为什么这个简单的连接不起作用？答案

【问题标题】：Why won't this simple join work?为什么这个简单的连接不起作用？
【发布时间】：2015-04-17 16:05:05
【问题描述】：

我正在尝试让内部连接工作，它从两个表中获取区号并将城市名称放在适当的 series_id 旁边

 CREATE VIEW medical As
 SELECT series_id AS City, FORMAT(AVG(value),2) AS Average_CPI,     SUBSTRING(series_id,5,4) as areacode, cuArea.city_name, cuArea.area_code
FROM CURRENT
WHERE 
(
(
SUBSTRING(series_id,5,4) = 'A311'
AND SUBSTRING(series_id,9,8) = 'SAM'
AND period = 'M13'
)
OR
(
SUBSTRING(series_id,5,4) = 'A316'
AND SUBSTRING(series_id,9,8) = 'SAM'
AND period = 'M13'
)
)
GROUP BY series_id
INNER JOIN cuArea
ON cuArea.area_code = areacode

【问题讨论】：

你的问题我不清楚。请你能解释一下究竟是什么错误？

标签： mysql join inner-join

【解决方案1】：

您的 JOIN 位置错误。它应该在 FROM 之后。

CREATE VIEW medical As
 SELECT series_id AS City, FORMAT(AVG(value),2) AS Average_CPI,     SUBSTRING(series_id,5,4) as areacode, cuArea.city_name, cuArea.area_code
FROM CURRENT
INNER JOIN cuArea
ON cuArea.area_code = current.areacode
WHERE 
(
(
SUBSTRING(series_id,5,4) = 'A311'
AND SUBSTRING(series_id,9,8) = 'SAM'
AND period = 'M13'
)
OR
(
SUBSTRING(series_id,5,4) = 'A316'
AND SUBSTRING(series_id,9,8) = 'SAM'
AND period = 'M13'
)
)
GROUP BY series_id

【讨论】：

谢谢@JoeStefanelli