I have table with folowing structure. The content in this table is duplicate. I want to show the record with min price.
GROUP IDX NAME PRICE
141003 6 0285-00 499.88000
141003 2 028500 519.13000
141003 1 0285-00 424.12000
141005 2 02851 559.13000
141005 1 0285-1 434.12000
RESULT TABLE
141003 1 0285-00 424.12000
141005 1 0285-1 434.12000
I trying this but returned result is:
141003 6 0285-00 499.88000
Query:
select group, idx, name, price
FROM table
GROUP BY group
ORDER BY price;
Use the following query if you want to know the minimum price in each group:
select `group`, `idx`, `name`, MIN(price) AS min_price
FROM `table`
GROUP BY `group`;
Documentation for 'MIN' function
SELECT group, idx, name, price
FROM table
WHERE group = 141003
ORDER BY price ASC
LIMIT 1;
Would do the trick.
First of all group is a reserved word in MySQL. You should use ` around it or rename the table. Like this:
select `group`, idx, name, price from test_2 where `group` = 141003 order by price desc limit 1;
Why do you want to do group by if the group is in the the where condition with one value? Do you want the cheapest item from all the groups?
Related
I have this table
i want to ignore productNo and sum all product count accordingly.
select sum(count), max(productNo)
from Table
where date between 117 and 118
group by product
this one gives wrong result...
I want to have sum of counts for each Product-ProductNo combination
try like below
select product,productno,sum(count) as result
from table_name
where productno='X1'
group by product,productno
seems you need the firts rows order by result
select product,productno,sum(count) as result
from table
group by product,productno
order by result
limit 1
Since you haven't tagged any DBMS so, i would use row_number():
select t.*
from (select product, productno, sum(count) as cnt,
row_number() over (partition by product order by sum(count) desc) as seq
from table t
group by product, productno
) t
where seq = 1;
You can also use LIMIT clause (but not for each product) :
select product, productno, sum(count) as cnt
from table t
group by product, productno
order by cnt desc
limit 1;
Some other DBMS requires TOP clause instead of LIMIT clause so, you can change accordingly but the idea would be same.
select sum(count), max(productNo)
from Table
where date between 117 and 118
group by product, productNo
with this it works :)
From the following four records, I want to select the OwnerId of second-latest record
ItemId OwnerId Date
11477 20981 2013-05-13
11477 1 2013-05-21
11477 21086 2013-05-22 #this is the one I'm talking about
11477 3868 2013-05-24
How to go about it?
This needs ItemID to be specified,
SELECT *
FROM TableName
WHERE ItemID = '11477'
ORDER BY DATE DESC
LIMIT 1,1
SQLFiddle Demo
However, if you don't want to specify the ItemID, and you want to get all second latest record for every ItemID, you can use a correlated subquery to generate a sequence number for every ItemID based on lastest DATE,
SELECT ItemId, OwnerID, Date
FROM
(
SELECT A.ItemId,
A.OwnerId,
A.Date,
(
SELECT COUNT(*)
FROM tableName c
WHERE c.ItemId = a.ItemId AND
c.Date >= a.Date) AS RowNumber
FROM TableName a
) x
WHERE RowNumber = 2
SQLFiddle Demo
select ownerid
from your_table
order by date desc
limit 1, 1
I think you can just to ORDER BY date descending, which will give you an order from newer to older, then LIMIT 1,1 to get only the second result, which should be the one you look for
SELECT *
FROM table
ORDER BY date DESC
LIMIT 1,1
I'm trying to select rows where a certain column does not have a certain value, such as 0, but I'm unable to get this to work.
SELECT *
FROM rentals
GROUP BY date, rooms, price
HAVING show_independently < 1
If show_independently is 1, then I don't want to group them. However, this statement shows no rows even though most rows have show_independently as 0.
