Have Users table, where users can have multiple accounts.
Table can look like this:
u_id | u_parent_d | date_added
1 | 1 | 2017-01-01
2 | 2 | 2017-01-04
3 | 1 | 2017-01-05
4 | 4 | 2017-01-06
5 | 2 | 2017-01-07
How can I order these records by date added but grouped connected accounts together
u_id | u_parent_d | date_added
5 | 2 | 2017-01-07
2 | 2 | 2017-01-04
4 | 4 | 2017-01-06
3 | 1 | 2017-01-05
1 | 1 | 2017-01-01
You can build your query in two steps. First of all get the maximum date for each u_parent_d
select u_parent_d, max(date_added) as max_date
from Users
group by u_parent_d
Then you can join this with the initial table, and use max_date for sorting
select t1.*
from Users t1
join (
select u_parent_d, max(date_added) as max_date
from Users
group by u_parent_d
) t2
on t1.u_parent_d = t2.u_parent_d
order by t2.max_date desc, t1.date_added desc
Order both by date and parent id:
SELECT * FROM users
ORDER BY u_parent_id, date_added DESC
Related
I have the following tables:
purchase_tbl
id | productId | purchaseQuantity
---+-----------+-----------------
1 | 1 | 30
2 | 2 | 30
3 | 1 | 10
4 | 2 | 10
sale_tbl
id | productId | saleQuantity
---+-----------+-------------
1 | 1 | 10
2 | 2 | 10
3 | 1 | 10
4 | 2 | 10
5 | 1 | 10
6 | 2 | 10
I need to get the output as this one:
productId | totalPurchasedQuantity| totalSaleQuantity
----------+-----------------------+------------------
1 | 40 | 30
2 | 40 | 30
I'm using this query and how to get the desired result?
SELECT purchase_tbl.productId
, SUM(purchase_tbl.purchaseQuantity) AS totalPurchaseQuantity
, SUM(sale_tbl.saleQuantity) AS totalSaleQuantity
FROM purchase_tbl
JOIN sale_tbl
ON purchase_tbl.productId = sale_tbl.productId
GROUP BY purchase_tbl.productId
Current output
productId | totalPurchaseQuantity | totalSaleQuantity
----------+-----------------------+------------------
1 | 120 | 60
2 | 120 | 60
You better group then in separate query, as table have multiple records for each product, which getting cross product.
SELECT purchase.productId, totalPurchaseQuantity, totalSaleQuantity
FROM
(SELECT purchase_tbl.productId
, SUM(purchase_tbl.purchaseQuantity) AS totalPurchaseQuantity
FROM purchase_tbl
GROUP BY purchase_tbl.productId) purchase
INNER JOIN
(SELECT sale_tbl.productId
, SUM(sale_tbl.saleQuantity) AS totalSaleQuantity
FROM sale_tbl
GROUP BY sale_tbl.productId
) sale ON sale.productId= purchase.productId;
To obtain your expected result you have to do the aggregation on the individual table before joining them. Your query with be like:
SELECT A.productId, A.totalpurchaseQuantity, B.totalsaleQuantity
FROM
(SELECT productId, SUM(purchaseQuantity)
totalpurchaseQuantity FROM purchase_tbl
GROUP BY productId) A JOIN
(SELECT productId, SUM(saleQuantity)
totalsaleQuantity FROM sale_tbl
GROUP BY productId) B ON
A.productId=B.productId;
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 want to join 2 tables:
source_table
----------------------------------
| source_id label |
|----------------------------------|
| 1 Contact Form |
| 2 E-Mail |
| 3 Inbound Call |
| 4 Referral |
----------------------------------
related_table
---------------------------------------
| id created_at source |
|---------------------------------------|
| 1 2013-12-26 2 |
| 2 2013-12-26 2 |
| 3 2013-12-26 4 |
| 4 2013-12-25 1 |
| 5 2013-12-18 2 |
| 6 2013-12-16 4 |
| 7 2013-11-30 2 |
---------------------------------------
So that it looks like this:
---------------------------------------
| created_at source amount |
|---------------------------------------|
| 2013-12-26 E-Mail 2 |
| 2013-12-26 Referral 1 |
| 2013-12-25 Contact Form 1 |
| 2013-12-18 E-Mail 1 |
| 2013-12-16 Referral 1 |
---------------------------------------
I want to count the occurrences of each source in related_table grouped by the source for each date in the range.
But I'm not sure how to write the query.
