我在mysql中有此表称为safespot
+---------+---------------+
| term_id | userid | safe |
+---------+--------|------+
| 1 | 1 | large number, unix timestamp here
| 1 | 2 | large number, unix timestamp here
| 1 | 3 | large number, unix timestamp here
| 1 | 4 | large number, unix timestamp here
+---------+--------+
这是表
users：+----+-------------+-------------+
| id | userid | cash |
+----+-------------+-------------+
| 1 | 1 | 100000 |
| 2 | 2 | 100000 |
| 3 | 3 | 100000 |
+----+-------------+-------------+
我该怎么做
SELECT * FROM `users` where `userid`=1 and `cash`>= 1000 and " userid do not exist in table safespot" or "if the user exists in the safestop table, check if the current timestamp is higher than the safe colum)
因此，基本上可以执行一个查询，如果用户ID在Safespot表中不存在，或者如果确实存在，则该时间戳大于safe_value，该查询也将返回该查询。
最佳答案
SELECT * FROM users u
LEFT JOIN safespot s ON s.userid = u.userid
WHERE
u.userid = 1
AND u.cash = 1000
AND (s.userid IS NULL OR s.safe > UNIX_TIMESTAMP())
这将使用户返回
给定用户名的
safespot中没有条目，或者safespot中有一个条目，其值safe大于当前时间戳。关于php - mysql-query其他表条件PHP，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/32441960/