SELECT *
FROM (SELECT id, user, MAX(score) FROM table_1 GROUP BY user) AS sub
ORDER BY 'sub.score' ASC;
This SQL query should select from a table only a score per user, and for accuracy, the highest.
The table structure is this:
+-----------------------+
| id | score | username |
+-----------------------+
| 1 | 15 | mike |
| 2 | 23 | tom |
| 3 | 16 | mike |
| 4 | 22 | jack |
etc..
The result should be like:
3 mike 16
2 tom 23
4 jack 22
And then reordered:
3 mike 16
4 jack 22
2 tom 23
But the query does not reorder the subquery by score. How to do so?
Let's look at what you are doing step by step:
SELECT id, user, MAX(score) FROM table_1 GROUP BY user
Here you are grouping by user name, so you get one result row per user name. In this result row you select the user name, the maximum score found for this user name (which is 16 for 'mike') and one of the IDs found for the user name (which can be 1 or 3 for 'mike', the DBMS is free to choose one). This is probably not what you want.
SELECT * FROM (...) AS sub ORDER BY 'sub.score' ASC;
'sub.score' is a string (single quotes). You want to order by the max score from your subquery instead. So first give the max(score) a name, e.g. max(score) as max_score, and then access that: ORDER BY sub.max_score ASC.
Anyway, if you want the record with the maximum score for a user name (so as to get the according ID, too), you could look for records for which not exists a record with the same user name and a higher score. Sorting is easy then: as there is no aggregation, you simply order by score:
select * from table_1 t1 where not exists
(select * from table_1 higher where higher.name = t1.name and higher.score > t1.score)
order by score;
Assuming user|score is unique..:
SELECT x.*
FROM table_1 x
JOIN ( SELECT user, MAX(score) score FROM table_1 GROUP BY user) y
ON y.user = x.user
AND y.score = x.score
ORDER BY x.score
No need to write sub queries.
Simply you can use this way:
SELECT id, `user`, MAX(score) FROM table_1 GROUP BY `user`
ORDER BY MAX(score);
If you want query with sub query:
SELECT * FROM (SELECT id, `user`, MAX(score) as max_score FROM table_1
GROUP BY `user`) AS sub ORDER BY max_score;
Related
I have the following table called 'players' which contains scores from 3 games played by Tim, Bob and Jon.
playid | name | score
1 | Tim | 10
1 | Bob | 5
2 | Tim | 5
2 | Bob | 10
3 | Tim | 5
3 | Bob | 10
3 | Jon | 4
I want to be able to count the number of times that Tim, Bob and Jon have come second i.e. Tim = 2, Bob = 1, Jon = 0.
I have the following query:
SELECT name FROM players WHERE playid = 1 ORDER BY score Desc LIMIT 1, 1
Which returns the name of the person in second place in the first game i.e. Bob, but I can't figure out how to extend this to cover all games and players. Eventually I also want to be able to count the number of times they come 3rd, 4th etc.
Thanks in advance
Try with following one:
SELECT count(playid), name, score
FROM `players`
WHERE score = (SELECT MAX(score) FROM players WHERE score < (SELECT MAX(score) FROM players))
GROUP BY score, name;
With multiple joins and groupings:
select pl.name, ifnull(counter, 0) counter from (
select distinct name from players) pl
left join (
select players.name, count(*) counter from players
inner join (
select p.playid, max(p.score) as secondscore from (
select players.* from players
left join (
select playid, max(score) as maxscore
from players
group by playid) p
on p.playid = players.playid and p.maxscore = players.score
where p.maxscore is null) p
group by p.playid
) p
on p.playid = players.playid and p.secondscore = players.score
group by players.name) p
on p.name = pl.name
See the demo
You can use this query below to find the secondly-ranked people :
SELECT q2.playid, q2.name
FROM
(
SELECT q1.* , if(switch1,#r:=#r+1,#r:=0) as switch2
FROM
(
SELECT playid, score, name,
if(playid=#p,#p:=0,#p:=playid+1) as switch1
FROM players
JOIN ( SELECT #p:=0, #r:=-1 ) p2
ORDER BY playid, score desc
) q1
) q2
WHERE q2.switch2 = 1
ORDER BY q2.playid
playid name
------ ----
1 Bob
2 Tim
3 Tim
4 George
Rextester Demo
As per the comment by Raymond Nijland, in MySQL 8.0+ you can use window functions to achieve this:
SELECT name, COUNT(*) AS second_place_count
FROM (
SELECT
name,
playid,
ROW_NUMBER() OVER (PARTITION BY playid ORDER BY score DESC) AS rn
FROM players
) AS ranks
WHERE ranks.rn = 2
GROUP BY name
...or if you want to extend this to all places:
SELECT
name,
rn AS place,
COUNT(*) AS place_count
FROM (
SELECT
name,
playid,
ROW_NUMBER() OVER (PARTITION BY playid ORDER BY score DESC) AS rn
FROM players
) AS ranks
GROUP BY
name,
rn
I have a table with multiple records in it for full name and surname.
