I have a table RacesTimes with columns RaceID and TotalTime, how I can retrieve the lowest TotalTime for every RaceID ?
I tried (few other variations too, but no result)
SELECT *
FROM `RacesTimes`
GROUP BY `RaceID`
ORDER BY `TotalTime` ASC
but it gives me the first inserted row from that RaceID, not the lowest TotalTime row from that RaceID.
SELECT MIN(TotalTime), RaceID
FROM RacesTimes
GROUP BY RaceID
ORDER BY MIN(TotalTime) ASC
Note that MySQL allows the syntax you did have: other RDBMS do not.
This says "give me minimum TotalTime per RacedID"
You need an aggregate function, specifically MIN:
SELECT RaceID, MIN(TotalTime)
FROM RaceTimes
GROUP BY RaceID
I have few others columns too in the table, how I can extract them all (using the MIN column grouped) ?
Related
I have table_schedule(id,instructorNumber,time,date,cpNumber,user_id);
Is it possible to output the values of instructorNumber,time, etc. with the same highest occurence with same values?
sorry I am new in sql, so I am hoping that someone can help me to the query
Just group by all the fields you need for "same value comparison", order by count desc (so the result with most occurences will be first), and take first.
select
instructorNumber, time, date, cpNumber
from table_schedule
group by instructorNumber, time, date, cpNumber
order by count(*) desc
LIMIT 1
you may use this as a join on a main query if you want more than one result.
First group by the values you want to compare by and count. Get the maximum count (in MySQL you can use LIMIT for that). Then do the same group-by query again and take only the results HAVING the maximum count. Use GROUP_CONCAT to get a list of IDs in a string.
select instructorNumber, time, date, cpNumber, user_id, group_concat(id) as ids
from table_schedule
group by instructorNumber, time, date, cpNumber, user_id
having count(*) =
(
select count(*)
from table_schedule
group by instructorNumber, time, date, cpNumber, user_id
order by count(*) desc limit 1
);
I have a database which consists of 5 columns
Each column is an INT
I want to find out which numbers occur most frequently in each column
I also wanted to know which sequence of numbers occurs more frequently
The same thing is true with the least common
I would like to use MYSQL or SQLITE
assuming schema as follows
tbl(..., int_col1, int_col2, int_col3, int_col4, int_col5, ...)
most frequently
do following query for each column
SELECT int_col1, COUNT(int_col1)
FROM tbl
GROUP BY int_col1
ORDER BY COUNT(int_col) DESC LIMIT 1
least frequently
do following query for each column
SELECT int_col1, COUNT(int_col1)
FROM tbl
GROUP BY int_col1
ORDER BY COUNT(int_col) ASC LIMIT 1
sequence of numbers occurs more frequently
SELECT int_col1, int_col2, int_col3, int_col4, int_col5, COUNT(*)
FROM tbl
GROUP BY int_col1, int_col2, int_col3, int_col4, int_col5
ORDER BY COUNT(*) DESC;
assuming column is userid
select UserID, count(UserID)
from myUsers
group by UserID
order by count(UserID) desc
OR
with cte as
(
select user_id,ROW_NUMBER() over (order by UserID) as rn
)
select user_id,count(user_id) as se_count from cte group by user_id
I have the following query:
SELECT t.ID, t.caseID, time
FROM tbl_test t
INNER JOIN (
SELECT ID, MAX( TIME )
FROM tbl_test
WHERE TIME <=1353143351
GROUP BY caseID
ORDER BY caseID DESC -- ERROR HERE!
) s
USING (ID)
It seems that I only get the correct result if I use the ORDER BY in the inner join. Why is that? I am using the ID for the join, so the order should take no effekt.
If I remove the order by, I get too old entries from the database.
ID is the primary key, the caseID is a kind of object with multiple entries with different timestamps.
This query is ambiguous:
SELECT ID, MAX( TIME )
FROM tbl_test
WHERE TIME <=1353143351
GROUP BY caseID
It's ambiguous because it does not guarantee that it returns the ID of the row where the MAX(TIME) occurs. It returns the MAX(TIME) for each distinct value of caseID, but the value of other columns (like ID) is chosen arbitrarily from members of the group.
In practice, MySQL chooses the row that it finds first in the group as it scans rows in storage order.
Example:
caseID ID time
1 10 15:00
1 12 18:00
1 14 13:00
The max time is 18:00, which is the row with ID 12. But the query will return ID 10, simply because it's the first one in the group. If you were to reverse the order with ORDER BY, it would return ID 14. Still not the row where the max time is found, but it's from the other end of the group of rows.
Your query works with ORDER BY caseID DESC because, by coincidence, your Time values increase with the increasing ID.
This sort of query is actually an error in standard SQL and most other brands of SQL database. MySQL permits it, trusting that you know how to form an unambiguous query.
The fix is to use columns in the select-list only if they are unambiguous, that is, if they are in the GROUP BY clause, then each group is guaranteed to have only one distinct value:
SELECT caseID, MAX( TIME )
FROM tbl_test
WHERE TIME <=1353143351
GROUP BY caseID
SELECT t.ID, t.caseID, time
FROM tbl_test t
INNER JOIN (
SELECT caseID, MAX( TIME ) maxtime
FROM tbl_test
WHERE TIME <=1353143351
GROUP BY caseID
) s
ON t.caseID = s.caseID and t.time = s.maxtime
You are seeing that issue because you are getting the MAX(TIME) per caseID, but since you are grouping by caseID and NOT ID, you are getting an arbitrary ID. That happens because when you use an aggregate function, like MAX, you must, for every non-grouped field in the select specify how you want to aggregate it. That means, if it's in the SELECT and NOT in the GROUP BY, you have to tell MySQL how to aggregate. If you don't then you get a RANDOM row (well, not random per se, but it's not going to be in an order that you necessarily expect).
The reason ORDER BY is working for you, is that it kind of tricks the query optimizer into sorting the results before grouping, which just so happens to produce the result you want, but be warned, that will not always be the case.
What you want is the ID that has the MAX(TIME) given a caseID. Which means your INNER join needs to connect by caseID (not ID) and time (which will give you 1 row per each 1 row in the outer table).
Barmar beat me to the actual query, but that's the way you want to go.
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