Select last value in SQL table with conditions - mysql

i have a sample data like this:
Now i want to get lastest data used to have user_id = 41 and highest price, output like this:
How can i do it with SQL command line ?
Thank for read


Try this query
select User_id,auction_id,price from tablename where price in(select price from tablename where id in(select max(id) from tablename group by user_id))


As per your output that you wish try this:
select user_id,auction_id,price
from table_name
where user_id=41
order by id desc;


If you have updated output incorrectly then try it as per your provided conditions-
select user_id,auction_id,price
from mytable
where user_id=41
order by auction_id;
But if as per your output you want 1 latest row of user_id=41 and then all rows of highest price then you can try it-
SELECT user_id,auction_id,price
FROM mytable
WHERE user_id=41
ORDER BY id DESC LIMIT 1
UNION
SELECT b.user_id,b.auction_id,price
FROM mytable b
JOIN (
SELECT MAX(price) AS price
FROM mytable
) a ON a.price=b.price;

Related

My sql select that has multiple row in the same criteria

I have mysql table like this
I want to get row that has minimum 2 or more than 2 (multiple) row only from this table, so the result would be like this
What do i do?
thank you

Use GROUP BY and HAVING clauses
SELECT t.* FROM my_table t
JOIN (
SELECT cust_id, MIN(transaction_no) AS transaction_no
FROM my_table
GROUP BY cust_id
HAVING COUNT(cust_id) > 1
) agg ON t.transaction_no = agg.transaction_no

how to find next max(id) in a table

lets say
My table contains
Id Status
1 0
2 1
3 0
4 0
5 1
6 0
I need output like
Id Status
5 1
I tried like Max(id) but it gives output as
id status
6 0

I can only suppose, that you want to know about the maximum id of those entries having Status=1, right? Then use
select max(id) from mytable where Status=1

Try this:
SELECT id, status
FROM myTable ORDER BY `id` DESC LIMIT 1 , 1
I have assumed that you are looking for the second highest Id record values.

I guess you would like to have the max(id) where Status=1?
Then use select max(id) from table where Status=1

Since you want to get the absolute max from both columns use :
SELECT GREATEST(MAX(id), MAX(status));

Your question is a bit unclear what it is you actually want. From your example, I guess you want the maximum ID for the max status.
This is the max id for every status:
select max(id), status from table
group by status;
This will result in 5,1 and 6,0.
You could then filter out what you don't need, e.g.
select * from (
select max(id), status from t
group by status
) maxidperstatus
where maxidperstatus.status = (select max(status) from t);

Try this also
select Id,Status
from table
where Id=(
select MAX(Id) from table where Id <> (select MAX(Id) from table)
)

I assume that you want the Id where the status is the biggest
Try this
select max(status),id from table group by id order by max(status) DESC

Mysql numbered rows in group

I have table
users|visits
--------------
user1|visit1
user1|visit2
user1|visit3
user2|visit1
user2|visit2
I want to get numbered rows in any group. How i can do this?
users|visits|visit number
-----------------------
user1|visit1|1
user1|visit2|2
user1|visit3|3
user2|visit1|1
user2|visit2|2

Try as below :
SELECT
users, visits, count(*) as `visit number`
FROM tableName group by visits,users

From what you have provide about your problem i can imagine that you have records something like this
users|visits
------------
user1|visit1
user1|visit2
user1|visit1
user1|visit3
And you want to get
users|visits|visit number
-------------------------
user1|visit1|2
user1|visit2|1
user1|visit3|1
To do that use query like this
SELECT
DISTINCT a.users,
a.visits,
(select count(*) from tableName b where b.users = a.users and b.visits = a.visits) as `visit number`
FROM tableName a

Query to fetch second highest salary from table [duplicate]

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.

UNION two ordered MySQL statements

I have a table with 3 columns (id, name, code) and 10 rows. Some of the rows don't have a code so that column is empty for some. What I'm trying to accomplish is SELECT the rows with code column not empty first ordered by last inserted followed by all rows with code column empty ordered by last inserted.
I have tried
(SELECT * from tablename WHERE code <> '' ORDER BY ID DESC) UNION
(SELECT * from tablename WHERE code = '' ORDER BY ID DESC)
The UNION works but the order does not. I have read here about other questions and found out adding ORDER BY like I added will not work and I should add it at the end but that would not help me accomplish what I want and will mix rows that have a code with rows that don't.
Is there a way to succeed with what I'm looking for?

I think you just need to put your sort logic in the ORDER BY clause
SELECT id, name, code
FROM tablename
ORDER BY code = '', ID desc;

Try this:
SELECT * FROM
(
(SELECT * from tablename WHERE code <> '' ORDER BY ID DESC)
UNION
(SELECT * from tablename WHERE code = '' ORDER BY ID DESC)
)tab ORDER BY ID DESC;
Or
SELECT * from tablename ORDER BY code DESC,ID DESC
Change ASC/DESC as per you want it to show