I have three tables.
For each "id" value, I would like the sum of the col1 values, the sum of col2 values & the sum of col3 values listed separately. I am not summing across tables.
table a
num | id | col1
================
1 100 0
2 100 1
3 100 0
1 101 1
2 101 1
3 101 0
table b
idx | id | col2
=================
1 100 20
2 100 20
3 100 20
4 101 100
5 101 100
table c
idx | id | col3
==============================
1 100 1
2 100 1
3 100 1
4 101 10
5 101 1
I would like the results to look like this,
ID | sum_col1 | sum_col2 | sum_col3
====================================
100 1 60 3
101 2 200 11
Here is my query which runs too long and then times out. My tables are about 25,000 rows.
SELECT a.id as id,
SUM(a.col1) as sum_col1,
SUM(b.col2) as sum_col2,
SUM(c.col3) as sum_col3
FROM a, b, c
WHERE a.id=b.id
AND a=id=c.id
GROUP by id
Order by id desc
The number of rows in each table may be different, but the range of "id" values in each table is the same.
This appears to be a similar question, but I can't make it work,
Mysql join two tables sum, where and group by
Here is a solution based on your data. Issue with your query is that you were joining tables on a non-unique column resulting in Cartesian product.
Data
DROP TABLE IF EXISTS A;
CREATE TABLE A
(num int,
id int,
col1 int);
INSERT INTO A VALUES (1, 100, 0);
INSERT INTO A VALUES (2, 100, 1);
INSERT INTO A VALUES (3, 100, 0);
INSERT INTO A VALUES (1, 101, 1);
INSERT INTO A VALUES (2, 101, 1);
INSERT INTO A VALUES (3 , 101, 0);
DROP TABLE IF EXISTS B;
CREATE TABLE B
(idx int,
id int,
col2 int);
INSERT INTO B VALUES (1, 100, 20);
INSERT INTO B VALUES (2, 100, 20);
INSERT INTO B VALUES (3, 100, 20);
INSERT INTO B VALUES (4, 101, 100);
INSERT INTO B VALUES (5, 101, 100);
DROP TABLE IF EXISTS C;
CREATE TABLE C
(idx int,
id int,
col3 int);
INSERT INTO C VALUES (1, 100, 1);
INSERT INTO C VALUES (2, 100, 1);
INSERT INTO C VALUES (3, 100, 1);
INSERT INTO C VALUES (4, 101, 10);
INSERT INTO C VALUES (5, 101, 1);
Solution
SELECT a_sum.id, col1_sum, col2_sum, col3_sum
FROM (SELECT id, SUM(col1) AS col1_sum
FROM a
GROUP BY id ) a_sum
JOIN
(SELECT id, SUM(col2) AS col2_sum
FROM b
GROUP BY id ) b_sum
ON (a_sum.id = b_sum.id)
JOIN
(SELECT id, SUM(col3) AS col3_sum
FROM c
GROUP BY id ) c_sum
ON (a_sum.id = c_sum.id);
Result is as expected
Note: Do outer joins if an id doesnt have to be present in all three tables.
Maybe this will do?
Haven't got a chance to run it, but i think it can do the job.
