Question

我对SQL很陌生，我需要你的帮助。

即使条件（001）在某些月份不存在，我希望在SELECT - YM范围内的201801列（年）中201804行pod11。

表YM code hour ------------------------------------ 201712 005 32 201712 002 16 201712 003 24 201712 007 112 201801 001 112 201801 003 12 201801 005 24 201801 007 64 201802 001 64 201802 002 128 201803 005 32 201803 002 16 201804 003 24 201804 007 112 201804 008 86：

SELECT YM,code,hour
FROM pod11
WHERE  YM >= '201801' AND YM <= '201804' and code = '001'
GROUP by YM,code,hour
ORDER BY YM

我用过这个：

YM  code    hour   
------------------------------------
201801  001 112
201802  001 64

我明白了：

YM  code    hour   
------------------------------------
201801  001 112
201802  001 64
201803  0   0
201804  0   0

有关如何编写查询以实现此目的的任何建议吗？

scrapy crawl swspider

Answer 1

我想知道这是否可以在没有子查询的情况下完成。我想它可以是：

select top (1) with ties YM,
       (case when code = '001' then hour else 0 end) as code,
       (case when code = '001' then code else 0 end) as hour
from pod11
where YM between '201801' and '201804' 
order by rank() over (partition by ym
                      order by (case when code = '001' then 1 else 2 end),
                               (case when code <> '001' then hour end)
                     )

Answer 2

一种选择是使用联合来引入没有匹配code记录的月份的记录。在我的输出中，有201712的数据，因为您没有指定它不应出现的原因，并且它没有匹配的代码数据。如果您不想在那里使用该记录，那么在WHERE子句中删除它就很容易了。

SELECT YM, code, hour
FROM pod11
WHERE YM BETWEEN '201801' AND '201804' AND code = '001'
GROUP BY YM, code, hour
UNION ALL
SELECT DISTINCT t1.YM, 0, 0
FROM pod11 t1
WHERE NOT EXISTS (SELECT 1 FROM pod11 t2 WHERE t1.YM = t2.YM AND t2.code = '001')
ORDER BY YM;

Demo

Answer 3

<强> MSSQL：

SELECT 
    tableYM.YM, 
    ISNULL(pod11.code, 0) AS code, 
    ISNULL(SUM(pod11.hour), 0) AS hour 
FROM (
    SELECT DISTINCT YM FROM pod11 WHERE YM >= '201801' AND YM <= '201804'
) tableYM LEFT JOIN pod11 ON tableYM.YM = pod11.YM AND pod11.code = '001'
GROUP BY pod11.code, tableYM.YM
ORDER BY tableYM.YM

演示： http://sqlfiddle.com/#!18/f068b/3/0

<强> MySQL的：

SELECT 
    tableYM.YM, 
    IFNULL(pod11.code, 0) AS code, 
    IFNULL(SUM(pod11.hour), 0) AS hour 
FROM (
    SELECT DISTINCT YM FROM pod11 WHERE YM >= '201801' AND YM <= '201804'
) tableYM LEFT JOIN pod11 ON tableYM.YM = pod11.YM AND pod11.code = '001'
GROUP BY pod11.code, tableYM.YM
ORDER BY tableYM.YM

演示： http://sqlfiddle.com/#!9/f068b/18/0

Answer 4

你可以自己加入你的桌子：

[pythondemo@panel python3.6]$ sudo yum install sqlite3 libsqlite3-dev
Loaded plugins: fastestmirror, universal-hooks
Setting up Install Process
Loading mirror speeds from cached hostfile
 * EA4: 69.72.212.11
 * cpanel-addons-production-feed: 69.72.212.11
 * base: mirror.jaleco.com
 * epel: mirror.texas3006.com
 * extras: mirror.jaleco.com
 * ius: ius.mirror.constant.com
 * updates: mirror.jaleco.com
No package sqlite3 available.
No package libsqlite3-dev available.

Answer 5

SELECT YM, COALESCE(code,0) code, COALESCE(hour,0) hour
(
    SELECT YM,code,hour
    FROM pod11
    WHERE  YM >= '201801' AND YM <= '201804' and code = '001'
    GROUP by YM,code,hour

    RIGHT JOIN

    (SELECT 201801 as YM
    UNION
    SELECT 201802
    UNION
    SELECT 201803
    UNION
    SELECT 201804
    ) as t
    ON pod11. YM = t.YM
)
    ORDER BY YM

您可以将年份存储在不同的表格中并从中查询，或者如果您知道pod11表格中的所有年份，即201801到201804，那么您可以从该表格查询范围和不同

即使条件不存在，如何显示行

5 个答案:

Demo