每个派生表都必须有自己的别名答案

【问题标题】：Every derived table must have its own alias每个派生表都必须有自己的别名
【发布时间】：2011-12-15 18:14:44
【问题描述】：

我有以下疑问：

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

在我添加最后一个 JOIN 之前它工作得很好。现在我得到错误：“每个派生表都必须有自己的别名”。我知道这是因为每个表都需要它的别名，我需要将“as S”或其他东西放在某个地方，但我无法在此查询中找到如何做到这一点。

【问题讨论】：

标签： mysql join random derived

【解决方案1】：

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

【讨论】：

是的，我成功了。真的真的很感谢！！顺便说一句，我还更改了 =>，因为这给出了另一个错误。

【解决方案2】：

正确的语法是：

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

【讨论】：