SELECT sumA.id, sumA.sumCol1, sumB.sumCol2, sumC.sumCol3
FROM
(SELECT id, SUM(col1) AS sumCol1 FROM a GROUP BY id ORDER BY id ASC) AS sumA
JOIN (SELECT id, SUM(col2) AS sumCol2 FROM b GROUP BY id ORDER BY id ASC) AS sumB ON sumB.id = sumA.id
JOIN (SELECT id, SUM(col3) AS sumCol3 FROM c GROUP BY id ORDER BY id ASC) AS sumC ON sumC.id = sumB.id
;
EDIT
SELECT IF(sumA.id IS NOT NULL, sumA.id, IF(sumB.id IS NOT NULL, sumB.id, IF(sumC.id IS NOT NULL, sumC.id,''))),,
sumA.sumCol1, sumB.sumCol2, sumC.sumCol3
FROM
(SELECT id, SUM(col1) AS sumCol1 FROM a GROUP BY id ORDER BY id ASC) AS sumA
OUTER JOIN (SELECT id, SUM(col2) AS sumCol2 FROM b GROUP BY id ORDER BY id ASC) AS sumB ON sumB.id = sumA.id
OUTER JOIN (SELECT id, SUM(col3) AS sumCol3 FROM c GROUP BY id ORDER BY id ASC) AS sumC ON sumC.id = sumB.id
;
I would do the summing first, then union the results, then pivot them round:
SELECT
id,
MAX(CASE WHEN which = 'a' then sumof end) as sum_a,
MAX(CASE WHEN which = 'b' then sumof end) as sum_b,
MAX(CASE WHEN which = 'c' then sumof end) as sum_c
FROM
(
SELECT id, sum(col1) as sumof, 'a' as which FROM a GROUP BY id
UNION ALL
SELECT id, sum(col2) as sumof, 'b' as which FROM b GROUP BY id
UNION ALL
SELECT id, sum(col3) as sumof, 'c' as which FROM c GROUP BY id
) a
GROUP BY id
You could also union, then sum:
SELECT
id,
SUM(CASE WHEN which = 'a' then v end) as sum_a,
SUM(CASE WHEN which = 'b' then v end) as sum_b,
SUM(CASE WHEN which = 'c' then v end) as sum_c
FROM
(
SELECT id, col1 as v, 'a' as which FROM a GROUP BY id
UNION ALL
SELECT id, col2 as v, 'b' as which FROM b GROUP BY id
UNION ALL
SELECT id, col3 as v, 'c' as which FROM c GROUP BY id
) a
GROUP BY id
You cant easily use a join, unless all tables have all values of ID, in which case I'd say you can sum them as subqueries and then join the results together.. But if one of your tables suddenly lacks an id value that the other two tables have, that row disappears from your results (unless you use full outer join and some really ugly coalescing in your ON clause)
Using union in this case will give you a more missing-value-tolerant result set, as it can cope with missing values of ID in any table. Thus, we union the tables together into one dataset, but use a constant to track which table the value came from, that way we can pick it out into its own summation later
If any id value is not present in any table, then the sum for that column will be null. If you want it to be 0, you can change the MAX to SUM or wrap the MAX in a COALESCE
Related
I have table like this
mapname id time
---------------------
cvjourney 1 65.24
cvjourney 3 69.45
cvjourney 2 64.25
cvjourney 8 75.45
dark_hop 2 65.87
dark_hop 1 61.51
coldrunss 1 12.50
I want to select data from that table to get something like this (for id 1, 2)
cvjourney 1 65.24
cvjourney 2 64.25
dark_hop 2 65.87
dark_hop 1 61.51
so I want select data where two players are both having records on same map, I dont want select maps where player[1] has record and player[2] doesnt, or the other way.
Here is my select, but I cant get rid of maps where one player has record and another doesnt.
SELECT *
FROM table
WHERE mapname IN (
SELECT mapname
FROM table
WHERE id IN ('1','2')
)
AND id IN ('1', '2')
ORDER BY mapname
Can you help me, please?
You can use aggregation to get only those mapnames which have both the ids present (which means after applying the id filter, distinct count of ids will be two) and then use that to get the relevant rows.
select t1.*
from table t1
join (
select mapname
from table
where id in (1, 2)
group by mapname
having count(distinct id) = 2
) t2 on t1.mapname = t2.mapname
where t1.id in (1, 2)
order by t1.mapname
Use can use IN too:
select t1.*
from table t1
where mapname in (
select mapname
from table
where id in (1, 2)
group by mapname
having count(distinct id) = 2
)
and id in (1, 2)
order by mapname
Demo
I have a table:
create table #t
(
ID int,
value nvarchar(5)
)
insert #t
values (1,'A'), (2, 'B'), (3, 'A'), (3, 'B')
Sample data:
ID value
------------
1 A
2 B
3 A
3 B
For my project I need the ID which has having both the values
Result :
ID
3
Kindly help me out.
