I have following table in MySql
Number
1
2
3
2
3
4
4
4
I would like to get numbers by count but with excluded highest and lowest count. This should be the result
Number Count
2 2
3 2
Please help me with this query.
Try this:
SELECT number, COUNT(*) cnt
FROM mytable
GROUP BY number
HAVING cnt <> (SELECT COUNT(*)
FROM mytable
GROUP BY number
ORDER BY COUNT(*) LIMIT 1)
AND
cnt <> (SELECT COUNT(*)
FROM mytable
GROUP BY number
ORDER BY COUNT(*) DESC LIMIT 1)
The two subqueries used in the HAVING clause return the minimum and maximum count. Hence the two predicates of the HAVING clause filter out the groups with the highest and lowest count.
A solution is using user variable and sub-query.
Query needed:
select Number, Count
from (
select Number, count(*) as Count, #rank := #rank+1 as rank
from t1 cross join (select #rank := 0) param
group by Number
order by Count ) t2
where rank not in (1, #rank);
Demo:
create table t1(Number int);
insert into t1 values(1),(2),(3),(2),(3),(4),(4),(4);
select Number, Count
from (
select Number, count(*) as Count, #rank := #rank+1 as rank
from t1 cross join (select #rank := 0) param
group by Number
order by Count ) t2
where rank not in (1, #rank);
Output:
mysql> select Number, Count
-> from (
-> select Number, count(*) as Count, #rank := #rank+1 as rank
-> from t1 cross join (select #rank := 0) param
-> group by Number
-> order by Count ) t2
-> where rank not in (1, #rank);
+--------+-------+
| Number | Count |
+--------+-------+
| 2 | 2 |
| 3 | 2 |
+--------+-------+
2 rows in set (0.00 sec)
select number,count(number)
from numbers
where number not in ((select max(number) from numbers),(select min(number) from numbers) )
group by number
sqlfiddle demo
First select all rows except the maximum and minimum numbers.
Then find the count(number) from the result set.
Query
select t.number, count(t.number) as `Count` from(
select * from your_table_name
where number not in (
select max(number) from your_table_name
)
and number not in (
select min(number) from your_table_name
)
)t
group by t.num;
SELECT column FROM tbl
LIMIT 10 OFFSET 20
SELECT MIN(taka) AS mintaka, MAX(taka) AS maxtaka
FROM (SELECT taka FROM tbl LIMIT 10 OFFSET 20);
Related
I currently need to order data by highest value down, and then lowest value up, in between.
My Query is close, but doesn't quite order by largest down, though it is inserting the lowest in between:
DEMO Fiddle
select users.*
from users CROSS JOIN (select #even := 0, #odd := 0) param
order by
IF(score > 1, 2*(#odd := #odd + 1), 2*(#even := #even + 1) + 1),
score DESC;
Current Results
Email Score
----- --------
foo1#gmail.com 42
foo5#gmail.com 1
foo2#gmail.com 49
foo6#gmail.com 0
foo3#gmail.com 37
foo4#gmail.com 7
foo#gmail.com 22
Desired Results
Email Score
----- --------
foo2#gmail.com 49
foo6#gmail.com 0
foo1#gmail.com 42
foo5#gmail.com 1
foo3#gmail.com 37
foo4#gmail.com 7
foo#gmail.com 22
You can achieve using MySQL but avoid such complex SQL statements as the same can be achieved using programming language very easily.
SET #totalRows := (CASE WHEN (SELECT COUNT(*) FROM users) IS NULL THEN 0 ELSE (SELECT COUNT(*) FROM users) END);
PREPARE stmt1 FROM '(SELECT t.* FROM (SELECT ROW_NUMBER() OVER (ORDER BY score DESC) AS row_num, email, score FROM users UNION ALL SELECT ROW_NUMBER() OVER (ORDER BY score ASC) AS row_num, email, score FROM users) AS t ORDER BY t.row_num, t.score DESC LIMIT 0,?)';
EXECUTE stmt1 USING #totalRows;
Here is the explanation to achieve it:
Set variable #totalRows contains total rows in users table as that many rows will be displayed as a final result set.
