问题描述
我需要在 WHERE 子句中使用别名，但它一直告诉我它是一个未知列.有没有办法解决这个问题?我需要选择评分高于 x 的记录.评分计算为以下别名:
I need to use an alias in the WHERE clause, but It keeps telling me that its an unknown column. Is there any way to get around this issue? I need to select records that have a rating higher than x. Rating is calculated as the following alias:
sum(reviews.rev_rating)/count(reviews.rev_id) as avg_rating
推荐答案
您可以使用 HAVING 子句，可以看到别名，例如
You could use a HAVING clause, which can see the aliases, e.g.
HAVING avg_rating>5
但是在 where 子句中，您需要重复您的表达式，例如
but in a where clause you'll need to repeat your expression, e.g.
WHERE (sum(reviews.rev_rating)/count(reviews.rev_id))>5
但是！并非所有表达式都被允许 - 使用 SUM 之类的聚合函数将不起作用，在这种情况下，您需要使用 HAVING 子句.
BUT! Not all expressions will be allowed - using an aggregating function like SUM will not work, in which case you'll need to use a HAVING clause.
来自 MySQL 手册:
不允许引用WHERE 子句中的列别名，因为列值可能还没有当 WHERE 子句被确定被执行.参见 B.1.5.4 节，列别名问题".
It is not allowable to refer to a column alias in a WHERE clause, because the column value might not yet be determined when the WHERE clause is executed. See Section B.1.5.4, "Problems with Column Aliases".
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