I have table called "users" and need to select 2 rows before and after specific row, sorted by users.score ASC
users table (structure):
id name score
1 John 2
2 Sara 1
3 san 3
4 test 2
5 jery 5
6 simon 6
7 bob2 7
8 jack 4
9 man 2
for example: need to select 2 rows before and after users.id = 5 order by users.score
result should be like:
id name score
3 san 3
8 jack 4
5 jery 5
6 simon 6
7 bob2 7
thanks,
Using union all and subqueries to limit the records should do it:
select * from users where id = 5
union all (
select * from users
where score < (select score from users where id = 5)
order by score desc limit 2
)
union all (
select * from users
where score > (select score from users where id = 5)
order by score asc limit 2
)
order by score
Sample SQL Fiddle
Edit: I think a better method is to number the rows according to score and then select the rows with number -2 and +2 from the rows of id 5:
select id, name, score
from (select
t.*, #rownum1 := #rownum1 + 1 as rank
from users t, (select #rownum1 := 0) r
order by score
) a,
(select rank from (
select t.*,
#rownum := #rownum + 1 as rank
from users t, (select #rownum := 0) r
order by score
) t
where id = 5
) b
where b.rank between a.rank -2 and a.rank+2
order by score;
Sample SQL Fiddle
Perhaps using union all
(
select * from users where id < 5 order by score limit 2
)
union all
(
select * from users where id > 5 order by score limit 2
)
(SELECT x.* FROM users x JOIN users y ON y.score <= x. score WHERE y.id = 5 ORDER BY score LIMIT 3)
UNION
(SELECT x.* FROM users x JOIN users y ON y.score >= x. score WHERE y.id = 5 ORDER BY score DESc LIMIT 3)
[ORDER BY score] ;
http://www.sqlfiddle.com/#!9/45c22/42
I just write the query, based on "jpw" solution (many thanks to him)
select * from users where id = 5
union all (
select * from users
where id in (select id from users where score < (select score from users u where u.id = 5) order by score ASC)
order by score desc limit 2
)
union all (
select * from users
where id in (select id from users where score > (select score from users u where u.id = 5) order by score ASC)
order by score ASC limit 2
)
order by score
Selecting arbitrarily ordered rows before and after a specific id
SET #j = 0;
SET #i = 0;
SELECT *
FROM (
SELECT id, col1, col2, ..., #j:=#j+1 AS pos
FROM `table`
WHERE col1=... ORDER BY col1 DESC, col2 ASC
) AS zz
WHERE (
SELECT position
FROM (
SELECT id AS id2, #i:=#i+1 AS position
FROM `table`
WHERE col1=... ORDER BY col1 DESC, col2 ASC
) AS zz
WHERE id2=$currId
)
IN (pos-5,pos-4,pos-3,pos-2,pos-1,pos,pos+1,pos+2,pos+3,pos+4,pos+5)
Related
For example, we have 1, 2 and 3 are the most frequent values at the same time, how to return them when it is a tie?
id
1
1
1
2
2
2
3
3
3
4
You could try:
SELECT id
FROM yourTable
GROUP BY id
HAVING COUNT(*) = (SELECT COUNT(*) FROM yourTable
GROUP BY id ORDER BY COUNT(*) DESC LIMIT 1);
On more recent versions of MySQL 8+, we can use RANK here:
WITH cte AS (
SELECT id, RANK() OVER (ORDER BY COUNT(*) DESC) rnk
FROM yourTable
GROUP BY id
)
SELECT id
FROM cte
WHERE rnk = 1;
I have a table like this:
id|name|points
1|Ralph|15
2|Dave|2
3|Mourphy|180
I need to get the user with id x and 5 users before and after him based on points rank:
I can retrive the user with
select *,rank() OVER (ORDER BY points DESC ) as rank from client where id = x;
How to retrive the others ?
Thank you
You already gave the answer inside your question.
