I have a users table with columns: user_id, teacher_id1, teacher_id2, teacher_id3
and
teachers table with id
Each user can have the same id's for teacher_id1, teacher_id2, teacher_id3
I would like to count how many users have same teacher.
User table
+----------------------------------------+
| user_Id teacher_id1 teacher_id2 teacher_id3 |
+----------------------------------------+
| 1 1 1 1 |
| 2 2 1 3 |
| 3 2 3 3 |
| 4 2 2 2 |
+----------------------------------------+
Teacher table
+------+
| id |
+------+
| 1 |
| 2 |
| 3 |
Count for $id1 is: 2
Count for $id2 is: 3
Count for $id3 is: 2
I tried something like this, but it is not correct!
SELECT COUNT(*) FROM users WHERE concat_ws('',teacher_id1 OR teacher_id2
OR teacher_id3) like '{$id}' ";
You have data in three different columns. You need to combine it into one column, to get the distinct counts that you want. For this, you can use union all. Then the count is simply count(distinct):
select teacher_id, COUNT(distinct USER_ID)
from ((select user_id, teacher_id1 as teacher_id
from t
) union all
(select user_id, teacher_id2
from t
) union all
(select user_id, teacher_id3
from t
)
) s
group by teacher_id;
Try this query
select b.id, count(*)
from
tbl1 a
inner join
tbl2 b
on b.id = teacher_id1 or b.id = teacher_id2 or b.id = teacher_id3
group by b.id
SQL FIDDLE:
| ID | COUNT(*) |
-----------------
| 1 | 2 |
| 2 | 3 |
| 3 | 2 |
Related
I need to find the sum of occurrences of ids in a join table. The id could be present in two different columns (id_type_1 and id_type_2).
Table types
id | name
1 | Test1
2 | Test2
3 | Test3
Table products
id | name | id_type_1 | id_type_2
1 | Product1 | 1 | 2
2 | Product2 | 3 | 1
3 | Product3 | 1 | 3
I need to get a result like this:
Type | Total
Test1 | 3
Test2 | 1
Test3 | 2
Here's my query, but it takes several seconds to execute:
SELECT t.name,
(SELECT COUNT(p.id) FROM products p WHERE p.id_type_1 = t.id || p.id_type_2 = t.id) AS total
FROM types t
WHERE 1
ORDER BY total DESC
Is there a more effective way to achieve the result?
Join the tables and aggregate:
select t.id, t.name,
sum((t.id = p.id_type_1) + (t.id = p.id_type_2)) Total
from types t inner join products p
on t.id in (p.id_type_1, p.id_type_2)
group by t.id, t.name
If there is no case for the id to exist in both id_type_1 and id_type_2 in the same row then:
select t.id, t.name,
count(*) Total
from types t inner join products p
on t.id in (p.id_type_1, p.id_type_2)
group by t.id, t.name
See the demo.
Results:
> id | name | Total
> -: | :---- | ----:
> 1 | Test1 | 3
> 2 | Test2 | 1
> 3 | Test3 | 2
I have a table with columns ID, Content and Day. I am trying to find all rows that have duplicate Content and Day values and display all rows
SELECT ID,Content, `Day`, Count(*)
FROM table
GROUP BY Content,`Day`
HAVING COUNT(*) > 1
The current code will return a list of duplicate Content and 'Day' values for instance:
ID|Content|Day
1 | a | 1
2 | a | 1
3 | a | 1
4 | b | 2
5 | b | 2
6 | c | 3
7 | c | 4
Will result in:
ID|Content|Day|Count
1 | a | 1 | 3
4 | b | 2 | 2
But I want to display all the unique IDs as well;
ID|Content|Day
1 | a | 1
2 | a | 1
3 | a | 1
4 | b | 2
5 | b | 2
Just make a Sub-Query
select *
from table
where `day` in
(
SELECT ID
FROM table
GROUP BY Content,`Day`
HAVING COUNT(*) > 1
) A
Use that query as a subquery to join against the table again:-
SELECT table.ID, table.Content, table.`Day`
FROM table
INNER JOIN
(
SELECT Content, `Day`, Count(*)
FROM table
GROUP BY Content,`Day`
HAVING COUNT(*) > 1
) sub0
ON sub0.Content = table.Content
AND sub0.`Day` = table.`Day`
I have the following table:
+----+-----------+-----------+
| id | teacherId | studentId |
+----+-----------+-----------+
| 1 | 1 | 4 |
| 2 | 1 | 2 |
| 3 | 1 | 1 |
| 4 | 1 | 3 |
| 5 | 2 | 2 |
| 6 | 2 | 1 |
| 7 | 2 | 3 |
| 8 | 3 | 9 |
| 9 | 3 | 6 |
| 10 | 1 | 6 |
+----+-----------+-----------+
I need a query to find two teacherId's with maximum number of common studentId's.