---------------------------
id | name | lastname
---------------------------
1 | A smith | smith
2 | B smith | smith
3 | c smith | smith
4 | A josh | josh
5 | B josh | josh
6 | C josh | josh
7 | D josh | josh
8 | A white | white
9 | D white | white
10| z white | white
And so so....more than 100k records. Now what i want to do is to retrieve latest 7 records for each surname up to 9 surnames. I have 500 surnames but i just want latest 9 surnames.. In my application "latest" means "largest value of id column."
This is the command that i tried to make but when i execute it. i am not getting any response from server. this is happening because of database size and my command is taking a lot of time. its just keep me waiting:
SELECT * FROM `queue` s WHERE ( SELECT COUNT(*) FROM `queue` f WHERE f.lastname = s.lastname AND f.id >= s.id LIMIT 0 , 7) <=7
Can someone suggest me better way of retrieving my goal.
Let's build this up from the basics.
Your first step is to create a subquery to get the latest nine surnames (http://sqlfiddle.com/#!9/aee62e/19/0). By that I mean the surnames with the highest id values.
SELECT lastname, MAX(id) namerank
FROM t
GROUP BY lastname
ORDER BY MAX(id) DESC
LIMIT 9
And, in MySQL, that was the easy part. Now you need to retrieve the seven highest ranked (largest id) rows for each selected surname. As a start, you could do this to get all records for the selected surnames, in descending order by id. (http://sqlfiddle.com/#!9/aee62e/18/0).
SELECT t.*, namerank
FROM t
JOIN (
SELECT lastname, MAX(id) namerank
FROM t
GROUP BY lastname
ORDER BY MAX(id) DESC
LIMIT 9
) h ON t.lastname = h.lastname
ORDER BY t.lastname, t.id DESC
This is correct, but contains too many rows. Next we need to get the ranking for each lastname's rows. A lower ranking means a higher id value. This is the nasty hack in MySQL. (Nasty because it mixes procedural operations on local variables with the inherently declarative nature of SQL.) (http://sqlfiddle.com/#!9/aee62e/17/0)
SELECT IF(detail.lastname = #prev_lastname, #rank := #rank+1, #rank :=1) rank,
namerank,
#prev_lastname := detail.lastname lastname,
id,
name
FROM (
SELECT t.*, namerank
FROM t
JOIN (
SELECT lastname, MAX(id) namerank
FROM t
GROUP BY lastname
ORDER BY MAX(id) DESC
LIMIT 9
) h ON t.lastname = h.lastname
ORDER BY t.lastname, t.id DESC
) detail
JOIN (SELECT #rank := 0, #prev_lastname := '') initializer
Finally we need to wrap that whole mess in an outer query to pick off the seven highest ranked rows for each lastname value. (http://sqlfiddle.com/#!9/aee62e/16/0)
SELECT *
FROM (
SELECT IF(detail.lastname = #prev_lastname, #rank := #rank+1, #rank :=1) rank,
namerank,
#prev_lastname := detail.lastname lastname,
id,
name
FROM (
SELECT t.*, namerank
FROM t
JOIN (
SELECT lastname, MAX(id) namerank
FROM t
GROUP BY lastname
ORDER BY MAX(id) DESC
LIMIT 9
) h ON t.lastname = h.lastname
ORDER BY t.lastname, t.id DESC
) detail
JOIN (SELECT #rank := 0, #prev_lastname := '') initializer
) ranked
WHERE rank <= 7
ORDER BY namerank DESC, rank
I believe the technical term for the complexity of your requirement and this solution is "hairball." It definitely puts the structured in Structured Query Language.
Is it possible to select the next lower number from a table without using limit.