Used ROW_NUMBER() function of MySQL to set ordering based on SOCRE field DESCENDING and ASCENDING for another result set.
Combined both the result set using UNION ALL statement of MySQL
Add the LIMIT keyword to make sure the final result must have a total number of rows that should not exceed #totalRows variable.
As in MySQL LIMIT statement we can't pass a dynamic value at a query level. I used the approach of prepare a statement.
You can ignore row_num column I used as final result set.
Hope my solution will help you.
For MySql 8.0+ you can use ROW_NUMBER() window function:
SELECT email, score
FROM (
SELECT *,
ROW_NUMBER() OVER (ORDER BY score DESC) rn1,
ROW_NUMBER() OVER (ORDER BY score ASC) rn2
FROM users
) t
ORDER BY LEAST(rn1, rn2), rn1;
For previous versions you can simulate ROW_NUMBER() with correlated subqueries (with the cost of poor performance for large datasets):
SELECT email, score
FROM (
SELECT u1.*,
(SELECT COUNT(*) FROM users u2 WHERE u2.score > u1.score) rn1,
(SELECT COUNT(*) FROM users u2 WHERE u2.score < u1.score) rn2
FROM users u1
) t
ORDER BY LEAST(rn1, rn2), rn1;
See the demo.
I'm attempting to create an SQL query that retrieves the total_cost for every row in a table. Alongside that, I also need to collect the most dominant value for both columnA and columnB, with their respective values.
For example, with the following table contents:
cost
columnA
columnB
target
250
Foo
Bar
XYZ
200
Foo
Bar
XYZ
150
Bar
Bar
ABC
250
Foo
Bar
ABC
The result would need to be:
total_cost
columnA_dominant
columnB_dominant
columnA_value
columnB_value
850
Foo
Bar
250
400
Now I can get as far as calculating the total cost - that's no issue. I can also get the most dominant value for columnA using this answer. But after this, I'm not sure how to also get the dominant value for columnB and the values too.
This is my current SQL:
SELECT
SUM(`cost`) AS `total_cost`,
COUNT(`columnA`) AS `columnA_dominant`
FROM `table`
GROUP BY `columnA_dominant`
ORDER BY `columnA_dominant` DESC
WHERE `target` = "ABC"
UPDATE: Thanks to #Barmar for the idea of using a subquery, I managed to get the dominant values for columnA and columnB:
SELECT
-- Retrieve total cost.
SUM(`cost`) AS `total_cost`,
-- Get dominant values.
(
SELECT `columnA`
FROM `table`
GROUP BY `columnA`
ORDER BY COUNT(*) DESC
LIMIT 1
) AS `columnA_dominant`,
(
SELECT `columnB`
FROM `table`
GROUP BY `columnB`
ORDER BY COUNT(*) DESC
LIMIT 1
) AS `columnB_dominant`
FROM `table`
WHERE `target` = "XYZ"
However, I'm still having issues figuring out how to calculate the respective values.
