我正在使用 Postgresql 8.2 进行快速查询,并且我之前已经进行过一千次这样的查询,但我无法弄清楚为什么会出现此错误。我可能遗漏了一些明显的东西,但它说我的“FROM 中的子查询必须有一个别名”。我的子查询“inner”确实有一个别名,但我不知道为什么会出现错误。
SELECT "Branch", "Zip_5", "CountofStops", avg("EarlyTime") As
"Average_Arrival_Time"
FROM
(SELECT branch_id as "Branch", substring(stop_zip_postal_code, 1, 5) as
"Zip_5", count(stop_name) as "CountofStops", min(actual_arrival_time) as
"EarlyTime"
FROM distribution_stop_information
WHERE company_no = '001' AND route_date > '3/13/2017'
GROUP BY branch_id, stop_zip_postal_code)
inner
GROUP BY "Branch", "Zip_5"
ORDER BY Zip_5
********** Error **********
ERROR: subquery in FROM must have an alias
SQL state: 42601
Hint: For example, FROM (SELECT ...) [AS] foo.
最佳答案
inner
是保留关键字。使用其他名称作为别名。
关于sql - PostgreSQL "subquery in FROM must have an alias"错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43987783/