To get IDs having 2 values
select id
from #t
group by id
having count(distinct value) >= 2
or to get all IDs having A and B
select id
from #t
where value in ('A','B')
group by id
having count(distinct value) = 2
or to make it more generic to get IDs having all values
select id
from #t
group by id
having count(distinct value) = (select count(distinct value) from #t)
I have the following question.
I have to columns and I want for each unique entry the first column the most frequent element of the second column. An Example would be:
COL A COL B
1 a
2 c
2 c
1 a
1 b
2 d
The query should output:
Col A COL B
1 a
2 c
Given this sample data:
CREATE TABLE t
(`a` int, `b` varchar(1))
;
INSERT INTO t
(`a`, `b`)
VALUES
(1, 'a'),
(2, 'c'),
(2, 'c'),
(1, 'a'),
(1, 'b'),
(2, 'd')
;
you first have to get the count for each with a query like this:
SELECT
a, b,
COUNT(*) AS amount
FROM
t
GROUP BY
a, b
Then you can use this query as a subquery to get the rows where a certain column holds the maximum. There's a nice article about this in the manual: The Rows Holding the Group-wise Maximum of a Certain Column
Choosing for example the last method described in said article, your final query would be this:
SELECT sq1.a, sq1.b FROM
(
SELECT
a, b,
COUNT(*) AS amount
FROM
t
GROUP BY
a, b
) sq1
LEFT JOIN
(
SELECT
a, b,
COUNT(*) AS amount
FROM
t
GROUP BY
a, b
) sq2 ON sq1.a = sq2.a AND sq1.amount < sq2.amount
WHERE sq2.amount IS NULL;
With this result:
+------+------+
| a | b |
+------+------+
| 1 | a |
| 2 | c |
+------+------+
Using a subselect in the SELECT clause is a simple way to solve your problem:
SELECT ColA, (SELECT ColB
FROM yourtable i
WHERE i.ColA = o.ColA
GROUP BY ColB
ORDER BY COUNT(*) DESC
LIMIT 1) AS ColB
FROM yourtable o
GROUP BY ColA;
o is just an alias for the outer query, i for the inner query. They're needed for the WHERE clause to work.
The above query is a result of the following query to find the most common occurence of ColB with a given ColA:
SELECT ColB
FROM yourtable
WHERE ColA = 1 -- Replace 1; this is where the magic happens in the above query
GROUP BY ColB
ORDER BY COUNT(*) DESC
LIMIT 1
you have to create My Sql function for this let say you function name is getMaxOccr so it will look like
CREATE FUNCTION `getMaxOccr`(val INT)
RETURNS varchar(25) CHARSET latin1
BEGIN
DECLARE answer VARCHAR(25) DEFAULT '';
SELECT colb FROM `tablename`
WHERE cola = val
ORDER BY COUNT(colb)
DESC INTO answer;
RETURN answer;
END
Once this function is created than you simply have to call
SELECT cola,getMaxOccr(cola) from tablename GROUP BY cola
this will give you the list with what you are looking for hope this helps
x = doesn't matter
table A
0 row0
1 row1
2 row2
table B
0 x
1 x
0 X
2 X
2 X
0 x
there are in this example 3 rows of 0, 2 rows of 2 and 1 row 1.
i want to get for example the two rows who has the highest count of rows.
desired result:
0 row0 ==> because 3 rows in b is the highest amount.
2 row2 ==> because 2 rows in b is the second highest amount.
my attempt so far:
SELECT Id, Name FROM A
WHERE Id =
(
SELECT IdB FROM B
GROUP BY IdB
ORDER BY count(IdB) DESC
LIMIT 2
)
edit: i use mysql
thanks
You want to JOIN your tables together by id:
SELECT
A.id,
A.name
FROM
A join
B on A.id = B.id
GROUP BY
A.name
ORDER BY
count(B.id) desc
LIMIT 2
SQLFIDDLE
That will return an output that matches your example above.