SELECT date, rooms, price
FROM rentals
WHERE show_independently < 1
GROUP BY date, rooms, price
If you only want to group some rows and leave others ungrouped, you can use a UNION:
SELECT *
FROM rentals
WHERE show_independently <> 1
GROUP BY date, rooms, price
UNION ALL
SELECT *
FROM rentals
WHERE show_independently = 1
This groups only those where show_independently is not 1, and includes the rest without grouping them.
A HAVING clause is used when you are using an aggregate function to filter data.
A typical query with HAVING:
SELECT yourColumn, aggregate_function(otherColumn)
FROM yourTable
WHERE yourColumn = someValue
GROUP BY yourColumn
HAVING aggregate_function(otherColumn) = someOtherValue
I think you want to be using a WHERE clause:
SELECT date, rooms, price
FROM rentals
WHERE show_independently < 1
GROUP BY date, rooms, price
I want to select data from a table in MySQL where a specific field has the minimum value, I've tried this:
SELECT * FROM pieces WHERE MIN(price)
Please any help?
this will give you result that has the minimum price on all records.
SELECT *
FROM pieces
WHERE price = ( SELECT MIN(price) FROM pieces )
SQLFiddle Demo
This is how I would do it, assuming I understand the question.
SELECT * FROM pieces ORDER BY price ASC LIMIT 1
If you are trying to select multiple rows where each of them may have the same minimum price, then #JohnWoo's answer should suffice.
Basically here we are just ordering the results by the price in ascending order (ASC) and taking the first row of the result.
This also works:
SELECT
pieces.*
FROM
pieces inner join (select min(price) as minprice from pieces) mn
on pieces.price = mn.minprice
(since this version doesn't have a where condition with a subquery, it could be used if you need to UPDATE the table, but if you just need to SELECT i would reccommend to use John Woo solution)
Use HAVING MIN(...)
Something like:
SELECT MIN(price) AS price, pricegroup
FROM articles_prices
WHERE articleID=10
GROUP BY pricegroup
HAVING MIN(price) > 0;
Efficient way (with any number of records):
SELECT id, name, MIN(price) FROM (select * from table order by price) as t group by id
In fact, depends what you want to get:
- Just the min value:
SELECT MIN(price) FROM pieces
A table (multiples rows) whith the min value: Is as John Woo said above.
But, if can be different rows with same min value, the best is ORDER them from another column, because after or later you will need to do it (starting from John Woo answere):
SELECT * FROM pieces
WHERE price = ( SELECT MIN(price) FROM pieces)
ORDER BY stock ASC
To improve #sberry's answer, if the column has a null value then simply doing ORDER BY would select a row with null value. Add a WHERE clause to get correct results:
SELECT * FROM pieces
WHERE price>0
ORDER BY price ASC
LIMIT 1;
Or if there is a chance of having negative values and/or VARCHAR, etc. do:
SELECT * FROM pieces
WHERE price IS NOT NULL
ORDER BY price ASC
LIMIT 1;
To make it simpler
SELECT *,MIN(price) FROM prod LIMIT 1
Put * so it will display the all record of the minimum value
How can I find the most frequent value in a given column in an SQL table?
For example, for this table it should return two since it is the most frequent value:
one
two
two
three
SELECT
<column_name>,
COUNT(<column_name>) AS `value_occurrence`
FROM
<my_table>
GROUP BY
<column_name>
ORDER BY
`value_occurrence` DESC
LIMIT 1;
Replace <column_name> and <my_table>. Increase 1 if you want to see the N most common values of the column.
Try something like:
SELECT `column`
FROM `your_table`
GROUP BY `column`
ORDER BY COUNT(*) DESC
LIMIT 1;
Let us consider table name as tblperson and column name as city. I want to retrieve the most repeated city from the city column:
select city,count(*) as nor from tblperson
group by city
having count(*) =(select max(nor) from
(select city,count(*) as nor from tblperson group by city) tblperson)
Here nor is an alias name.
Below query seems to work good for me in SQL Server database:
select column, COUNT(column) AS MOST_FREQUENT
from TABLE_NAME
GROUP BY column
ORDER BY COUNT(column) DESC
Result:
column MOST_FREQUENT
item1 highest count
item2 second highest
item3 third higest
..
