在SQL内部连接中使用别名

Question

我已经查看了一些使用连接来获取组中最高值的示例，但我尝试的方法不喜欢使用内部连接之外的别名。

SELECT f.year, f.name, f.date_start, f.date_end, f.max_year FROM ( SELECT EXTRACT(year FROM date_start) AS year, MAX(DATEDIFF(date_end,date_start)) AS max_year FROM mytable GROUP BY year ) AS x inner join mytable AS f on f.year = x.year and f.max_year = x.max_year;

所以，如果我有一个表：

Name date_start date_end John 1950-04-05 1960-07-08 Jack 1950-04-06 1960-12-31 Mark 1954-01-01 1970-01-01 Jane 1954-10-10 1978-10-01

然后，我希望它弹出有两个日期之间的最大差距的条目，每一年的起始日期为：

Year Name date_start date_end max_year 1950 Jack 1950-04-06 1960-12-31 3922 1954 Jane 1954-10-10 1978-10-01 8758

任何的如何解决建议 这个？

Answer 1

它看起来像你使用你的SELECT列表错误的别名，这应该是更好：

SELECT x.year, f.name, f.date_start, f.date_end, x.max_year FROM 
(
    SELECT 
     EXTRACT(year FROM date_start) AS year, 
     MAX(DATEDIFF(date_end,date_start)) AS max_year 
    FROM mytable GROUP BY year 
) AS x inner join 
mytable AS f on EXTRACT(year FROM f.date_start) = x.year 
AND DATEDIFF(f.date_end, f.date_start) = x.max_year; 

不过，我会做这样的：

SELECT name, date_start, date_end 
FROM mytable f 
WHERE NOT EXISTS (
    SELECT * FROM mytable 
    WHERE 
    EXTRACT(year FROM date_start) = EXTRACT(year FROM f.date_start) AND 
    DATEDIFF(date_end, date_start) > DATEDIFF(f.date_end, f.date_start) 
) 

Answer 2

你可以用这个试试：

SELECT * FROM 
(
    SELECT *, EXTRACT(year FROM date_start) AS year, DATEDIFF(date_end,date_start) AS diff 
    FROM mytable 
    ORDER BY diff DESC 
) sq 
GROUP BY year 
ORDER BY year ASC 

小提琴演示：http://sqlfiddle.com/#!2/4d0da/11