Retrieving the last record in each group - MySQL

Question

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?

Answer 1

In my quest for a universal groupwise-max, I've seen many answers and blog posts on the subject. Even my favorite (actually part of a fantastic series on the subject) failed to identify a portable solution, instead diving deep into specifics per RDMBS.

Luckily, a portable solution does exist!

The secondary index you need for this is name. (name, id would be identical, as the primary key is always included implicitly.)

Create groups of the messages, and use a dependent subquery to get the latest row for each group.

SELECT m.*

-- Step 1: Start by obtaining the groupwise maximums
FROM
(
    SELECT (
        -- Step 1b: Find the ID of the group maximum by seeking in the index
        SELECT id
        FROM messages m
        WHERE m.name = groups.name
        ORDER BY m.name DESC, m.id DESC -- Match the index EXACTLY, and indicate direction
        LIMIT 1
    ) AS id

    -- Step 1a: Find the groups by seeking through the index
    FROM messages AS groups
    GROUP BY groups.name
) AS maxes

-- Step 2: For each group, join the max row by ID
-- This neatly separates any potential followup SQL from the groupwise-max tactics
INNER JOIN messages m ON m.id = maxes.id
;

This is portable because it requires only the following combination of building blocks:

• Indexed GROUP BY.
• Indexed SELECT with ORDER BY [ASC/DESC] and LIMIT/TOP.
• Dependent subqueries.

Just be sure to have the correct index: GroupKeyColumn(s), GroupWinnerColumn(s), PrimaryKeyColumn(s).

In OP's case, the group key is name, the group's winner is determined by id, and the primary key is already covered by that, so: name, id.

Many have suggested solutions involving subqueries, but the most overlooked aspect is the highly specific set of ordering clauses that causes the correct index to be used - in the right traversal direction, no less.

Additional Advantages

• Easily adjustable for min (ASC) vs. max (DESC).
• Winner per group can be composite, e.g. timestamp, id. (This also allows us to disambiguate non-unique winners, such as "latest timestamp".)
• Group key can be composite, e.g. company_id, department_name.
• Easily extended with a WHERE on which groups to select.
• Easily extended with a WHERE on what items to ignore, both indexed (id >= 1000) and non-indexed (is_deleted = 0).

Why does this work [optimally]?

Imagine leafing through the physical phone book, finding the last entry for each town, i.e. the entry with that town's alphabetically greatest name. How would you do it?

You would start at the very end. The very last entry in the book is the group maximum of the last town. It is the first result row that you encounter.

For each subsequent desired result row, you would binary search backwards, to the next-greatest town. At the point where the current town transitions into its predecessor, there is the predecessor's last row (alphabetically greatest name), i.e. your next result row. Repeat until no more towns.

Loosely speaking, the phone book is like a secondary index on { Town, Name, PhoneNumber }, with PhoneNumber serving as the primary key. (I'm simplifying things for agument's sake, pretending phone numbers are assigned to one person and names form a single column.)

You are effectively doing a reverse seek through the index. By repeatedly jumping to the next town efficiently (thanks to binary search or a B-tree structure), the work is constrained by the number of result rows rather than the total number of rows. This is asymptotically optimal. And thanks to the reverse traversal direction, each town you encounter "starts" with its greatest row, your target. That is important: imagine the absurd amount of needless work if you'd have to scan all rows for a town.

Changing the solution to a groupwise-min is as trivial as changing the traversal direction, i.e. from DESC to ASC.

RDBMS Notes

• Whereas MySQL 8 correctly shows Using index for this, MySQL 5.7 shows a worrisome Using where; Using index, but it actually performs correctly. (Tested on a huge data set involving very large groups. Tens of thousands of results spread through hundreds of millions of records were obtained in ~3 seconds.)
• For SQL Server, the syntax is SELECT TOP 1 instead of SELECT ... LIMIT 1.

