I am really confused about the query that needing to return top N rows having biggest values on particular column.
For example, if the rows N-1, N, N + 1 have same values. Must I return just top N or top N + 1 rows.
If you do:
select *
from t
order by value desc
limit N
You will get the top N rows.
If you do:
select *
from t join
(select min(value) as cutoff
from (select value
from t
order by value
limit N
) tlim
) tlim
on t.value >= tlim;
Or you could phrase this a bit more simply as:
select *
from t join
(select value
from t
order by value
limit N
) tlim
on t.value = tlim.value;
The following is conceptually what you want to do, but it might not work in MySQL:
select *
from t
where t.value >= ANY (select value from t order by value limit N)
Use the following SQL query.
SELECT salary FROM salesperson
ORDER BY salary DESC
LIMIT 2,1
You should use self join for this.
first find the top (n) possible values for a perticular column
join it with same table based on the primary key
For E.g. on below sample table
CREATE TABLE `employee` (
`ID` INT(11) AUTO_INCREMENT PRIMARY KEY,
`NAME` VARCHAR(50) NOT NULL,
`SALARY` INT(11) NOT NULL ,
JOINING_DATE TIMESTAMP
) ENGINE=MYISAM
INSERT INTO employee (NAME,salary,joining_date) VALUES('JAMES',50000,'2010-02-02'),
('GARGI',60000,'2010-02-02'),('DAN',30000,'2010-02-02'),('JOHN',10000,'2010-02-02'),('MICHEL',70000,'2010-02-02'),
('STIEVE',50000,'2010-02-02'),('CALRK',20000,'2010-02-02'),('BINNY',50000,'2010-02-02'),('SMITH',40000,'2010-02-02'),
('ROBIN',60000,'2010-02-02'),('CRIS',80000,'2010-02-02');
With the above table-data set up Query to find employees having top 3 salaries would be :
SELECT e1.* FROM
(SELECT DISTINCT salary FROM Employee ORDER BY salary DESC LIMIT 3 ) S1
JOIN employee e1
ON e1.salary = s1.salary
ORDER BY e1.salary DESC
TIP:-
If you need top 4 then just change LIMIT 3 to LIMIT 4
Related
Structure is:
CREATE TABLE current
(
id BIGINT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
symbol VARCHAR(5),
UNIQUE (id), INDEX (symbol)
) ENGINE MyISAM;
id
symbol
1
A
2
B
3
C
4
C
5
B
6
A
7
C
8
C
9
A
10
B
I am using the following
SELECT *
FROM current
WHERE id
IN
(
SELECT MAX(id)
FROM current
GROUP BY symbol
)
to return the last records in a table.
id
symbol
8
C
9
A
10
B
How can I return the next-to-last results in a similar fashion?
I know that I need
ORDER BY id DESC LIMIT 1,1
somewhere, but my foo is weak.
I would want to return
id
symbol
5
B
6
A
7
C
For versions of MySql prior to 8.0, use a subquery in the WHERE clause to filter out the max id of each symbol and then aggregate:
SELECT MAX(id) id, symbol
FROM current
WHERE id NOT IN (SELECT MAX(id) FROM current GROUP BY symbol)
GROUP BY symbol
ORDER BY id;
See the demo.
SELECT *
FROM current
WHERE id IN (
SELECT DISTINCT T.id FROM current AS T
WHERE id=(
SELECT id FROM current
WHERE symbol=T.symbol
ORDER BY id DESC LIMIT 1,1
)
)
Easy if your MySql can use ROW_NUMBER. (MySql 8)
Just make it sort descending, then take the 2nd.
WITH CTE AS (
SELECT *
, ROW_NUMBER() OVER (PARTITION BY symbol ORDER BY id DESC) AS symbol_rn
FROM current
)
SELECT id, symbol
FROM CTE
WHERE symbol_rn = 2
ORDER BY id;
In MySql 7.5 you can simply self-join on the symbol, and group by.
Then the 2nd last will have 1 higher id.
SELECT c1.id, c1.symbol
FROM current c1
LEFT JOIN current c2
ON c2.symbol = c1.symbol
AND c2.id >= c1.id
GROUP BY c1.id, c1.symbol
HAVING COUNT(c2.id) = 2
ORDER BY c1.id;
id
symbol
5
B
6
A
7
C
db<>fiddle here
The performance will really benefit from an index on symbol.
You can try this;
SELECT *
FROM current
WHERE id
IN (SELECT MAX(id)
FROM current
GROUP BY symbol)
ORDER BY id DESC LIMIT 1,3
limit 1,3 says; get the last 3 results excluding the last result. You can change the numbers.
Is there way of sorting by frequency that a value occurs? If a value appears in multiple rows, would we just use the WHERE clause? Is it just about making the query more specific?
As a simple example:
CREATE TABLE mytable
( id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY
, val VARCHAR(15) NOT NULL
);
INSERT INTO mytable (id, val) VALUES
(1,'one')
,(2,'prime')
,(3,'prime')
,(4,'square')
,(5,'prime')
,(6,'six')
,(7,'prime')
,(8,'cube')
,(9,'square')
;
We can write a simple query to return the rows
SELECT t.val
, t.id
FROM mytable t
ORDER BY t.val
But what query do we use to get the most frequently occurring values listed first? To return a result like this:
freq val id
---- ------ --
4 prime 2
4 prime 3
4 prime 5
4 prime 7
2 square 4
2 square 9
1 cube 8
1 one 1
1 six 6
where freq is the frequency (the count of the number of rows) that a value appears in the val column. The value 'prime' appears in four rows, so freq has a value of 4.