It would be (Sql Server style)
DECLARE #myRank int
SELECT #myRank = rank() OVER (Order BY points DESC) FROM client WHERE id = x;
Select *, rank() OVER (Order BY points DESC) as rank
FROM client
HAVING rank between (#myRank - 5) and (#myRank +5);
If you want it in pure SQL, you'll have to work a little extra, but it's the same idea (just with sub-queries).
One method is to calculate the rank for "x" and to compare that to the rank for each row:
select c.*
from (select max(case when id = #x then rank end) over () as x_rank
from (select c.*, rank() OVER (ORDER BY score DESC ) as rank
from client c
) c
) c
where rank >= x_rank - 5 and rank <= x_rank + 5;
Note that this may not return exactly 11 rows if you have ties.
If you want exactly 5 before and after plus all rows with the same score:
with c as (
select max(case when id = #x then rank end) over () as x_rank
from (select c.*, rank() OVER (ORDER BY score DESC ) as rank
from client c
) c
)
(select c.*
from c
where rank < x_rank
order by rank desc
limit 5
) union all
(select c.*
from c
where rank = x_rank
) union all
(select c.*
from c
where rank > x_rank
order by rank asc
limit 5
) ;
Need assistance with simplifying this SQL query to possibly a single SELECT:
(SELECT * FROM `deals`
WHERE category_id = 1
ORDER BY id desc
LIMIT 10)
UNION
(SELECT * FROM `deals`
WHERE category_id = 2
ORDER BY id desc
LIMIT 10)
UNION
(SELECT * FROM `deals`
WHERE category_id = 4
ORDER BY id desc
LIMIT 10)
UNION
(SELECT * FROM `deals`
WHERE category_id = 5
ORDER BY id desc
LIMIT 10)
UNION
(SELECT * FROM `deals`
WHERE category_id = 6
ORDER BY id desc
LIMIT 10)
UNION
(SELECT * FROM `deals`
WHERE category_id = 8
ORDER BY id desc
LIMIT 10)
UNION
(SELECT * FROM `deals`
WHERE category_id = 9
ORDER BY id desc
LIMIT 10)
UNION
(SELECT * FROM `deals`
WHERE category_id = 10
ORDER BY id desc
LIMIT 10)
UNION
(SELECT * FROM `deals`
WHERE category_id = 17
ORDER BY id desc
LIMIT 10)
I've been told to try using GROUP BY and HAVING. However, any query I tried didn't work in the slightest...
Any help will be greatly appreciated!
EDIT - apologies, forgot to mention database engine is MySQL
You can condense this down with a window function to limit each group bucket to 10.
SELECT
*
FROM
(
SELECT
*,
ROW_NUMBER() OVER PARTITION BY(category_id ORDER BY id DESC) AS GroupOrder
FROM `deals`
WHERE category_id BETWEEN 1 AND 10
)AS X
WHERE
GroupOrder<=10
I'm not sure, I need to know if you need that limit of 10, is this like take top 10 of all those things?
if not then
SELECT * FROM `deals`
WHERE category_id between 0 and 10 or category_id=17
ORDER BY category_id asc, id desc
For older versions of MySQL without the windowing functions, here is the code.
SELECT T1.ID, T1.Category_ID, T1.Name
FROM (
SELECT #row_num := IF(#prev_value=concat_ws('',t.Category_ID),#row_num+1,1) AS RowNumber
,t.*
,#prev_value := concat_ws('',t.Category_ID)
FROM data t,
(SELECT #row_num := 1) x,
(SELECT #prev_value := '') y
ORDER BY t.Category_ID
) T1
WHERE T1.RowNumber < 10
AND T1.Category_ID IN (1,2,3,4,5,6,7,8,9,10)
You will need to add the necessary field names to the other select.