Answer 2

We will look at how you can use MySQL at getting the last record in a Group By of records. For example if you have this result set of posts.

id	category_id	post_title
1	1	Title 1
2	1	Title 2
3	1	Title 3
4	2	Title 4
5	2	Title 5
6	3	Title 6

I want to be able to get the last post in each category which are Title 3, Title 5 and Title 6. To get the posts by the category you will use the MySQL Group By keyboard.

select * from posts group by category_id

But the results we get back from this query is.

id	category_id	post_title
1	1	Title 1
4	2	Title 4
6	3	Title 6

The group by will always return the first record in the group on the result set.

SELECT id, category_id, post_title
FROM posts
WHERE id IN (
    SELECT MAX(id)
    FROM posts
    GROUP BY category_id );

This will return the posts with the highest IDs in each group.

id	category_id	post_title
3	1	Title 3
5	2	Title 5
6	3	Title 6

Reference Click Here

Answer 3

I had a similar issue

Subquery and join to the rescue

SELECT p."Date"
        ,p."Symbol"
        ,p."ratio_roll_qtr_ret"
    FROM PUBLIC."prices_vw" AS p
    JOIN (
        SELECT "Symbol"
            ,max("Date")
        FROM PUBLIC."prices_vw"
        GROUP BY "Symbol"
        ) AS sq ON p."Date" = sq."max"
        AND p."Symbol" = sq."Symbol"
    WHERE p."ratio_roll_qtr_ret" IS NOT NULL
    ORDER BY "ratio_roll_qtr_ret" DESC;

Answer 4

Yet another option without subqueries.

This solution uses MySQL LAST_VALUE window function, exploiting Window Function Frame available MySQL tool from .

SELECT DISTINCT 
    LAST_VALUE(Id)            
        OVER(PARTITION BY Name 
             ORDER     BY Id 
             ROWS BETWEEN 0 PRECEDING 
                      AND UNBOUNDED FOLLOWING),
    Name,
    LAST_VALUE(Other_Columns)            
        OVER(PARTITION BY Name 
             ORDER     BY Id 
             ROWS BETWEEN 0 PRECEDING 
                      AND UNBOUNDED FOLLOWING)
FROM   
    tab

Try it here.

Answer 5

MySQL 8.0 now supports windowing functions, like almost all popular SQL implementations. With this standard syntax, we can write greatest-n-per-group queries:

WITH ranked_messages AS (
  SELECT m.*, ROW_NUMBER() OVER (PARTITION BY name ORDER BY id DESC) AS rn
  FROM messages AS m
)
SELECT * FROM ranked_messages WHERE rn = 1;

This and other approaches to finding groupwise maximal rows are illustrated in the MySQL manual.

Below is the original answer I wrote for this question in 2009:

---

I write the solution this way:

SELECT m1.*
FROM messages m1 LEFT JOIN messages m2
 ON (m1.name = m2.name AND m1.id < m2.id)
WHERE m2.id IS NULL;

Regarding performance, one solution or the other can be better, depending on the nature of your data. So you should test both queries and use the one that is better at performance given your database.

For example, I have a copy of the StackOverflow August data dump. I'll use that for benchmarking. There are 1,114,357 rows in the Posts table. This is running on MySQL 5.0.75 on my Macbook Pro 2.40GHz.

I'll write a query to find the most recent post for a given user ID (mine).

First using the technique shown by @Eric with the GROUP BY in a subquery:

SELECT p1.postid
FROM Posts p1
INNER JOIN (SELECT pi.owneruserid, MAX(pi.postid) AS maxpostid
            FROM Posts pi GROUP BY pi.owneruserid) p2
  ON (p1.postid = p2.maxpostid)
WHERE p1.owneruserid = 20860;

1 row in set (1 min 17.89 sec)

Even the EXPLAIN analysis takes over 16 seconds:

+----+-------------+------------+--------+----------------------------+-------------+---------+--------------+---------+-------------+
| id | select_type | table      | type   | possible_keys              | key         | key_len | ref          | rows    | Extra       |
+----+-------------+------------+--------+----------------------------+-------------+---------+--------------+---------+-------------+
|  1 | PRIMARY     | <derived2> | ALL    | NULL                       | NULL        | NULL    | NULL         |   76756 |             | 
|  1 | PRIMARY     | p1         | eq_ref | PRIMARY,PostId,OwnerUserId | PRIMARY     | 8       | p2.maxpostid |       1 | Using where | 
|  2 | DERIVED     | pi         | index  | NULL                       | OwnerUserId | 8       | NULL         | 1151268 | Using index | 
+----+-------------+------------+--------+----------------------------+-------------+---------+--------------+---------+-------------+
3 rows in set (16.09 sec)

Now produce the same query result using my technique with LEFT JOIN:

SELECT p1.postid
FROM Posts p1 LEFT JOIN posts p2
  ON (p1.owneruserid = p2.owneruserid AND p1.postid < p2.postid)
WHERE p2.postid IS NULL AND p1.owneruserid = 20860;

1 row in set (0.28 sec)

The EXPLAIN analysis shows that both tables are able to use their indexes:

+----+-------------+-------+------+----------------------------+-------------+---------+-------+------+--------------------------------------+
| id | select_type | table | type | possible_keys              | key         | key_len | ref   | rows | Extra                                |
+----+-------------+-------+------+----------------------------+-------------+---------+-------+------+--------------------------------------+
|  1 | SIMPLE      | p1    | ref  | OwnerUserId                | OwnerUserId | 8       | const | 1384 | Using index                          | 
|  1 | SIMPLE      | p2    | ref  | PRIMARY,PostId,OwnerUserId | OwnerUserId | 8       | const | 1384 | Using where; Using index; Not exists | 
+----+-------------+-------+------+----------------------------+-------------+---------+-------+------+--------------------------------------+
2 rows in set (0.00 sec)

---

Here's the DDL for my Posts table:

CREATE TABLE `posts` (
  `PostId` bigint(20) unsigned NOT NULL auto_increment,
  `PostTypeId` bigint(20) unsigned NOT NULL,
  `AcceptedAnswerId` bigint(20) unsigned default NULL,
  `ParentId` bigint(20) unsigned default NULL,
  `CreationDate` datetime NOT NULL,
  `Score` int(11) NOT NULL default '0',
  `ViewCount` int(11) NOT NULL default '0',
  `Body` text NOT NULL,
  `OwnerUserId` bigint(20) unsigned NOT NULL,
  `OwnerDisplayName` varchar(40) default NULL,
  `LastEditorUserId` bigint(20) unsigned default NULL,
  `LastEditDate` datetime default NULL,
  `LastActivityDate` datetime default NULL,
  `Title` varchar(250) NOT NULL default '',
  `Tags` varchar(150) NOT NULL default '',
  `AnswerCount` int(11) NOT NULL default '0',
  `CommentCount` int(11) NOT NULL default '0',
  `FavoriteCount` int(11) NOT NULL default '0',
  `ClosedDate` datetime default NULL,
  PRIMARY KEY  (`PostId`),
  UNIQUE KEY `PostId` (`PostId`),
  KEY `PostTypeId` (`PostTypeId`),
  KEY `AcceptedAnswerId` (`AcceptedAnswerId`),
  KEY `OwnerUserId` (`OwnerUserId`),
  KEY `LastEditorUserId` (`LastEditorUserId`),
  KEY `ParentId` (`ParentId`),
  CONSTRAINT `posts_ibfk_1` FOREIGN KEY (`PostTypeId`) REFERENCES `posttypes` (`PostTypeId`)
) ENGINE=InnoDB;

---

Note to commenters: If you want another benchmark with a different version of MySQL, a different dataset, or different table design, feel free to do it yourself. I have shown the technique above. Stack Overflow is here to show you how to do software development work, not to do all the work for you.

Answer 6

Here is a more efficient version in 1 line, works as long as the table has a time stamp column.