..
For use with SQL Server.
As there is no limit command support in that.
Yo can use the top 1 command to find the maximum occurring value in the particular column in this case (value)
SELECT top1
`value`,
COUNT(`value`) AS `value_occurrence`
FROM
`my_table`
GROUP BY
`value`
ORDER BY
`value_occurrence` DESC;
Assuming Table is 'SalesLT.Customer' and the Column you are trying to figure out is 'CompanyName' and AggCompanyName is an Alias.
Select CompanyName, Count(CompanyName) as AggCompanyName from SalesLT.Customer
group by CompanyName
Order By Count(CompanyName) Desc;
If you can't use LIMIT or LIMIT is not an option for your query tool. You can use "ROWNUM" instead, but you will need a sub query:
SELECT FIELD_1, ALIAS1
FROM(SELECT FIELD_1, COUNT(FIELD_1) ALIAS1
FROM TABLENAME
GROUP BY FIELD_1
ORDER BY COUNT(FIELD_1) DESC)
WHERE ROWNUM = 1
If you have an ID column and you want to find most repetitive category from another column for each ID then you can use below query,
Table:
Query:
SELECT ID, CATEGORY, COUNT(*) AS FREQ
FROM TABLE
GROUP BY 1,2
QUALIFY ROW_NUMBER() OVER(PARTITION BY ID ORDER BY FREQ DESC) = 1;
Result:
Return all most frequent rows in case of tie
Find the most frequent value in mysql,display all in case of a tie gives two possible approaches:
Scalar subquery:
SELECT
"country",
COUNT(country) AS "cnt"
FROM "Sales"
GROUP BY "country"
HAVING
COUNT("country") = (
SELECT COUNT("country") AS "cnt"
FROM "Sales"
GROUP BY "country"
ORDER BY "cnt" DESC,
LIMIT 1
)
ORDER BY "country" ASC
With the RANK window function, available since MySQL 8+:
SELECT "country", "cnt"
FROM (
SELECT
"country",
COUNT("country") AS "cnt",
RANK() OVER (ORDER BY COUNT(*) DESC) "rnk"
FROM "Sales"
GROUP BY "country"
) AS "sub"
WHERE "rnk" = 1
ORDER BY "country" ASC
This method might save a second recount compared to the first one.
RANK works by ranking all rows, such that if two rows are at the top, both get rank 1. So it basically directly solves this type of use case.
RANK is also available on SQLite and PostgreSQL, I think it might be SQL standard, not sure.
In the above queries I also sorted by country to have more deterministic results.
Tested on SQLite 3.34.0, PostgreSQL 14.3, GitHub upstream.
Most frequent for each GROUP BY group
MySQL: MySQL SELECT most frequent by group
PostgreSQL:
Get most common value for each value of another column in SQL
https://dba.stackexchange.com/questions/193307/find-most-frequent-values-for-a-given-column
SQLite: SQL query for finding the most frequent value of a grouped by value
SELECT TOP 20 WITH TIES COUNT(Counted_Column) AS Count, OtherColumn1,
OtherColumn2, OtherColumn3, OtherColumn4
FROM Table_or_View_Name
WHERE
(Date_Column >= '01/01/2023') AND
(Date_Column <= '03/01/2023') AND
(Counted_Column = 'Desired_Text')
GROUP BY OtherColumn1, OtherColumn2, OtherColumn3, OtherColumn4
ORDER BY COUNT(Counted_Column) DESC
20 can be changed to any desired number
WITH TIES allows all ties in the count to be displayed
Date range used if date/time column exists and can be modified to search a date range as desired
Counted_Column 'Desired_Text' can be modified to only count certain entries in that column
Works in INSQL for my instance
One way I like to use is:
select *<given_column>*,COUNT(*<given_column>*)as VAR1 from Table_Name
group by *<given_column>*
order by VAR1 desc
limit 1