I have a table like this
I want to check the all rows in Column A with column B and get the count of duplicates.
For example, I want to get the
count of 12 as 3(2 times in A+1 time in B)
count of 11 as 2(2 times in A+0 time in B)
count of 13 as 2(1 time in A+0 time in B)
How can I acheive it?
You can calculate the total occurrences from a union all. A where clause can show only the values that occur in the A column:
select nr
, count(*)
from (
select A as nr
from YourTable
union all
select B
from YourTable
) sub
where nr in -- only values that occur at least once in the A column
(
select A
from YourTable
)
group by
nr
having count(*) > 1 -- show only duplicates
You can combine all values in A and B then do the group by.
Then only select those values found in column A.
Select A, count(A) as cnt
From (
Select A
from yourTable
Union All
Select B
from yourTable) t
Where t.A in
(select distinct A from yourTable)
Group by t.A
Order by t.A;
Result:
A cnt
11 2
12 3
13 1
See demo: http://sqlfiddle.com/#!9/9fcfe9/3
How can you select the top n max values from a table?
For a table like this:
column1 column2
1 foo
2 foo
3 foo
4 foo
5 bar
6 bar
7 bar
8 bar
For n=2, the result needs to be:
3
4
7
8
The approach below selects only the max value for each group.
SELECT max(column1) FROM table GROUP BY column2
Returns:
4
8
For n=2 you could
SELECT max(column1) m
FROM table t
GROUP BY column2
UNION
SELECT max(column1) m
FROM table t
WHERE column1 NOT IN (SELECT max(column1)
WHERE column2 = t.column2)
for any n you could use approaches described here to simulate rank over partition.
EDIT:
Actually this article will give you exactly what you need.
Basically it is something like this
SELECT t.*
FROM
(SELECT grouper,
(SELECT val
FROM table li
WHERE li.grouper = dlo.grouper
ORDER BY
li.grouper, li.val DESC
LIMIT 2,1) AS mid
FROM
(
SELECT DISTINCT grouper
FROM table
) dlo
) lo, table t
WHERE t.grouper = lo.grouper
AND t.val > lo.mid
Replace grouper with the name of the column you want to group by and val with the name of the column that hold the values.
To work out how exactly it functions go step-by-step from the most inner query and run them.
Also, there is a slight simplification - the subquery that finds the mid can return NULL if certain category does not have enough values so there should be COALESCE of that to some constant that would make sense in the comparison (in your case it would be MIN of domain of the val, in article it is MAX).
EDIT2:
I forgot to mention that it is the LIMIT 2,1 that determines the n (LIMIT n,1).
If you are using mySQl, why don't you use the LIMIT functionality?
Sort the records in descending order and limit the top n i.e. :
SELECT yourColumnName FROM yourTableName
ORDER BY Id desc
LIMIT 0,3
Starting from MySQL 8.0/MariaDB support window functions which are designed for this kind of operations:
SELECT *
FROM (SELECT *,ROW_NUMBER() OVER(PARTITION BY column2 ORDER BY column1 DESC) AS r
FROM tab) s
WHERE r <= 2
ORDER BY column2 DESC, r DESC;
DB-Fiddle.com Demo
This is how I'm getting the N max rows per group in MySQL
SELECT co.id, co.person, co.country
FROM person co
WHERE (
SELECT COUNT(*)
FROM person ci
WHERE co.country = ci.country AND co.id < ci.id
) < 1
;
how it works:
self join to the table
groups are done by co.country = ci.country
N elements per group are controlled by ) < 1 so for 3 elements - ) < 3
to get max or min depends on: co.id < ci.id
co.id < ci.id - max
co.id > ci.id - min
Full example here:
mysql select n max values per group/
mysql select max and return multiple values
Note: Have in mind that additional constraints like gender = 0 should be done in both places. So if you want to get males only, then you should apply constraint on the inner and the outer select
I have two queries one will return data ordered by likes and in the user city the other one return data by the distance .
so if query 1 return : id 1,2,3 (order by likes)
and query 2 return : id 4,5,6 (order by distance)
i need the final set results to be 1,2,3,4,5,6
i've tried to do union between the two queries but it's not working. any other suggestions ?
You can use left join or union according to this link.
Union ALL also works like you can see here.
Example: SELECT 1 UNION ALL SELECT 2
the solution was to put a limit to each query then the union will work correct :
(SELECT DISTINCT ID, 'a' as type,... FROM table1 GROUP BY ID ORDER BY likesDESC limit 50) union all( SELECT DISTINCT ID, 'b' as type,....FROM table1 GROUP BY ID ORDER BY distance limit 50) order by type asc.