SELECT Id, Name, SUBSTRING_INDEX(MAX(CONCAT(TimeStamp, ',', Other_Columns)), ',', -1)
FROM Messages
ORDER BY id DESC GROUP BY Name 

This will return the latest record for the group on "Other_Columns"

Answer 7

As of MySQL 8.0.14, this can also be achieved using Lateral Derived Tables:

SELECT t.*
FROM messages t
JOIN LATERAL (
  SELECT name, MAX(id) AS id 
  FROM messages t1
  WHERE t.name = t1.name
  GROUP BY name
) trn ON t.name = trn.name AND t.id = trn.id

db<>fiddle

Answer 8

If you need the most recent or oldest record of a text column in a grouped query, and you would rather not use a subquery, you can do this...

Ex. You have a list of movies and need to get the count in the series and the latest movie

id	series	name
1	Star Wars	A New hope
2	Star Wars	The Empire Strikes Back
3	Star Wars	Return of The Jedi

SELECT COUNT(id), series, SUBSTRING(MAX(CONCAT(id, name)), LENGTH(id) + 1), 
FROM Movies
GROUP BY series

This returns...

id	series	name
3	Star Wars	Return of The Jedi

MAX will return the row with the highest value, so by concatenating the id to the name, you now will get the newest record, then just strip off the id for your final result.

More efficient than using a subquery.

So for the given example:

SELECT MAX(Id), Name, SUBSTRING(MAX(CONCAT(Id, Other_Columns)), LENGTH(Id) + 1), 
FROM messages
GROUP BY Name

Happy coding, and "May The Force Be With You" :)

Answer 9

MariaDB 10.3 and newer using GROUP_CONCAT.

The idea is to use ORDER BY + LIMIT:

SELECT GROUP_CONCAT(id ORDER BY id DESC LIMIT 1) AS id,
       name,
       GROUP_CONCAT(Other_columns ORDER BY id DESC LIMIT 1) AS Other_columns
FROM t
GROUP BY name;

db<>fiddle demo

Answer 10

What about:

select *, max(id) from messages group by name 

I have tested it on sqlite and it returns all columns and max id value for all names.

Answer 11

i find best solution in https://dzone.com/articles/get-last-record-in-each-mysql-group

select * from `data` where `id` in (select max(`id`) from `data` group by `name_id`)

Answer 12

Solution by sub query fiddle Link

select * from messages where id in
(select max(id) from messages group by Name)

Solution By join condition fiddle link

select m1.* from messages m1 
left outer join messages m2 
on ( m1.id<m2.id and m1.name=m2.name )
where m2.id is null

Reason for this post is to give fiddle link only. Same SQL is already provided in other answers.

Answer 13

Another approach :

Find the propertie with the max m2_price withing each program (n properties in 1 program) :

select * from properties p
join (
    select max(m2_price) as max_price 
    from properties 
    group by program_id
) p2 on (p.program_id = p2.program_id)
having p.m2_price = max_price

Answer 14

UPD: 2017-03-31, the version 5.7.5 of MySQL made the ONLY_FULL_GROUP_BY switch enabled by default (hence, non-deterministic GROUP BY queries became disabled). Moreover, they updated the GROUP BY implementation and the solution might not work as expected anymore even with the disabled switch. One needs to check.

Bill Karwin's solution above works fine when item count within groups is rather small, but the performance of the query becomes bad when the groups are rather large, since the solution requires about n*n/2 + n/2 of only IS NULL comparisons.

I made my tests on a InnoDB table of 18684446 rows with 1182 groups. The table contains testresults for functional tests and has the (test_id, request_id) as the primary key. Thus, test_id is a group and I was searching for the last request_id for each test_id.

Bill's solution has already been running for several hours on my dell e4310 and I do not know when it is going to finish even though it operates on a coverage index (hence using index in EXPLAIN).

I have a couple of other solutions that are based on the same ideas:

• if the underlying index is BTREE index (which is usually the case), the largest (group_id, item_value) pair is the last value within each group_id, that is the first for each group_id if we walk through the index in descending order;
• if we read the values which are covered by an index, the values are read in the order of the index;
• each index implicitly contains primary key columns appended to that (that is the primary key is in the coverage index). In solutions below I operate directly on the primary key, in you case, you will just need to add primary key columns in the result.
• in many cases it is much cheaper to collect the required row ids in the required order in a subquery and join the result of the subquery on the id. Since for each row in the subquery result MySQL will need a single fetch based on primary key, the subquery will be put first in the join and the rows will be output in the order of the ids in the subquery (if we omit explicit ORDER BY for the join)

3 ways MySQL uses indexes is a great article to understand some details.