What MySQL SELECT query would I use to return a result like this?
Try this:
SELECT A.Freq , A.val , A.id
FROM ( SELECT COUNT(*) AS Freq , val , id
FROM mytable
GROUP BY val , id ) A
ORDER BY Freq DESC ;
EDIT:
As suggested by spencer7593, the id is defined as auto-increment in the table and hence the GROUP BY should not include it. Still, if that would be the case, it is not clear how the result could be as shown. I'm adding here an alternative SELECT that, supposedly, should yield the shown output:
SELECT B.Freq , A.val , A.id
FROM mytable A
INNER JOINT ( SELECT val , COUNT(*) AS Freq
FROM mytable
GROUP BY val) B
ON A.val = B.val
ORDER BY B.Freq DESC ;
[NOTE: This was NOT tested!!!!]
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.
We read values from a set of sensors, occasionally a reading or two is lost for a particular sensor , so now and again I run a query to see if all sensors have the same record count.
GROUP BY sensor_id HAVING COUNT(*) != xxx;
So I run a query once to visually get a value of xxx and then run it again to see if any vary.
But is there any clever way of doing this automatically in a single query?
You could do:
HAVING COUNT(*) != (SELECT MAX(count) FROM (
SELECT COUNT(*) AS count FROM my_table GROUP BY sensor_id
) t)
Or else group again by the count in each group (and ignore the first result):
SELECT count, GROUP_CONCAT(sensor_id) AS sensors
FROM (
SELECT sensor_id, COUNT(*) AS count FROM my_table GROUP BY sensor_id
) t
GROUP BY count
ORDER BY count DESC
LIMIT 1, 18446744073709551615
SELECT sensor_id,COUNT(*) AS count
FROM table
GROUP BY sensor_id
ORDER BY count
Will show a list of the sensor_id along with a count of all the records it has, you can then manually check to see if any vary.
SELECT * FROM (
SELECT sensor_id,COUNT(*) AS count
FROM table
GROUP BY sensor_id
) AS t1
GROUP BY count
Will show all the counts that vary, but the group by will lose information about which sensor_ids have which counts.
---EDIT---
Taken a bit from both mine and eggyal's answer and created this, for the count that is most frequent I call the id default, and then for any values that stand out I have given them separate rows. This way you maintain the readability of a table if you have many results Multi Row, but also have a simple one row column if all counts are the same One Row. If however you are happy with the concocted strings then go with eggyal's answer.
Might be a bit over the top but here goes:
select 'default' as id,t5.c1 as count from(
select id,count(*) as c1 from your_table group by id having count(*)=
(select t4.count from
(
select max(t3.count2) as max,t3.count as count from
(
select count(*) as count2,t2.count from
(
SELECT id,COUNT(*) AS count
FROM your_table
GROUP BY id
) as t2
GROUP BY count
) as t3
) as t4)) as t5 group by count
union all
select t5.id as id,t5.c1 as count from(
select id,count(*) as c1 from your_table group by id having count(*)<>
(select t4.count from
(
select max(t3.count2) as max,t3.count as count from
(
select count(*) as count2,t2.count from
(
SELECT id,COUNT(*) AS count
FROM your_table
GROUP BY id
) as t2
GROUP BY count
) as t3
) as t4)) as t5
select sum(value) as 'Value',max(value)
from table_name where sum(value)=max(sum(value)) group by id_name;
The error is: Invalid use of group function (ErrorNr. 1111)
Any idea?
Thanks.
Can you maybe try
SELECT Value, MXValue
FROM (
select sum(value) as 'Value',max(value) MXValue
from table_name
group by id_name
) as t1
order by value desc
LIMIT 0,1
From MySQL Forums :: General :: selecting MAX(SUM())
Or you could try something like
SELECT id_name,
Value
FROM (
select id_name,sum(value) as 'Value'
from table_name
group by id_name
) t
WHERE Value = (
SELECT TOP 1 SUM(Value) Mx
FROM table_name
GROUP BY id_name
ORDER BY SUM(Value) DESC
)
Or even with an Inner join
SELECT id_name,
Value
FROM (
select id_name,sum(value) as Value
from table_name
group by id_name
) t INNER JOIN
(
SELECT TOP 1 SUM(Value) Mx
FROM table_name
GROUP BY id_name
ORDER BY SUM(Value) DESC
) m ON Value = Mx
The =max(sum(value)) part requires comparing the results of two grouped selects, not just one. (The max of the sum.)
Let's step back, though: What information are you actually trying to get? Because the sum of the values in the table is unique; there is no minimum or maximum (or, depending on your viewpoint, there is -- the value is its own minimum and maximum). You'd need to apply some further criteria in there for the results to be meaningful, and in doing so you'd probably need to be doing a join or a subselect with some criteria.