Not sure how you want to handle ties.
What this does is JOIN (learn more about joins) the two tables together based on your ID's then aggregate the results counting the occurrences of B.IDB and displaying that count along with the name from table A
SELECT A.Name, count(B.IDB) cnt
FROM A
INNER JOIN IDB
on A.ID = B.IDB
GROUP BY A.Name
ORDER BY count(B.IDB) desc
LIMIT 2
But with your example this should return
Name cnt
row0 3
row2 2
-- Sample table
-- table A schema
create table A (id int, name varchar(10))
insert into A
values (0, 'row0')
insert into A
values (1, 'row1')
insert into A
values (2, 'row2')
select * from A
-- table A schema
create table B (id int, name varchar(10))
insert into B
values (0, 'X')
insert into B
values (0, 'X')
insert into B
values (0, 'X')
insert into B
values (0, 'X')
insert into B
values (1, 'X')
insert into B
values (2, 'X')
insert into B
values (2, 'X')
insert into B
values (2, 'X')
insert into B
values (2, 'X')
insert into B
values (2, 'X')
insert into B
values (2, 'X')
select * from B
--Query
select Top 2 A.name, count(B.id) from A
inner join B on A.id = b.id
group by A.name
order by count(A.name) desc
You can try ;
select a.name, b.count
from A a,
(select idB, count(idB) count from B group by idB) b
where a.id = b.idB
and count > 1
I have the following mysql table:
id | member |
1 | abc
1 | pqr
2 | xyz
3 | pqr
3 | abc
I have been trying to write a query which would return the id which has exact same members as a given id. For example, if given id is 1 then the query should return 3 because both id 1 and id 3 have exact same members viz. {abc, pqr}. Any pointers? Appreciate it.
EDIT: The table may have duplicates, e.g. id 3 may have members {abc, abc} instead of {pqr, abc}, in which case the query should not return id 3.
Here's a solution that finds matching pairs for the entire table - you can add a where clause to filter as needed. Basically it does a self-join based on equal "member" and unequal "id". It then compares the resulting count grouped by the 2 ids and compares them to the total count of those ids from the original table. If they both match, it means they have the same exact members.
select
t1.id, t2.id
from
table t1
inner join table t2
on t1.member = t2.member
and t1.id < t2.id
inner join (select id, count(1) as cnt from table group by id) c1
on t1.id = c1.id
inner join (select id, count(1) as cnt from table group by id) c2
on t2.id = c2.id
group by
t1.id, t2.id, c1.cnt, c2.cnt
having
count(1) = c1.cnt
and count(1) = c2.cnt
order by
t1.id, t2.id
This is some sample data I used which returned matches of (1,3) and (6,7)
insert into table
values
(1, 'abc'), (1, 'pqr'), (2, 'xyz'), (3, 'pqr'), (3, 'abc'), (4, 'abc'), (5, 'pqr'),
(6, 'abc'), (6, 'def'), (6, 'ghi'), (7, 'abc'), (7, 'def'), (7, 'ghi')
similar (to Derek Kromm's) approach using sub-queries:
SELECT id
FROM mc a
WHERE
id != 1 AND
member IN (
SELECT member FROM mc WHERE id=1)
GROUP BY id
HAVING
COUNT(*) IN (
SELECT COUNT(*) FROM mc WHERE id=1) AND
COUNT(*) IN (
SELECT COUNT(*) FROM mc where id=a.id);
a logic here is we need all ids that match following 2 conditions:
member is among those that belong to id 1
total number of members is same as number of those that belong to id 1
total number of selected members equal to total number of members for current id
try this:
declare #id int
set #id=1
select a.id from
(select id,COUNT(*) cnt from sample_table
where member in (select member from sample_table where id=#id)
and id <>#id
group by id)a
join
(select count(distinct member) cnt from sample_table where id=#id)b
on a.cnt=b.cnt