Solution 1

This one is incredibly fast, it takes about 0,8 secs on my 18M+ rows:

SELECT test_id, MAX(request_id) AS request_id
FROM testresults
GROUP BY test_id DESC;

If you want to change the order to ASC, put it in a subquery, return the ids only and use that as the subquery to join to the rest of the columns:

SELECT test_id, request_id
FROM (
    SELECT test_id, MAX(request_id) AS request_id
    FROM testresults
    GROUP BY test_id DESC) as ids
ORDER BY test_id;

This one takes about 1,2 secs on my data.

Solution 2

Here is another solution that takes about 19 seconds for my table:

SELECT test_id, request_id
FROM testresults, (SELECT @group:=NULL) as init
WHERE IF(IFNULL(@group, -1)=@group:=test_id, 0, 1)
ORDER BY test_id DESC, request_id DESC

It returns tests in descending order as well. It is much slower since it does a full index scan but it is here to give you an idea how to output N max rows for each group.

The disadvantage of the query is that its result cannot be cached by the query cache.

Answer 15

Hope below Oracle query can help:

WITH Temp_table AS
(
    Select id, name, othercolumns, ROW_NUMBER() over (PARTITION BY name ORDER BY ID 
    desc)as rank from messages
)
Select id, name,othercolumns from Temp_table where rank=1

Answer 16

**

Hi, this query might help :

**

SELECT 
  *
FROM 
  message 

WHERE 
  `Id` IN (
    SELECT 
      MAX(`Id`) 
    FROM 
      message 
    GROUP BY 
      `Name`
  ) 
ORDER BY 
   `Id` DESC

Answer 17

SELECT * FROM table_name WHERE primary_key IN (SELECT MAX(primary_key) FROM table_name GROUP BY column_name )

Answer 18

You can group by counting and also get the last item of group like:

SELECT 
    user,
    COUNT(user) AS count,
    MAX(id) as last
FROM request 
GROUP BY user

Answer 19

If performance is really your concern you can introduce a new column on the table called IsLastInGroup of type BIT.

Set it to true on the columns which are last and maintain it with every row insert/update/delete. Writes will be slower, but you'll benefit on reads. It depends on your use case and I recommend it only if you're read-focused.

So your query will look like:

SELECT * FROM Messages WHERE IsLastInGroup = 1

Answer 20

Clearly there are lots of different ways of getting the same results, your question seems to be what is an efficient way of getting the last results in each group in MySQL. If you are working with huge amounts of data and assuming you are using InnoDB with even the latest versions of MySQL (such as 5.7.21 and 8.0.4-rc) then there might not be an efficient way of doing this.

We sometimes need to do this with tables with even more than 60 million rows.

For these examples I will use data with only about 1.5 million rows where the queries would need to find results for all groups in the data. In our actual cases we would often need to return back data from about 2,000 groups (which hypothetically would not require examining very much of the data).

I will use the following tables:

CREATE TABLE temperature(
  id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
  groupID INT UNSIGNED NOT NULL, 
  recordedTimestamp TIMESTAMP NOT NULL, 
  recordedValue INT NOT NULL,
  INDEX groupIndex(groupID, recordedTimestamp), 
  PRIMARY KEY (id)
);

CREATE TEMPORARY TABLE selected_group(id INT UNSIGNED NOT NULL, PRIMARY KEY(id)); 

The temperature table is populated with about 1.5 million random records, and with 100 different groups. The selected_group is populated with those 100 groups (in our cases this would normally be less than 20% for all of the groups).

As this data is random it means that multiple rows can have the same recordedTimestamps. What we want is to get a list of all of the selected groups in order of groupID with the last recordedTimestamp for each group, and if the same group has more than one matching row like that then the last matching id of those rows.