This uses the technique described here
I've the following table structure:
id |name |date
1 a 2012-01-01
2 a 2011-01-01
3 a 2010-01-01
4 a 2014-01-01
5 a 2011-01-01
I'd like to perform a select order by date (desc), and after select the first 3 rows from the results by a condition which would be where id = 1. So the second part of the query would be "give me the first 3 rows starting from the row whose id equals to 1"
EDIT:
After the first "part" the result would be:
SELECT id, name, date FROM table ORDER BY date DESC
id |name |date
4 a 2014-01-01
1 a 2012-01-01
2 a 2011-01-01
5 a 2011-01-01
3 a 2010-01-01
After the second part it should look like this (so the first 3 after the row whose id is 1):
id |name |date
2 a 2011-01-01
5 a 2011-01-01
3 a 2010-01-01
I have no any idea how could I solve it, please help me.
EDIT:
This is the concrete code I'd like to re-write:
SELECT `id`, `questions`.`userid`, `categories`.`name`, `user`.`username`, `title`,
`details`, `date` FROM `questions`
LEFT JOIN `user`
ON `questions`.`userid` = `user`.`userid`
LEFT JOIN `categories`
ON `questions`.`categoryid` = `categories`.`categoryid`
ORDER BY `date` DESC LIMIT 10
SELECT *
FROM table
WHERE date < (SELECT date FROM table WHERE id = 1)
ORDER BY date DESC
LIMIT 3
This isn't pretty because MySQL doesn't support row_number() or common table expressions, but it should work. Basically, get the row number ordered by the date, then select those whose row number is greater than an arbitrary value (in this case 1). Finally use limit to select the number of records you want.
SELECT id, name, mydate
FROM (
SELECT id, name, mydate, #rn:=#rn+1 rn
FROM mytable, (select #rn:=0) t
ORDER BY mydate DESC
) t2
WHERE rn > (
select rn
from (
SELECT id, name, mydate, #rn:=#rn+1 rn
FROM mytable, (select #rn:=0) t
ORDER BY mydate DESC
) t2
where id = 1
)
LIMIT 3
SQL Fiddle Demo
This is what you want to do... if finds the first id thats equal to 4 and then selects those out. then limit the offset to go to the next row and pull out 3
SELECT id, name, m_date from(
SELECT id, name, m_date, #a := id, if(#a = 4, #b := 1, #b) AS join_id
FROM test
join(SELECT #a := 0, #b := 0) t
ORDER BY m_date DESC
) AS tt
WHERE join_id = 1
LIMIT 1,3
SELECT temp.`id`, temp.`userid`, `categories`.`name`, `user`.`username`, temp.`title`,
temp.`details`, temp.`date` FROM (
SELECT `id`, `categoryid`, `details`, `title`, `userid`, `date`, #a := id, if(#a = 11, #b := 1, #b) AS join_id
FROM `questions`
join(SELECT #a := 0, #b := 0) t
ORDER BY `date` DESC
) as temp
LEFT JOIN `user`
ON temp.`userid` = `user`.`userid`
LEFT JOIN `categories`
ON temp.`categoryid` = `categories`.`categoryid`
WHERE join_id = 1
LIMIT 1,10;
SEE FIDDLE for clarification
This question already has answers here:
How to get next/previous record in MySQL?
(23 answers)
Closed 4 years ago.
I have the following table, named Example:
id(int 11) //not autoincriment
value (varchar 100)
It has the following rows of data:
0 100
2 150
3 200
6 250
7 300
Note that id values are not contiguous.
I've written this SQL so far:
SELECT * FROM Example WHERE id = 3
However, I don't know how to get the value of previous id and value of the next id...
Please help me get previous value and next value if id = 3 ?
P.S.: in my example it will be: previous - 150, next - 250.
Select the next row below:
SELECT * FROM Example WHERE id < 3 ORDER BY id DESC LIMIT 1
Select the next row above:
SELECT * FROM Example WHERE id > 3 ORDER BY id LIMIT 1
Select both in one query, e.g. use UNION:
(SELECT * FROM Example WHERE id < 3 ORDER BY id DESC LIMIT 1)
UNION
(SELECT * FROM Example WHERE id > 3 ORDER BY id LIMIT 1)
That what you mean?
