us.is_first_time = '0'
              ) AND
      u.id IN (SELECT us.user_id FROM users_score us WHERE us.score = '20' AND us.is_first_time = '1'
              );

select us.user_id
from users_score us
where score = 20 and us.is_first_time in (0, 1)
group by us.user_id
having count(distinct us.is_first_time) = 2;