If hypothetically MySQL had a last() function which returned values from the last row in a special ORDER BY clause then we could simply do:

SELECT 
  last(t1.id) AS id, 
  t1.groupID, 
  last(t1.recordedTimestamp) AS recordedTimestamp, 
  last(t1.recordedValue) AS recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.groupID = g.id
ORDER BY t1.recordedTimestamp, t1.id
GROUP BY t1.groupID;

which would only need to examine a few 100 rows in this case as it doesn't use any of the normal GROUP BY functions. This would execute in 0 seconds and hence be highly efficient. Note that normally in MySQL we would see an ORDER BY clause following the GROUP BY clause however this ORDER BY clause is used to determine the ORDER for the last() function, if it was after the GROUP BY then it would be ordering the GROUPS. If no GROUP BY clause is present then the last values will be the same in all of the returned rows.

However MySQL does not have this so let's look at different ideas of what it does have and prove that none of these are efficient.

Example 1

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.id = (
  SELECT t2.id
  FROM temperature t2 
  WHERE t2.groupID = g.id
  ORDER BY t2.recordedTimestamp DESC, t2.id DESC
  LIMIT 1
);

This examined 3,009,254 rows and took ~0.859 seconds on 5.7.21 and slightly longer on 8.0.4-rc

Example 2

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM temperature t1
INNER JOIN ( 
  SELECT max(t2.id) AS id   
  FROM temperature t2
  INNER JOIN (
    SELECT t3.groupID, max(t3.recordedTimestamp) AS recordedTimestamp
    FROM selected_group g
    INNER JOIN temperature t3 ON t3.groupID = g.id
    GROUP BY t3.groupID
  ) t4 ON t4.groupID = t2.groupID AND t4.recordedTimestamp = t2.recordedTimestamp
  GROUP BY t2.groupID
) t5 ON t5.id = t1.id;

This examined 1,505,331 rows and took ~1.25 seconds on 5.7.21 and slightly longer on 8.0.4-rc

Example 3

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM temperature t1
WHERE t1.id IN ( 
  SELECT max(t2.id) AS id   
  FROM temperature t2
  INNER JOIN (
    SELECT t3.groupID, max(t3.recordedTimestamp) AS recordedTimestamp
    FROM selected_group g
    INNER JOIN temperature t3 ON t3.groupID = g.id
    GROUP BY t3.groupID
  ) t4 ON t4.groupID = t2.groupID AND t4.recordedTimestamp = t2.recordedTimestamp
  GROUP BY t2.groupID
)
ORDER BY t1.groupID;

This examined 3,009,685 rows and took ~1.95 seconds on 5.7.21 and slightly longer on 8.0.4-rc

Example 4

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.id = (
  SELECT max(t2.id)
  FROM temperature t2 
  WHERE t2.groupID = g.id AND t2.recordedTimestamp = (
      SELECT max(t3.recordedTimestamp)
      FROM temperature t3 
      WHERE t3.groupID = g.id
    )
);

This examined 6,137,810 rows and took ~2.2 seconds on 5.7.21 and slightly longer on 8.0.4-rc

Example 5

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM (
  SELECT 
    t2.id, 
    t2.groupID, 
    t2.recordedTimestamp, 
    t2.recordedValue, 
    row_number() OVER (
      PARTITION BY t2.groupID ORDER BY t2.recordedTimestamp DESC, t2.id DESC
    ) AS rowNumber
  FROM selected_group g 
  INNER JOIN temperature t2 ON t2.groupID = g.id
) t1 WHERE t1.rowNumber = 1;

This examined 6,017,808 rows and took ~4.2 seconds on 8.0.4-rc

Example 6

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM (
  SELECT 
    last_value(t2.id) OVER w AS id, 
    t2.groupID, 
    last_value(t2.recordedTimestamp) OVER w AS recordedTimestamp, 
    last_value(t2.recordedValue) OVER w AS recordedValue
  FROM selected_group g
  INNER JOIN temperature t2 ON t2.groupID = g.id
  WINDOW w AS (
    PARTITION BY t2.groupID 
    ORDER BY t2.recordedTimestamp, t2.id 
    RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
  )
) t1
GROUP BY t1.groupID;

