SQL从"自定义"post_type中获取X最后一个条目,计算用户个人自定义post_type的数量

Question

我想,如果可能的话,可以进入一个查询:

• 最后4个不同的用户
• 排除 'ID' = '1'
• WITH post_type = 'custom'
• 由dateor或ID(DESC)订购
• 计算每个用户的"自定义"post_type总数(COUNT())

这是一个数据示例:

Table Name: 'post'

   ID   |  user |           Date            |    title      |     status    |    post_type
        |       |                           |               |               |   
"2785"  |   "1" |   "2016-05-24 18:49:15"   |   "Title"     |   "published" |   "page_post"
"2783"  |   "5" |   "2016-05-24 11:24:08"   |   "Title"     |   "published" |   "custom"
"2781"  |   "1" |   "2016-05-18 20:40:11"   |   "Title"     |   "published" |   "custom"
"2759"  |   "3" |   "2016-05-07 14:00:22"   |   "Title"     |   "published" |   "custom"
"2757"  |  "12" |   "2016-05-02 12:41:00"   |   "Title"     |   "published" |   "custom"
"2756"  |   "1" |   "2016-04-30 22:47:07"   |   "Title"     |   "published" |   "custom"
"2755"  |   "5" |   "2016-04-29 13:54:21"   |   "Title"     |   "published" |   "blog_post"
"2754"  |   "1" |   "2016-04-29 11:33:36"   |   "Title"     |   "published" |   "page_post"
"2738"  |   "3" |   "2016-05-06 12:45:58"   |   "Title"     |   "published" |   "custom"
"2736"  |  "12" |   "2016-04-24 17:17:04"   |   "Title"     |   "published" |   "custom"
"2683"  |  "15" |   "2016-04-22 20:27:45"   |   "Title"     |   "published" |   "custom"
"2681"  |  "18" |   "2016-04-21 00:20:55"   |   "Title"     |   "published" |   "custom"
"2671"  |   "1" |   "2016-04-11 18:38:57"   |   "Title"     |   "published" |   "other_post"
"2652"  |   "4" |   "2016-04-02 17:43:41"   |   "Title"     |   "published" |   "custom"
"2651"  |   "5" |   "2016-03-28 17:12:00"   |   "Title"     |   "published" |   "custom"
"2639"  |  "18" |   "2016-03-22 14:58:00"   |   "Title"     |   "published" |   "custom"
"2630"  |  "19" |   "2016-03-21 15:27:00"   |   "Title"     |   "published" |   "custom"
"2617"  |  "14" |   "2016-03-17 12:22:06"   |   "Title"     |   "published" |   "custom"
"2616"  |   "5" |   "2016-03-16 15:23:00"   |   "Title"     |   "published" |   "page_post"
"2598"  |   "4" |   "2016-03-14 15:27:29"   |   "Title"     |   "published" |   "custom"
"2596"  |   "2" |   "2016-03-10 17:43:00"   |   "Title"     |   "published" |   "custom"
"2571"  |   "1" |   "2016-03-09 14:19:31"   |   "Title"     |   "published" |   "blog_post"
"2250"  |  "19" |   "2016-02-29 12:15:48"   |   "Title"     |   "published" |   "custom"
"2249"  |  "15" |   "2016-02-29 09:45:35"   |   "Title"     |   "published" |   "custom"
"2215"  |  "13" |   "2016-02-22 18:21:54"   |   "Title"     |   "published" |   "custom"
"2201"  |   "3" |   "2016-02-15 17:40:00"   |   "Title"     |   "published" |   "custom"
"1914"  |   "2" |   "2015-11-13 12:08:00"   |   "Title"     |   "published" |   "other_post"

我的不完整查询:

SELECT *
FROM 'posts' 
WHERE 'user' != 1 AND 'post_type' = 'custom_type' 
GROUP BY 'user' 
ORDER BY 'ID' DESC LIMIT 4

订购方式ID与订购方式类似date.没有GROUP BY 'user'这个工作,但问题是我想避免在这个选择2'自定义'的帖子为一个用户:我需要选择4个不同的用户.所以我的问题是GROUP BY.

我该如何解决这个问题？

最后一件事:
是否有可能COUNT()计算此选择中每个用户的总"自定义"帖子,并在新列中返回值？

Answer 1

试试这个;)

查询1:

select t1.*, t2.userCnt
from `posts` t1
inner join (
    select max(`ID`) as `ID`, `user`, count(1) as userCnt
    from `posts`
    where `user` != '1'
    and `post_type` = 'custom'
    group by `user`
) t2 on t1.`ID` = t2.`ID` and t1.`user` = t2.`user`
order by t1.`ID` desc limit 4

检查这个SqlFiddle 结果:

|   ID | user |                Date | title |    status | post_type | userCnt |
|------|------|---------------------|-------|-----------|-----------|---------|
| 2783 |    5 | 2016-05-24 11:24:08 | Title | published |    custom |       2 |
| 2759 |    3 | 2016-05-07 14:00:22 | Title | published |    custom |       3 |
| 2757 |   12 | 2016-05-02 12:41:00 | Title | published |    custom |       2 |
| 2683 |   15 | 2016-04-22 20:27:45 | Title | published |    custom |       2 |

子查询t2将ID在每个用户获得最大值时user != '1' and post_type = 'custom',然后inner joint1和t2 t1.ID = t2.ID and t1.user = t2.user将得到我们记录,其中ID每个user表都有最大值post.如:"2783","2759","2757","2683","2681","2652","2630","2617","2596","2215".

而在去年同order by和limit,当然你也可以得到"2783","2759","2757","2683".希望我没有弄错你的问题.