In this case teachers with teacherIds 1,2 have common students with studentIds 2, 1, 3, which is greater than 1,3 having common students 6.
Thanks in Advance!
[Edit]: After several hours I've had the following solution:
SELECT * FROM (
SELECT r1tid, r2tid, COUNT(r2tid) AS cnt
FROM (
SELECT r1.teacherId AS r1tid, r2.teacherId AS r2tid
FROM table r1
INNER JOIN table r2 ON r1.studentId=r2.studentId AND r1.teacherId!=r2.teacherId
ORDER BY r1tid
) t
GROUP BY r1tid, r2tid
ORDER BY cnt DESC
) t GROUP BY cnt ORDER BY cnt DESC LIMIT 1;
I was sure that there must exist more short and elegant solution, but I could not find it.
You would do this with a self-join. Assuming no duplicates in the table:
select t.teacherid, t2.teacherid, count(*) as NumStudentsInCommon
from table t join
table t2
on t.studentid = t2.studentid and
t.teacherid < t2.teacherid
group by t.teacherid, t2.teacherid
order by NumStudentsInCommon desc
limit 1;
If you had duplicates, you would just replace count(*) with count(distinct studentid), but count(distinct) requires a bit more work.
select t.teacherId, t2.teacherId, sum(t.studentId) as NumStudentsInCommon
from table1 t join
table1 t2
on t.studentId = t2.studentId and
t.teacherId < t2.teacherId
group by t.teacherId, t2.teacherId
order by NumStudentsInCommon desc
I'm struggling with this SQL query. Say I have these two tables
**USERS**
+----+-------+
| id | name |
+----+-------+
| 1 | james |
| 2 | tom |
| 3 | kate |
+----+-------+
**PHOTOS**
+-----------+-----------+---------+
| name | sent_from | sent_to |
+-----------+-----------+---------+
| beach.jpg | 1 | 2 |
| trees.jpg | 3 | 1 |
| earth.jpg | 2 | 1 |
+-----------+-----------+---------+
How could I get, using one SQL query, all the users that have more sent_to associated with their id than sent_from ?
I think of this as aggregating the data twice and then doing the comparison:
select sf.sent_from
from (select sent_from, count(*) as numsent
from photos
group by sent_from
) sf left outer join
(select sent_to, count(*) as numrecv
from photos
group by sent_to
) st
on sf.sent_from, st.sent_to
where numsent > numrecv;
If you want user information, then join that in.
An alternative way restructures the data first and then does the aggregation:
select who
from (select sent_from as who, 1 as sent_from, 0 as sent_to
from photos
union all
select sent_to as who, 0, 1
from photos
) p
group by who
having sum(sent_from) > sum(sent_to);
I think here is something that might help you:
SELECT * FROM (
SELECT `id`, `name`,
IFNULL((SELECT count(*) FROM `photos` WHERE `sent_from` = `users`.`id`),0) AS `sent_from_count`,
IFNULL((SELECT count(*) FROM `photos` WHERE `sent_t`o = `users`.`id`),0) AS `sent_to_count`
FROM `users`) AS `t1`
WHERE `t1`.`sent_to_count` > `t1`.`sent_to_count`
Consider this table:
I need to make a SQL query which returns highlighted rows. In other words: rows which are linked to each other by idContactTo.
Example:
1 has contact to 2, 2 has contact to 1 - they are linked and should be in result table. But even 1 has contact to 3 it doesn't mean that 3 has contact to 1 - they aren't linked.
You can do it via INNER JOIN,
SELECT a.*
FROM tableName a
INNER JOIN tableName b
ON a.idContantOwner = b.idContactTo AND
b.idContantOwner = a.idContactTo
SQLFiddle Demo
Another way to do it
SELECT t.*
FROM
(
SELECT MiN(id) min_id, MAX(id) max_id
FROM Table1
GROUP BY LEAST(idContactOwner, idContactTo),
GREATEST(idContactOwner, idContactTo)
HAVING COUNT(*) = 2
) q JOIN Table1 t
ON t.id IN(q.min_id, q.max_id)
Output:
| ID | IDCONTACTOWNER | IDCONTACTTO |
|----|----------------|-------------|
| 1 | 1 | 2 |
| 2 | 2 | 1 |
| 4 | 3 | 4 |
| 5 | 4 | 3 |
Here is SQLFiddle demo