I need to assign jobs to users based on a score (number of "chances") calculated from previous jobs they have done. Here's my table of users:
user chances
Anna 6
Barry 4
Steve 3
Jackson 3
Helga 3
Maureen 3
Paul 3
Karen 2
Anita 2
Samson 2
Frank 2
Jean 1
Lilly 1
Boris 1
In another table, I have 100 rows of unassigned jobs (with currently NULL user), e.g.
id title user
1 Sort filing NULL
2 Clean office NULL
3 Order stationery NULL
I want to assign these jobs to the users above using a weighting based on the number of "chances" they have. For example, Anna will have 6 chances to be assigned one of these jobs, while Boris will have 1.
I've been playing around with a CASE which will assign a user to jobs, but nothing is satisfactory.
What's the best way for me to achieve this? Thanks
Presumably, you're after something like this...
SELECT user
FROM my_table
ORDER BY RAND() * chances * (SELECT SUM(chances) FROM my_table) DESC ;
If the changes are a small number and integers, then the easiest way might be:
update anothertable at
set user = (select user
from chances c cross join
(select 1 as n union all select 2 union all select 3 union all
select 4 union all select 5 union all select 6
) n
on c.chances <= n.n
where at.user is null
order by rand()
limit 1
);
The where clause is just so MySQL doesn't get the (brilliant) idea of optimizing away the subquery and only calling it once.
Related
People travel in pairs. How to find the maximum and minimum number of days between trips every user?
People:
id
user
1
Harry
2
George
3
Thomas
4
Jacob
5
Jack
6
Oliver
Travels:
id
date
user1
user2
1
2005-10-03
2
3
2
2005-10-04
1
4
3
2005-10-05
5
6
4
2005-10-06
1
3
5
2005-10-07
2
4
6
2005-10-08
3
5
7
2005-10-10
1
4
8
2005-10-11
5
2
9
2005-10-15
6
4
I tried to solve this problem in the following way, but I still do not understand how to solve this problem:
select People.id,People.user, count(*)
from People
INNER join
(SELECT MIN(TIMESTAMPDIFF(day, t1.date, t2.date)) as mintime,max(TIMESTAMPDIFF(day, t1.date, t2.date))
from Travels as t1
join Travels as t2 on t1.PERSON_1 = t2.PERSON_1
WHERE t1.date< t2.date
GROUP BY t1.PERSON_1) as t3
group by People.id
There is an idea to use the position function to iterate over each user, and then, as a result, look at the dates and find the minimum and maximum, but I still don't understand how to do this
Best is to do it in steps with subqueries, as below (comments are in the query):
select user, max(dateDiff)
from (
select
user,
-- get the diff between previous and current row date to get diff between trips
datediff(date, lag(date) over (partition by user order by date)) dateDiff
from (
-- full flat list of all users and trip dates
select date, user1 `user`
from testtbl
union all
select date, user2
from testtbl
) a
) a group by user
SQL fiddle
Note that I used windowed function which are not avaiable in MySql 5 and below.
I'm trying to create an sql (mariadb) request that select multiples columns but need two columns to be a unique pair but making sure the pair selected has its created_at value the least than the other duplicata pairs.
Here is what my table approximately looks like :
id
from_user_id
to_user_id
created_at
1
1
2
1000000005
2
2
1
1000000002
3
2
3
1000000008
4
5
6
999999999
5
6
5
100000006
I made this table precise to explain the request I want.
So I want to select the distinct pair (from_user_id, to_user_id) implying that the couple (1,2) which could also be (2,1) should be unique. The second rule is it should pick the couple with the minimum created_at value.
So the result table I want is :
id
from_user_id
to_user_id
created_at
2
2
1
1000000002
3
2
3
1000000008
4
5
6
999999999
2,1,1000000002 because the created_at is lesser than the other same couple case (1,2,1000000005).
In this case if I want only the values above created_at:999999999 to be selected I just have to add one condition.
I really hope my question is clear. I'm struggling to make distinct pairs work with other columns.
Thanks in advance for your answers.
WITH
cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY GREATEST(from_user_id,to_user_id),
LEAST(from_user_id,to_user_id)
ORDER BY created_at) rn
FROM table
)
SELECT *
FROM cte
WHERE rn = 1
I did some research and learned about the COALESCE(sum(num), 0) function. The issue is the example I found only related to using one table.
I am calculating a sum from a second table, and if there are no records for an item in the second table, I still want it to show up in my query and have a sum of zero.
SELECT note.user, note.product, note.noteID, note.note, COALESCE(sum(noteTable.Score), 0) as points
FROM note, noteTable
WHERE note.user <> 3 AND note.noteID = noteTable.noteID
I am only recieving results if there is an entry in the second table noteTable. If there are scores added for a note, I still want them to show up in the result with a points value of zero.