You might get close, if we want to get percentage values we can try to add COUNT(*) at subquery to get max count by columnA and columnB then do division by total count
SELECT
SUM(cost),
(
SELECT tt.columnA
FROM T tt
GROUP BY tt.columnA
ORDER BY COUNT(*) DESC
LIMIT 1
) AS columnA_dominant,
(
SELECT tt.columnB
FROM T tt
GROUP BY tt.columnB
ORDER BY COUNT(*) DESC
LIMIT 1
) AS columnB_dominant,
(
SELECT COUNT(*)
FROM T tt
GROUP BY tt.columnA
ORDER BY COUNT(*) DESC
LIMIT 1
) / COUNT(*) AS columnA_percentage,
(
SELECT COUNT(*)
FROM T tt
GROUP BY tt.columnB
ORDER BY COUNT(*) DESC
LIMIT 1
) / COUNT(*) AS columnB_percentage
FROM T t1
If your MySQL version supports the window function, there is another way which reduce table scan might get better performance than a correlated subquery
SELECT SUM(cost) OVER(),
FIRST_VALUE(columnA) OVER (ORDER BY counter1 DESC) columnA_dominant,
FIRST_VALUE(columnB) OVER (ORDER BY counter2 DESC) columnB_dominant,
FIRST_VALUE(counter1) OVER (ORDER BY counter1 DESC) / COUNT(*) OVER() columnA_percentage,
FIRST_VALUE(counter2) OVER (ORDER BY counter2 DESC) / COUNT(*) OVER() columnB_percentage
FROM (
SELECT *,
COUNT(*) OVER (PARTITION BY columnA) counter1,
COUNT(*) OVER (PARTITION BY columnB) counter2
FROM T
) t1
LIMIT 1
sqlfiddle
try this query
select sum(cost) as total_cost,p.columnA,q.columnB,p.columnA_percentage,q.columnB_percentage
from get_common,(
select top 1 columnA,columnA_percentage
from(
select columnA,count(columnA) as count_columnA,cast(count(columnA) as float)/(select count(columnA) from get_common) as columnA_percentage
from get_common
group by columnA)s
order by count_columnA desc
)p,
(select top 1 columnB,columnB_percentage
from (
select columnB,count(columnB) as count_columnB, cast(count(columnB) as float)/(select count(columnB) from get_common) as columnB_percentage
from get_common
group by columnB) t
order by count_columnB desc)q
group by p.columnA,q.columnB,p.columnA_percentage,q.columnB_percentage
so if you want to get the percent and dominant value you must make their own query like this
select top 1 columnA,columnA_percentage
from(
select columnA,count(columnA) as count_columnA,cast(count(columnA) as float)/(select count(columnA) from get_common) as columnA_percentage
from get_common
group by columnA)s
order by count_columnA desc
then you can join with the sum query to get all value you want
hope this can help you
**castID**
nm0000116
nm0000116
nm0000116
nm0000116
nm0000116
nm0634240
nm0634240
nm0798899
This is my table (created as a view). Now I want to list the castID which has the most count (in this case which is nm0000116, and how many occurences/count it has in this table ( should be 5 times) and I'm not quite sure which query to use
try
Select CastId, count(*) countOfCastId
From table
Group By CastId
Having count(*)
= (Select Max(cnt)
From (Select count(*) cnt
From table
Group By CastId) z)
SELECT
MAX(E),
castId
FROM
(SELECT COUNT(castId) AS E,castId FROM [directors winning movies list] GROUP BY castId) AS t
You could just return the topmost count using LIMIT:
SELECT castID,
COUNT(*) AS Cnt
FROM atable
GROUP BY castID
ORDER BY Cnt DESC
LIMIT 1
;
However, if there can be ties, the above query would return only one row. If you want all the "winners", you could take the count from the above query as a scalar result and compare it against all the counts to return only those that match:
SELECT castID,
COUNT(*) AS Cnt
FROM atable
GROUP BY castID
HAVING COUNT(*) = (
SELECT COUNT(*)
FROM atable
GROUP BY castID
ORDER BY Cnt DESC
LIMIT 1
)
;
(Basically, same as Charles Bretana's approach, it just derives the top count differently.)
Alternatively, you could use a variable to rank all the counts and then return only those that have the ranking of 1:
SELECT castID,
Cnt
FROM (
SELECT castID,
COUNT(*) AS Cnt,
#r := IFNULL(#r, 0) + 1 AS r
FROM atable
GROUP BY castID
ORDER BY Cnt DESC
) AS s
WHERE r = 1
;
Please note that with the above method the variable must either not exist or be pre-initialised with a 0 or NULL prior to running the query. To be on the safe side, you could initialise the variable directly in your query:
SELECT s.castID,
s.Cnt
FROM (SELECT #r := 0) AS x
CROSS JOIN
(
SELECT castID,
COUNT(*) AS Cnt,
#r := #r + 1 AS r
FROM atable
GROUP BY castID
ORDER BY Cnt DESC
) AS s
WHERE s.r = 1
;
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.