Eg: If my table had 10, 3, 2 , 1 I'm trying to select * from table where col > 10.
The result I'm expecting is 3. I know I can use limit 1, but can it be done without that?
Try
SELECT MAX(no) no
FROM table1
WHERE no < 10
Output:
| NO |
------
| 3 |
SQLFiddle
Try this query
SELECT
*
FROM
(SELECT
#rid:=#rid+1 as rId,
a.*
FROM
tbl a
JOIN
(SELECT #rid:=0) b
ORDER BY
id DESC)tmp
WHERE rId=2;
SQL FIDDLE:
| RID | ID | TYPE | DETAILS |
------------------------------------
| 2 | 28 | Twitter | #sqlfiddle5 |
Another approach
select a.* from supportContacts a inner join
(select max(id) as id
from supportContacts
where
id in (select id from supportContacts where id not in
(select max(id) from supportContacts)))b
on a.id=b.id
SQL FIDDLE:
| ID | TYPE | DETAILS |
------------------------------
| 28 | Twitter | #sqlfiddle5 |
Alternatively, this query will always get the second highest number based on the inner where clause.
SELECT *
FROM
(
SELECT t.col,
(
SELECT COUNT(distinct t2.col)
FROM tableName t2
WHERE t2.col >= t.col
) as rank
FROM tablename t
WHERE col <= 10
) xx
WHERE rank = 2 -- <<== means second highest
SQLFiddle Demo
SQLFiddle Demo (supports duplicate values)
If you want to get next lower number from table
you can get it with this query:
SELECT distinct col FROM table1 a
WHERE 2 = (SELECT count(DISTINCT(b.col)) FROM table1 b WHERE a.col >= b.col);
later again if you want to get third lower number you can just pass 3 in place of 2 in where clause
again if you want to get second higher number, just change the condition of where clause in inner query with
a.col <= b.col
I have a query that returns data in the following format:
id | name | number
1 John 12545
1 John 50496
2 Mary 23443
3 Mark 54
3 Mark 5600
3 Mark 50206
I would like to find out the number of distinct ids that appear in the result set. For example, for the result above. I would like to obtain the value 3.
Is there any way to add a column so the result looks like this instead?
count | id | name | number
3 1 John 12545
3 1 John 50496
3 2 Mary 23443
3 3 Mark 54
3 3 Mark 5600
3 3 Mark 50206
My query is:
SELECT * FROM (
SELECT id FROM tableA
WHERE xyz
) as t1
JOIN tableB using (id)
SELECT (SELECT COUNT(DISTINCT id) FROM tableName) totalCount,
id,name,number
FROM tableName
or by using CROSS JOIN
SELECT x.totalCount,
a.id, a.name, a.number
FROM tableName a, (SELECT COUNT(DISTINCT id) totalCount
FROM tableName) x
You should try :
SELECT id,name,number, (SELECT COUNT(DISTINCT name) FROM YourTableName) FROM YourTableName
Good luck
SELECT COUNT(DISTINCT id) would be faster than using column name.
SELECT (SELECT COUNT(DISTINCT id) FROM tableName) as 'count',
id,name,number
FROM tableName
SELECT COUNT(id) AS count , id, name, number
FROM
(
SELECT id
FROM tableA
WHERE xyz
) as t1
JOIN tableB using (id)
GROUP BY id, name, number
I have 2 tables. One of tables has all mails from users and other table has all calls from users.
Table 1
call_id, | call_date | user_id
1 | 10/01/12| 3
2 | 9/01/12 | 3
Table 2
mail_id, | mail_date | user_id
1 | 8/01/12 | 3
2 7/01/12 | 3
I need to get last last 3 calls and mails :
10/01/12 - call
9/01/12 - call
8/01/12 - mail
Assuming MySQL, and for just one User_ID...
SELECT
*
FROM
(
SELECT 'call' AS type, id, call_date AS event_date, user_id FROM table_1
UNION ALL
SELECT 'mail' AS type, id, mail_date AS event_date, user_id FROM table_2
)
data
WHERE
user_id = 3
ORDER BY
event_date DESC
LIMIT
3
EDIT: Ooops, forgot to specify DESC in the ORDER BY, sorry.
Declare #userID int;
SET #userID=3;
select call_id FROM table_1
where user_id=#userID
order by call_date desc limit 2
UNION ALL
select mail_id FROM table_2
where user_id=#userID
order by mail_date desc limit 1