Question

一个相当复杂的SQL查询，我可能会变得更加困难：我有两张桌子：

新闻： newsid，datetime，newstext

照片： pictureid，datetime，imgPath

这两者没有关系，我只是在

上创建新闻/图片的日期加入到目前为止

SQL：

SELECT * FROM news as n LEFT OUTER JOIN (SELECT count(pictureid), datetime 
FROM picture GROUP BY DATE(datetime)) as p ON DATE(n.datetime) = DATE(p.datetime) 
UNION 
SELECT * FROM news as n RIGHT OUTER JOIN (SELECT count(pictureid), 
datetime FROM picture GROUP BY DATE(datetime)) as p ON 
DATE(n.datetime) = DATE(p.datetime)

我必须使用union来模拟MySQL中的完整外连接。结果：

newsid     text     datetime  count()   datetime 
1       sometext   2011-01-16   1       2011-01-16 
2         moo2    2011-01-19  NULL        NULL 
3        mooo3    2011-01-19  NULL        NULL 
NULL      NULL      NULL       4         2011-01-14

问题在于我显然最终得到了两个日期列 - 一个来自新闻，另一个来自图片，这意味着我无法按日期排序并按正确顺序排列！有任何想法吗？即使这意味着重组数据库！我需要将日期放在一个列中。

答案来自SeRPRo 完成的工作代码是：

SELECT `newsid`, `text`,
    CASE 
    WHEN `datetime` IS NULL 
    THEN `pdate` 
    ELSE `datetime` 
    END 
    as `datetime`, 

`pcount` FROM 
(
    (SELECT * FROM news as n LEFT OUTER JOIN 
        (SELECT count(pictureid) as pcount, datetime as pdate FROM picture GROUP BY DATE(datetime)) as p 
        ON DATE(n.datetime) = DATE(p.pdate) ORDER BY datetime
    )
    UNION
    (SELECT * FROM news as n RIGHT OUTER JOIN 
        (SELECT count(pictureid) as pcount, datetime as pdate FROM picture GROUP BY DATE(datetime)) as p 
        ON DATE(n.datetime) = DATE(p.pdate) ORDER BY datetime
    ) 

) as x
ORDER BY datetime

Answer 1

只使用你的数据库结构和你的查询，因为在MySQL中没有FULL OUTER JOIN，我认为解决方案可能就是这样：

SELECT
    `newsid`,
    `text`,
    CASE
        WHEN `datetime` IS NULL THEN `pdate`
        ELSE `datetime`
    END as `datetime,
    `pcount`
(
SELECT *
FROM `news` as `n`
    LEFT OUTER JOIN (
                        SELECT count(pictureid) as `pcount`, datetime as `pdate`
                        FROM picture GROUP BY DATE(datetime)
                    ) as p ON DATE(n.datetime) = DATE(p.datetime) 
    UNION
SELECT * 
    FROM `news` as `n`
    RIGHT OUTER JOIN (
                        SELECT count(pictureid) as `pcount`, datetime as `pdate`
                        FROM picture GROUP BY DATE(datetime)
                    ) as p ON DATE(n.datetime) = DATE(p.datetime) 

)

Answer 2

我认为你的代码最终会解决这个问题，你能否详细解释日期时间排序部分

SELECT *  
FROM   News FULL OUTER JOIN Picture 
ON News.Datetime = Picture.Datetime ORDER BY DateTime Asc

Answer 3

您始终可以使用其他查询包装查询：

SELECT ..., IF(datetime1 IS NULL, datetime2, datetime1) as datetime, ... FROM
(
   ... your query ...
)
ORDER BY datetime

MySql完全连接（联合）和多个日期列的排序

3 个答案: