我有以下查询:
SELECT `snap`.`ID`, `user`.`username`, `vote`.`type`
FROM (`snap`) JOIN `user` as u ON `u`.`ID` = `snap`.`user`
LEFT JOIN (select * from vote where user = "18") as vote ON `snap`.`ID` = `vote`.`snap`
JOIN (SELECT CEIL(MAX(ID)*RAND()) AS ID FROM snap)) AS x ON `snap`.`ID` => `x`.`ID`
WHERE `snap`.`active` = 0 LIMIT 1
它工作得很好，直到我添加了最后一个 JOIN。现在我收到错误:“每个派生表必须有自己的别名”。我知道这是因为每个表都需要它的别名，并且我需要将“as S”或其他内容放在某处，但我不知道如何在此查询中执行此操作。
最佳答案
好像你在snap之后有额外的右括号。它应该是 1 而不是 2 个右括号。
SELECT `snap`.`ID`, `user`.`username`, `vote`.`type`
FROM (`snap`) JOIN `user` as u ON `u`.`ID` = `snap`.`user`
LEFT JOIN (select * from vote where user = "18") as vote ON `snap`.`ID` = `vote`.`snap`
JOIN (SELECT CEIL(MAX(ID)*RAND()) AS ID FROM snap) AS x ON `snap`.`ID` = `x`.`ID`
WHERE `snap`.`active` = 0 LIMIT 1
关于mysql - 每个派生表必须有自己的别名，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/8524703/