This examined 6,017,908 rows and took ~17.5 seconds on 8.0.4-rc

Example 7

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM selected_group g
INNER JOIN temperature t1 ON t1.groupID = g.id
LEFT JOIN temperature t2 
  ON t2.groupID = g.id 
  AND (
    t2.recordedTimestamp > t1.recordedTimestamp 
    OR (t2.recordedTimestamp = t1.recordedTimestamp AND t2.id > t1.id)
  )
WHERE t2.id IS NULL
ORDER BY t1.groupID;

This one was taking forever so I had to kill it.

Answer 21

An approach with considerable speed is as follows.

SELECT * 
FROM messages a
WHERE Id = (SELECT MAX(Id) FROM messages WHERE a.Name = Name)

Result

Id  Name    Other_Columns
3   A   A_data_3
5   B   B_data_2
6   C   C_data_1

Answer 22

Here is my solution:

SELECT 
  DISTINCT NAME,
  MAX(MESSAGES) OVER(PARTITION BY NAME) MESSAGES 
FROM MESSAGE;

Answer 23

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

Answer 24

If you want the last row for each Name, then you can give a row number to each row group by the Name and order by Id in descending order.

QUERY

SELECT t1.Id, 
       t1.Name, 
       t1.Other_Columns
FROM 
(
     SELECT Id, 
            Name, 
            Other_Columns,
    (
        CASE Name WHEN @curA 
        THEN @curRow := @curRow + 1 
        ELSE @curRow := 1 AND @curA := Name END 
    ) + 1 AS rn 
    FROM messages t, 
    (SELECT @curRow := 0, @curA := '') r 
    ORDER BY Name,Id DESC 
)t1
WHERE t1.rn = 1
ORDER BY t1.Id;

SQL Fiddle

Answer 25

You can take view from here as well.

http://sqlfiddle.com/#!9/ef42b/9

FIRST SOLUTION

SELECT d1.ID,Name,City FROM Demo_User d1
INNER JOIN
(SELECT MAX(ID) AS ID FROM Demo_User GROUP By NAME) AS P ON (d1.ID=P.ID);

SECOND SOLUTION

SELECT * FROM (SELECT * FROM Demo_User ORDER BY ID DESC) AS T GROUP BY NAME ;

Answer 26

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

Answer 27

SELECT 
  column1,
  column2 
FROM
  table_name 
WHERE id IN 
  (SELECT 
    MAX(id) 
  FROM
    table_name 
  GROUP BY column1) 
ORDER BY column1 ;

Answer 28

Here is another way to get the last related record using GROUP_CONCAT with order by and SUBSTRING_INDEX to pick one of the record from the list

SELECT 
  `Id`,
  `Name`,
  SUBSTRING_INDEX(
    GROUP_CONCAT(
      `Other_Columns` 
      ORDER BY `Id` DESC 
      SEPARATOR '||'
    ),
    '||',
    1
  ) Other_Columns 
FROM
  messages 
GROUP BY `Name` 

Above query will group the all the Other_Columns that are in same Name group and using ORDER BY id DESC will join all the Other_Columns in a specific group in descending order with the provided separator in my case i have used || ,using SUBSTRING_INDEX over this list will pick the first one

Fiddle Demo

Answer 29

I've not yet tested with large DB but I think this could be faster than joining tables:

SELECT *, Max(Id) FROM messages GROUP BY Name

Answer 30

I arrived at a different solution, which is to get the IDs for the last post within each group, then select from the messages table using the result from the first query as the argument for a WHERE x IN construct:

SELECT id, name, other_columns
FROM messages
WHERE id IN (
    SELECT MAX(id)
    FROM messages
    GROUP BY name
);

I don't know how this performs compared to some of the other solutions, but it worked spectacularly for my table with 3+ million rows. (4 second execution with 1200+ results)

This should work both on MySQL and SQL Server.