Table Examples:
Note
user | product | noteID |note
3 1 1 Great
3 2 2 Awesome
4 1 3 Sweet
NoteTable
noteID | score
1 5
The query should show me this:
user | noteID | sum(points)
3 1 5
3 2 0
4 3 0
But I am only getting this:
user | noteID | sum(points)
3 1 5
http://sqlfiddle.com/#!9/aae812/2
SELECT
note.user,
note.product,
note.noteID, note.note,
COALESCE(sum(noteTable.Score),0) as points
FROM note
LEFT JOIN noteTable
ON note.noteID = noteTable.noteID
WHERE note.user <> 3
and I guess you should add:
GROUP BY note.noteid
if you expect to get SUM for every user. So you want to get more then 1 record back.
First, learn to use proper JOIN syntax and table aliases. The answer to your question is SUM() along with COALESCE():
SELECT n.user, n.product, n.noteID, n.note,
COALESCE(sum(nt.Score), 0) as points
FROM note n LEFT JOIN
noteTable nt
ON n.noteID = nt.noteID
WHERE n.user <> 3
GROUP BY n.user, n.product, n.noteID, n.note;
You also need a GROUP BY.
I have a "resources" table that contains information about how resources of a specific weight are placed inside a territory by an user.
territory_id user_id weight
1 1 1
1 1 4
1 1 2
1 2 2
2 3 2
2 2 3
2 2 3
3 1 1
4 1 1
4 1 1
4 2 2
4 3 3
4 3 1
4 3 2
5 3 2
5 3 3
5 2 1
4 3 1
I want to calculate, for each existing territory, which user has the highest total weight of resources (and what is this value).
So this should be an expected outcome for the previous data:
territory_id best_user_id best_user_total_weight_of_resources
1 1 7
2 2 6
3 1 1
4 3 6
5 3 5
I have already tried several nested queries with SUM, MAX, GROUP BY but I really didn't find the proper way to calculate this.
I have found a lot of similiar question, but not solving this exact problem.
Any help? Thanks in advance!
EDIT:
I found out right now that the double GROUP BY (i.e. "GROUP BY territory_id, user_id") with double ORDER BY partially solves my problem, but it shows also information that I don't want (not only the best user, but each single user that placed at least one resource).
SELECT territory_id, user_id AS best_user_id, SUM( weight ) AS best_user_total_weight
FROM resources
GROUP BY territory_id, user_id
ORDER BY territory_id ASC, best_user_total_weight DESC;
You can run a first query to determine SUM(weight) for each couple (territory_id,user_id) and then run a second SELECT query on that result set to retrieve the row corresponding to max summ value:
SELECT territory_id, user_id, MAX(summ)
FROM (
SELECT territory_id, user_id, SUM(weight) AS summ
FROM resources
GROUP BY territory_id, user_id
) AS t1
GROUP BY territory_id
I have a table with columns similar to below , but with about 30 date columns and 500+ records
id | forcast_date | actual_date
1 10/01/2013 12/01/2013
2 03/01/2013 06/01/2013
3 05/01/2013 05/01/2013
4 10/01/2013 09/01/2013
and what I need to do is get a query with output similar to
week_no | count_forcast | count_actual
1 4 6
2 5 7
3 2 1
etc
My query is
SELECT weekofyear(forcast_date) as week_num,
COUNT(forcast_date) AS count_forcast ,
COUNT(actual_date) AS count_actual
FROM
table
GROUP BY
week_num
but what I am getting is the forcast_date counts repeated in each column, i.e.
week_no | count_forcast | count_actual
1 4 4
2 5 5
3 2 2
Can any one please tell me the best way to formulate the query to get what I need??
Thanks
try:
SELECT weekofyear(forcast_date) AS week_forcast,
COUNT(forcast_date) AS count_forcast, t2.count_actual
FROM
t t1 LEFT JOIN (
SELECT weekofyear(actual_date) AS week_actual,
COUNT(forcast_date) AS count_actual
FROM t
GROUP BY weekOfYear(actual_date)
) AS t2 ON weekofyear(forcast_date)=week_actual
GROUP BY
weekofyear(forcast_date), t2.count_actual
sqlFiddle
You have to write about 30 (your date columns) left join, and the requirement is that your first date column shouldn'd have empty week (with a count of 0) or the joins will miss.
Try:
SELECT WeekInYear, ForecastCount, ActualCount
FROM ( SELECT A.WeekInYear, A.ForecastCount, B.ActualCount FROM (
SELECT weekofyear(forecast_date) as WeekInYear,
COUNT(forecast_date) as ForecastCount, 0 as ActualCount
FROM TableWeeks
GROUP BY weekofyear(forecast_date)
) A
INNER JOIN
( SELECT * FROM
(
SELECT weekofyear(forecast_date) as WeekInYear,
0 as ForecastCount, COUNT(actual_date) as ActualCount
FROM TableWeeks
GROUP BY weekofyear(actual_date)
) ActualTable ) B
ON A.WeekInYear = B.WeekInYear)
AllTable
GROUP BY WeekInYear;
Here's my Fiddle Demo
Just in case someone else comes along with the same question:
Instead of trying to use some amazing query, I ended up creating an array of date_columns_names and a loop in the program that was calling this query, and for each date_column_name, performing teh asme query. It is a bit slower, but it does work