I'm a beginner at SQL, how do I get a query which returns the most prevalent column value? Probably there is an answer somewhere but I don't know how to google it.
For example in the user_id column the query should return the value 1 because this is the most prevalent number.
One approach is to do a GROUP BY aggregation and then apply a LIMIT trick:
SELECT user_id, COUNT(*) AS cnt
FROM yourTable
GROUP BY user_id
ORDER BY COUNT(*) DESC
LIMIT 1;
If you want something more complex, then you would be getting into the realm of rank functionality. MySQL (at least as of the current release) does not support built-in rank support, so it can be tricky to perform such queries.
SELECT top 1 user_id, COUNT(*) AS cnt
FROM yourTable
GROUP BY user_id
ORDER BY COUNT(*) DESC
Have a common table expression that counts each user_id. Select user_id where the count is the max count. Will return both user_id's in case of a tie.
with cte as
(
SELECT user_id, COUNT(*) AS cnt
FROM yourTable
GROUP BY user_id
)
select user_id
from cte
where cnt = (select max(cnt) from cte)
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 :)
I am having trouble writing a query for the following problem. I have tried some existing queries but cannot get the results I need.
I have a results table like this:
userid score timestamp
1 50 5000
1 100 5000
1 400 5000
1 500 5000
2 100 5000
3 1000 4000
The expected output of the query is like this:
userid score
3 1000
1 1000
2 100
I want to select a top list where I have n best scores summed for each user and if there is a draw the user with the lowest timestamp is highest. I really tried to look at all old posts but could not find one that helped me.
Here is what I have tried:
SELECT sum(score) FROM (
SELECT score
FROM results
WHERE userid=1 ORDER BY score DESC LIMIT 3
) as subquery
This gives me the results for one user, but I would like to have one query that fetches all in order.
This is a pretty typical greatest-n-per-group problem. When I see those, I usually use a correlated subquery like this:
SELECT *
FROM myTable m
WHERE(
SELECT COUNT(*)
FROM myTable mT
WHERE mT.userId = m.userId AND mT.score >= m.score) <= 3;
This is not the whole solution, as it only gives you the top three scores for each user in its own row. To get the total, you can use SUM() wrapped around that subquery like this:
SELECT userId, SUM(score) AS totalScore
FROM(
SELECT userId, score
FROM myTable m
WHERE(
SELECT COUNT(*)
FROM myTable mT
WHERE mT.userId = m.userId AND mT.score >= m.score) <= 3) tmp
GROUP BY userId;
Here is an SQL Fiddle example.
EDIT
Regarding the ordering (which I forgot the first time through), you can just order by totalScore in descending order, and then by MIN(timestamp) in ascending order so that users with the lowest timestamp appears first in the list. Here is the updated query:
SELECT userId, SUM(score) AS totalScore
FROM(
SELECT userId, score, timeCol
FROM myTable m
WHERE(
SELECT COUNT(*)
FROM myTable mT
WHERE mT.userId = m.userId AND mT.score >= m.score) <= 3) tmp
GROUP BY userId
ORDER BY totalScore DESC, MIN(timeCol) ASC;
and here is an updated Fiddle link.
EDIT 2
As JPW pointed out in the comments, this query will not work if the user has the same score for multiple questions. To settle this, you can add an additional condition inside the subquery to order the users three rows by timestamp as well, like this:
SELECT userId, SUM(score) AS totalScore
FROM(
SELECT userId, score, timeCol
FROM myTable m
WHERE(
SELECT COUNT(*)
FROM myTable mT
WHERE mT.userId = m.userId AND mT.score >= m.score
AND mT.timeCol <= m.timeCol) <= 3) tmp
GROUP BY userId
ORDER BY totalScore DESC, MIN(timeCol) ASC;
I am still working on a solution to find out how to handle the scenario where the userid, score, and timestamp are all the same. In that case, you will have to find another tiebreaker. Perhaps you have a primary key column, and you can choose to take a higher/lower primary key?
Query for selecting top three scores from table.
SELECT score FROM result
GROUP BY id
ORDER BY score DESC
LIMIT 3;
Can you please try this?
SELECT score FROM result GROUP BY id ORDER BY score DESC, timestamp ASC LIMIT 3;
if 2 users have same score then it will set order depends on time.
You can use a subquery
SELECT r.userid,
( SELECT sum(r2.score)
FROM results r2
WHERE r2.userid = r.userid
ORDER BY score DESC
LIMIT 3
) as sub
FROM result r
GROUP BY r.userid
ORDER BY sub desc
You should do it like this
SELECT SUM(score) as total, min(timestamp) as first, userid FROM scores
GROUP BY userid
ORDER BY total DESC, first ASC
This is way more efficient than sub queries. If you want to extract more fields than userid, then you need to add them to the group by.
This will of cause not limit the number of scores pr user, which indeed seems to require a subquery to solve.
please forgive me if this has been answered, but could not find it using the search tool or a basic google query.
I am trying to return a value that indicates the maximum number of rows any distinct value in a column in SQL.
For example, I'd like to use something like
SELECT MAX(COUNT(DISTINCT person_id) AS MAX_NUM_PERS_ROW
FROM mytable
and if the person with most rows in the table had 5 rows, the value returned would be 5...
Any and all help is appreciated!
You can do this with nested aggregation:
select max(cnt)
from (select person_id, count(*) as cnt
from mytable
group by person_id
) p;
If you actually want the person, you can also do:
select person_id, count(*) as cnt
from mytable
group by person_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 need a MySql query to do the following:
Select all the items in a table, group them by a type, order them by count type and also sum all the count.
This is what I've done so far:
select type, count(*) as cnt from $tbl_name group by type order by count(*) desc;
This only gives me the count for each group. What should I add so that this code will also show the total count (sum the counts from every group).
select * from (
select type as type,
count(*) as cnt
from $tbl_name
group by type WITH ROLLUP) as inner_table
order by cnt desc;
Note that the first row will be the rolled up total sum.
ROLLUP reference
Try this:
select type, SUM(count(*)) as cnt from $tbl_name group by type order by count(*) desc;