A solution would be to use temporary variables:
select
#prev as previous,
e.id,
#prev := e.value as current
from
(
select
#prev := null
) as i,
example as e
order by
e.id
To get the "next" value, repeat the procedure. Here is an example:
select
id, previous, current, next
from
(
select
#next as next,
#next := current as current,
previous,
id
from
(
select #next := null
) as init,
(
select
#prev as previous,
#prev := e.value as current,
e.id
from
(
select #prev := null
) as init,
example as e
order by e.id
) as a
order by
a.id desc
) as b
order by
id
Check the example on SQL Fiddle
May be overkill, but it may help you
please try this sqlFiddle
SELECT value,
(SELECT value FROM example e2
WHERE e2.value < e1.value
ORDER BY value DESC LIMIT 1) as previous_value,
(SELECT value FROM example e3
WHERE e3.value > e1.value
ORDER BY value ASC LIMIT 1) as next_value
FROM example e1
WHERE id = 3
Edit: OP mentioned to grab value of previous id and value of next id in one of the comments so the code is here SQLFiddle
SELECT value,
(SELECT value FROM example e2
WHERE e2.id < e1.id
ORDER BY id DESC LIMIT 1) as previous_value,
(SELECT value FROM example e3
WHERE e3.id > e1.id
ORDER BY id ASC LIMIT 1) as next_value
FROM example e1
WHERE id = 3
SELECT *,
(SELECT value FROM example e1 WHERE e1.id < e.id ORDER BY id DESC LIMIT 1 OFFSET 0) as prev_value,
(SELECT value FROM example e2 WHERE e2.id > e.id ORDER BY id ASC LIMIT 1 OFFSET 0) as next_value
FROM example e
WHERE id=3;
And you can place your own offset after OFFSET keyword if you want to select records with higher offsets for next and previous values from the selected record.
Here's my solution may suit you:
SELECT * FROM Example
WHERE id IN (
(SELECT MIN(id) FROM Example WHERE id > 3),(SELECT MAX(id) FROM Example WHERE id < 3)
)
Demo: http://sqlfiddle.com/#!9/36c1d/2
A possible solution if you need it all in one row
SELECT t.id, t.value, prev_id, p.value prev_value, next_id, n.value next_value
FROM
(
SELECT t.id, t.value,
(
SELECT id
FROM table1
WHERE id < t.id
ORDER BY id DESC
LIMIT 1
) prev_id,
(
SELECT id
FROM table1
WHERE id > t.id
ORDER BY id
LIMIT 1
) next_id
FROM table1 t
WHERE t.id = 3
) t LEFT JOIN table1 p
ON t.prev_id = p.id LEFT JOIN table1 n
ON t.next_id = n.id
Sample output:
| ID | VALUE | PREV_ID | PREV_VALUE | NEXT_ID | NEXT_VALUE |
|----|-------|---------|------------|---------|------------|
| 3 | 200 | 2 | 150 | 4 | 250 |
Here is SQLFiddle demo
This query uses a user defined variable to calculate the distance from the target id, and a series of wrapper queries to get the results you want. Only one pass is made over the table, so it should perform well.
select * from (
select id, value from (
select *, (#x := ifnull(#x, 0) + if(id > 3, -1, 1)) row from (
select * from mytable order by id
) x
) y
order by row desc
limit 3
) z
order by id
See an SQLFiddle
If you don't care about the final row order you can omit the outer-most wrapper query.
If you do not have an ID this has worked for me.
Next:
SELECT * FROM table_name
WHERE column_name > current_column_data
ORDER BY column_name ASC
LIMIT 1
Previous:
SELECT * FROM table_name
WHERE column_name < current_column_data
ORDER BY column_name DESC
LIMIT 1
I use this for a membership list where the search is on the last name of the member. As long as you have the data from the current record it works fine.