Here's what I have so far:
SELECT DISTINCT
source_table.source_id,
source_table.label AS source,
related_table.created_at,
COUNT(*) AS amount
FROM source_table
INNER JOIN related_table
ON related_table.source=source_table.source_id AND
related_table.created_at>='2013-12-01' AND
related_table.created_at<='2013-12-31'
GROUP BY `source`
ORDER BY `created_at` ASC
I'm not very good with SQL, so the above query might be far off from what I need to have. All I know is that it doesn't work as expected.
My implementation:
select created_at, s.label, amount
from
(
select count(r.Source) as amount, r.source, r.created_at
from related_table r
group by r.source, r.created_at) a inner join source_table s
on a.source = s.source_id
where created_at between '2013-12-01' and '2013-12-31'
order by amount desc, created_at desc
http://sqlfiddle.com/#!2/841bd/2
adjusted demo to your example...
SELECT
created_at
,label as source
,COUNT(*) AS amount
FROM source_table
INNER JOIN related_table
ON source_table.source_id = related_table.source
GROUP BY label, created_at
ORDER BY created_at DESC
I have a table:
ID | User | Amount
1 | 1 | 50
2 | 1 | 80
3 | 2 | 80
4 | 2 | 100
5 | 1 | 90
6 | 1 | 120
7 | 2 | 120
8 | 1 | 150
9 | 2 | 300
I do a query:
SELECT * FROM TABLE ORDER BY amount DESC group by userid
I'm getting this:
ID | User | Amount
1 | 1 | 50
2 | 1 | 80
But I was expecting:
ID | User | Amount
9 | 2 | 300
8 | 1 | 150
What is wrong with my sql?
When grouping you have to use aggregate functions like max() for all columns that are not grouped by
select t.*
from table t
inner join
(
SELECT userid, max(amount) as total
FROM TABLE
group by userid
) x on x.userid = t.userid and x.total = t.amount
ORDER BY t.amount DESC
Another solution.Check SQL Fiddle
Using FIND_IN_SET clause
SELECT
ua.*
FROM user_amount ua
WHERE FIND_IN_SET(ua.amount,(SELECT
MAX(ua1.amount)
FROM user_amount ua1
WHERE ua1.user = ua.user)) > 0
ORDER BY amount desc;
Using IN clause
SELECT
ua.*
FROM user_amount ua
WHERE ua.amount IN (SELECT
MAX(ua1.amount)
FROM user_amount ua1
WHERE ua1.user = ua.user)
ORDER BY amount desc
My table looks like this:
id | user_id | period_id | completed_on
----------------------------------------
1 | 1 | 1 | 2010-01-01
2 | 2 | 1 | 2010-01-10
3 | 3 | 1 | 2010-01-13
4 | 1 | 2 | 2011-01-01
5 | 2 | 2 | 2011-01-03
6 | 2 | 3 | 2012-01-13
... | ... | ... | ...
I want to select only the latest users periods entries, bearing in mind that users will not all have the same period entries.
Essentially (assuming all I have is the above table) I want to get this:
id | user_id | period_id | completed_on
----------------------------------------
3 | 3 | 1 | 2010-01-13
4 | 1 | 2 | 2011-01-01
6 | 2 | 3 | 2012-01-13
Both of the below queries always resulted with the first user_id occurance being selected, not the latest (because the ordering happens after the rows are selected from what I understand):
SELECT
DISTINCT user_id,
period_id,
completed_on
FROM my_table
ORDER BY
user_id ASC,
period_id DESC
SELECT *
FROM my_table
GROUP BY user_id
ORDER BY
user_id ASC,
period_id DESC
Seems like this should work using MAX and a subquery:
SELECT t.Id, t.User_Id, t.Period_Id, t.Completed_On
FROM my_table t
JOIN (SELECT Max(completed_on) Max_Completed_On, t.User_Id
FROM my_table
GROUP BY t.User_ID
) t2 ON
t.User_Id = t2.User_Id AND t.Completed_On = t2.Max_Completed_On
However, if you potentially have multiple records where the completed_on date is the same per user, then this could return multiple records. Depending on your needs, potentially adding a MAX(Id) in your subquery and joining on that would work.
try this:
SELECT t.Id, t.User_Id, t.Period_Id, t.Completed_On
FROM table1 t
JOIN (SELECT Max(completed_on) Max_Completed_On, t.User_Id
FROM table1 t
GROUP BY t.User_ID) t2 ON t.User_Id = t2.User_Id AND t.Completed_On = t2.Max_Completed_On
DEMO HERE