Question

我有这张桌子。

 id  |  item_id  | created_at 
-----+-----------+------------
  1  |   Apple   | 2017-03-21
  2  |   Grape   | 2017-03-23
  3  |   Grape   | 2017-03-24
  4  |   Apple   | 2017-03-25

我想通过 created_at 订购，同时也可以通过 item_id 订购：

 id  |  item_id  | created_at 
-----+-----------+------------
  4  |   Apple   | 2017-03-25
  1  |   Apple   | 2017-03-21
  3  |   Grape   | 2017-03-24
  2  |   Grape   | 2017-03-23

因此，如果我为 item_id添加新行：Grape ，我的新结果应如下所示：

 id  |  item_id  | created_at 
-----+-----------+------------
  5  |   Grape   | 2017-03-28  (NEW)
  3  |   Grape   | 2017-03-24
  2  |   Grape   | 2017-03-23
  4  |   Apple   | 2017-03-25
  1  |   Apple   | 2017-03-21

然后如果我为 item_id添加新行：Apple ，它应该是这样的：

 id  |  item_id  | created_at 
-----+-----------+------------
  6  |   Apple   | 2017-03-28  (NEW)
  4  |   Apple   | 2017-03-25
  1  |   Apple   | 2017-03-21
  5  |   Grape   | 2017-03-27
  3  |   Grape   | 2017-03-24
  2  |   Grape   | 2017-03-23

...所以它按最新的 created_at 排序，并在其下方显示其他行 item_id

我已经尝试了ORDER BY created_at, item_id DESC，但它不起作用而是给我这个：

 id  |  item_id  | created_at 
-----+-----------+------------
  6  |   Apple   | 2017-03-28
  5  |   Grape   | 2017-03-27
  4  |   Apple   | 2017-03-25
  3  |   Grape   | 2017-03-24
  2  |   Grape   | 2017-03-23
  1  |   Apple   | 2017-03-21

Answer 1

您需要首先确定您的第一条规则，即created_at每item_id个item_id，因此添加了row_number()并添加了SELECT tt.id, tt.item_id, tt.created_at FROM test_table tt INNER JOIN (SELECT item_id, MAX(created_at), ROW_NUMBER() OVER (ORDER BY MAX(created_at) DESC) rn FROM test_table GROUP BY item_id ORDER BY 3) ord ON tt.item_id = ord.item_id ORDER BY ord.rn, tt.created_at DESC;的子查询，以便它可以知道哪种水果首先出现。然后加入你的桌子。

googlenews <- WebCorpus(GoogleNewsSource("Microsoft"))
Unknown IO errorfailed to load external entity "http://news.google.com/news?hl=en&q=Microsoft&ie=utf-8&num=100&output=rss"
Error: 1: Unknown IO error2: failed to load external entity "http://news.google.com/news?hl=en&q=Microsoft&ie=utf-8&num=100&output=rss"

library(tm.plugin.webmining)
library(purrr)

company <- c("Microsoft", "Apple", "Google", "Amazon", "Facebook",
             "Twitter", "IBM", "Yahoo", "Netflix")
symbol <- c("MSFT", "AAPL", "GOOG", "AMZN", "FB", "TWTR", "IBM", "YHOO", "NFLX")

download_articles <- function(symbol) {
  WebCorpus(GoogleFinanceSource(paste0("NASDAQ:", symbol)))
}

stock_articles <- data_frame(company = company,
                             symbol = symbol) %>%
  mutate(corpus = map(symbol, download_articles))
failed to load HTTP resource
Error in mutate_impl(.data, dots) : 1: failed to load HTTP resource

Answer 2

SQLFiddle: PostgreSQL

WITH grouped AS (
  select   item_id, max(created_at) as max_dt
  from     tbl 
  group by item_id
)
SELECT     tbl.*
FROM       grouped
LEFT JOIN  tbl      USING (item_id)
ORDER BY   grouped.max_dt desc, 
           grouped.item_id,      -- important if apple and grape both have same max dt
           tbl.created_at desc;

子查询以按项目（项目分组）获取最新日期
将实际表上的结果连接起来以获得所需的记录

Answer 3

对于mysql，试试这个：选择物品。* 来自项目左连接（选择item_id，max（created_at）maxdate 来自项目按item_id分组）t on item.item_id = t.item_id 按t.maxdate desc命令，item.created_at desc SQLFiddle中的Demo1 SQLFiddle中的Demo2

Answer 4

请试试 - SELECT id， ITEM_ID， created_at 从tblOrders ORDER BY item_ID， item_ID =（SELECT item_ID 从tblOrders WHERE id =（SELECT MAX（id）FROM tblOrders））， created_at;

Answer 5

这是一个mysql版本

set @ordr = 0;
select t.*
from tbl t
join (
    select o.item_id, o.created_at, @ordr := @ordr + 1 as `order`
    from (
        select tbl.item_id, max(tbl.created_at) as created_at
        from tbl
        group by tbl.item_id
        order by created_at desc
    ) as o
) sub
on t.item_id = sub.item_id
order by sub.`order`, sub.created_at desc

按2列排序

5 个答案:

SQLFiddle: PostgreSQL