I have a table with the following structure:
id name
1 X
1 X
1 Y
2 A
2 A
2 B
Basically what I am trying to do is to write a query that returns X for 1 because X has repeated more than Y (2 times) and returns A for 2. So if a value occurs more than the other one my query should return that. Sorry if the title is confusing but I could not find a better explanation. This is what I have tried so far:
SELECT MAX(counted) FROM(
SELECT COUNT(B) AS counted
FROM table
GROUP BY A
) AS counts;
The problem is that my query should return the actual value other than the count of it.
Thanks
This should work:
SELECT count(B) as occurrence, A, B
FROM table
GROUP BY B
ORDER BY occurrence DESC
LIMIT 1;
Please check: http://sqlfiddle.com/#!9/dfa09/3
You can try like this using a GROUP BY clause. See a Demo Here
select *, max(occurence) as Maximum_Occurence from
(
select B, count(B) as occurence
from table1
group by B
) xxx
This is how I finally handled my problem. Not the most efficient way but get the job done:
select A,B from
(select A,B, max(cnt) from
(select A ,B ,count(B) as cnt
from myTable
group by A,B
order by cnt desc
) as x group by A
) as xx
What is the simplest SQL query to find the second largest integer value in a specific column?
There are maybe duplicate values in the column.
SELECT MAX( col )
FROM table
WHERE col < ( SELECT MAX( col )
FROM table )
SELECT MAX(col)
FROM table
WHERE col NOT IN ( SELECT MAX(col)
FROM table
);
In T-Sql there are two ways:
--filter out the max
select max( col )
from [table]
where col < (
select max( col )
from [table] )
--sort top two then bottom one
select top 1 col
from (
select top 2 col
from [table]
order by col) topTwo
order by col desc
In Microsoft SQL the first way is twice as fast as the second, even if the column in question is clustered.
This is because the sort operation is relatively slow compared to the table or index scan that the max aggregation uses.
Alternatively, in Microsoft SQL 2005 and above you can use the ROW_NUMBER() function:
select col
from (
select ROW_NUMBER() over (order by col asc) as 'rowNum', col
from [table] ) withRowNum
where rowNum = 2
I see both some SQL Server specific and some MySQL specific solutions here, so you might want to clarify which database you need. Though if I had to guess I'd say SQL Server since this is trivial in MySQL.
I also see some solutions that won't work because they fail to take into account the possibility for duplicates, so be careful which ones you accept. Finally, I see a few that will work but that will make two complete scans of the table. You want to make sure the 2nd scan is only looking at 2 values.
SQL Server (pre-2012):
SELECT MIN([column]) AS [column]
FROM (
SELECT TOP 2 [column]
FROM [Table]
GROUP BY [column]
ORDER BY [column] DESC
) a
MySQL:
SELECT `column`
FROM `table`
GROUP BY `column`
ORDER BY `column` DESC
LIMIT 1,1
Update:
SQL Server 2012 now supports a much cleaner (and standard) OFFSET/FETCH syntax:
SELECT [column]
FROM [Table]
GROUP BY [column]
ORDER BY [column] DESC
OFFSET 1 ROWS
FETCH NEXT 1 ROWS ONLY;
I suppose you can do something like:
SELECT *
FROM Table
ORDER BY NumericalColumn DESC
LIMIT 1 OFFSET 1
or
SELECT *
FROM Table ORDER BY NumericalColumn DESC
LIMIT (1, 1)
depending on your database server. Hint: SQL Server doesn't do LIMIT.
The easiest would be to get the second value from this result set in the application:
SELECT DISTINCT value
FROM Table
ORDER BY value DESC
LIMIT 2
But if you must select the second value using SQL, how about:
SELECT MIN(value)
FROM ( SELECT DISTINCT value
FROM Table
ORDER BY value DESC
LIMIT 2
) AS t
you can find the second largest value of column by using the following query
SELECT *
FROM TableName a
WHERE
2 = (SELECT count(DISTINCT(b.ColumnName))
FROM TableName b WHERE
a.ColumnName <= b.ColumnName);
you can find more details on the following link
http://www.abhishekbpatel.com/2012/12/how-to-get-nth-maximum-and-minimun.html
MSSQL
SELECT *
FROM [Users]
order by UserId desc OFFSET 1 ROW
FETCH NEXT 1 ROW ONLY;
MySQL
SELECT *
FROM Users
order by UserId desc LIMIT 1 OFFSET 1
No need of sub queries ... just skip one row and select second rows after order by descending
A very simple query to find the second largest value
SELECT `Column`
FROM `Table`
ORDER BY `Column` DESC
LIMIT 1,1;
SELECT MAX(Salary)
FROM Employee
WHERE Salary NOT IN ( SELECT MAX(Salary)
FROM Employee
)
This query will return the maximum salary, from the result - which not contains maximum salary from overall table.
