Retrieving the last record in each group - MySQL

2018-12-30 22:26发布

There is a table messages that contains data as shown below:

Id   Name   Other_Columns
-------------------------
1    A       A_data_1
2    A       A_data_2
3    A       A_data_3
4    B       B_data_1
5    B       B_data_2
6    C       C_data_1

If I run a query select * from messages group by name, I will get the result as:

1    A       A_data_1
4    B       B_data_1
6    C       C_data_1

What query will return the following result?

3    A       A_data_3
5    B       B_data_2
6    C       C_data_1

That is, the last record in each group should be returned.

At present, this is the query that I use:

SELECT
  *
FROM (SELECT
  *
FROM messages
ORDER BY id DESC) AS x
GROUP BY name

But this looks highly inefficient. Any other ways to achieve the same result?

21条回答
素衣白纱
2楼-- · 2018-12-30 22:33

Here is my solution:

SELECT 
  DISTINCT NAME,
  MAX(MESSAGES) OVER(PARTITION BY NAME) MESSAGES 
FROM MESSAGE;
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零度萤火
3楼-- · 2018-12-30 22:34

Use your subquery to return the correct grouping, because you're halfway there.

Try this:

select
    a.*
from
    messages a
    inner join 
        (select name, max(id) as maxid from messages group by name) as b on
        a.id = b.maxid

If it's not id you want the max of:

select
    a.*
from
    messages a
    inner join 
        (select name, max(other_col) as other_col 
         from messages group by name) as b on
        a.name = b.name
        and a.other_col = b.other_col

This way, you avoid correlated subqueries and/or ordering in your subqueries, which tend to be very slow/inefficient.

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十年一品温如言
4楼-- · 2018-12-30 22:34
SELECT 
  column1,
  column2 
FROM
  table_name 
WHERE id IN 
  (SELECT 
    MAX(id) 
  FROM
    table_name 
  GROUP BY column1) 
ORDER BY column1 ;
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心情的温度
5楼-- · 2018-12-30 22:38

Here are two suggestions. First, if mysql supports ROW_NUMBER(), it's very simple:

WITH Ranked AS (
  SELECT Id, Name, OtherColumns,
    ROW_NUMBER() OVER (
      PARTITION BY Name
      ORDER BY Id DESC
    ) AS rk
  FROM messages
)
  SELECT Id, Name, OtherColumns
  FROM messages
  WHERE rk = 1;

I'm assuming by "last" you mean last in Id order. If not, change the ORDER BY clause of the ROW_NUMBER() window accordingly. If ROW_NUMBER() isn't available, this is another solution:

Second, if it doesn't, this is often a good way to proceed:

SELECT
  Id, Name, OtherColumns
FROM messages
WHERE NOT EXISTS (
  SELECT * FROM messages as M2
  WHERE M2.Name = messages.Name
  AND M2.Id > messages.Id
)

In other words, select messages where there is no later-Id message with the same Name.

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弹指情弦暗扣
6楼-- · 2018-12-30 22:38

How about this:

SELECT DISTINCT ON (name) *
FROM messages
ORDER BY name, id DESC;

I had similar issue (on postgresql tough) and on a 1M records table. This solution takes 1.7s vs 44s produced by the one with LEFT JOIN. In my case I had to filter the corrispondant of your name field against NULL values, resulting in even better performances by 0.2 secs

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只靠听说
7楼-- · 2018-12-30 22:39

Hi @Vijay Dev if your table messages contains Id which is auto increment primary key then to fetch the latest record basis on the primary key your query should read as below:

SELECT m1.* FROM messages m1 INNER JOIN (SELECT max(Id) as lastmsgId FROM messages GROUP BY Name) m2 ON m1.Id=m2.lastmsgId
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