Old question I know, but this gave me a better exec plan:
SELECT TOP 1 LEAD(MAX (column)) OVER (ORDER BY column desc)
FROM TABLE
GROUP BY column
This is very simple code, you can try this :-
ex :
Table name = test
salary
1000
1500
1450
7500
MSSQL Code to get 2nd largest value
select salary from test order by salary desc offset 1 rows fetch next 1 rows only;
here 'offset 1 rows' means 2nd row of table and 'fetch next 1 rows only' is for show only that 1 row. if you dont use 'fetch next 1 rows only' then it shows all the rows from the second row.
Simplest of all
select sal
from salary
order by sal desc
limit 1 offset 1
select * from (select ROW_NUMBER() over (Order by Col_x desc) as Row, Col_1
from table_1)as table_new tn inner join table_1 t1
on tn.col_1 = t1.col_1
where row = 2
Hope this help to get the value for any row.....
Use this query.
SELECT MAX( colname )
FROM Tablename
where colname < (
SELECT MAX( colname )
FROM Tablename)
select min(sal) from emp where sal in
(select TOP 2 (sal) from emp order by sal desc)
Note
sal is col name
emp is table name
select col_name
from (
select dense_rank() over (order by col_name desc) as 'rank', col_name
from table_name ) withrank
where rank = 2
SELECT
*
FROM
table
WHERE
column < (SELECT max(columnq) FROM table)
ORDER BY
column DESC LIMIT 1
It is the most esiest way:
SELECT
Column name
FROM
Table name
ORDER BY
Column name DESC
LIMIT 1,1
As you mentioned duplicate values . In such case you may use DISTINCT and GROUP BY to find out second highest value
Here is a table
salary
:
GROUP BY
SELECT amount FROM salary
GROUP by amount
ORDER BY amount DESC
LIMIT 1 , 1
DISTINCT
SELECT DISTINCT amount
FROM salary
ORDER BY amount DESC
LIMIT 1 , 1
First portion of LIMIT = starting index
Second portion of LIMIT = how many value
Tom, believe this will fail when there is more than one value returned in select max([COLUMN_NAME]) from [TABLE_NAME] section. i.e. where there are more than 2 values in the data set.
Slight modification to your query will work -
select max([COLUMN_NAME])
from [TABLE_NAME]
where [COLUMN_NAME] IN ( select max([COLUMN_NAME])
from [TABLE_NAME]
)
select max(COL_NAME)
from TABLE_NAME
where COL_NAME in ( select COL_NAME
from TABLE_NAME
where COL_NAME < ( select max(COL_NAME)
from TABLE_NAME
)
);
subquery returns all values other than the largest.
select the max value from the returned list.
This is an another way to find the second largest value of a column.Consider the table 'Student' and column 'Age'.Then the query is,
select top 1 Age
from Student
where Age in ( select distinct top 2 Age
from Student order by Age desc
) order by Age asc
select age
from student
group by id having age< ( select max(age)
from student
)
order by age
limit 1
SELECT MAX(sal)
FROM emp
WHERE sal NOT IN ( SELECT top 3 sal
FROM emp order by sal desc
)
this will return the third highest sal of emp table
select max(column_name)
from table_name
where column_name not in ( select max(column_name)
from table_name
);
not in is a condition that exclude the highest value of column_name.
Reference : programmer interview
Something like this? I haven't tested it, though:
select top 1 x
from (
select top 2 distinct x
from y
order by x desc
) z
order by x
See How to select the nth row in a SQL database table?.
Sybase SQL Anywhere supports:
SELECT TOP 1 START AT 2 value from table ORDER BY value
Using a correlated query:
Select * from x x1 where 1 = (select count(*) from x where x1.a < a)
select * from emp e where 3>=(select count(distinct salary)
from emp where s.salary<=salary)
This query selects the maximum three salaries. If two emp get the same salary this does not affect the query.