I have two tables cars and usage. I create a record in usage once a month for some of cars.
Now I want to get distinct list of cars with their latest usage that I saved.
first of all look at the tables please
cars:
| id | model | reseller_id |
|----|-------------|-------------|
| 1 | Samand Sall | 324228 |
| 2 | Saba 141 | 92933 |
usages:
| id | car_id | year | month | gas |
|----|--------|------|-------|-----|
| 1 | 2 | 2020 | 2 | 68 |
| 2 | 2 | 2020 | 3 | 94 |
| 3 | 2 | 2020 | 4 | 33 |
| 4 | 2 | 2020 | 5 | 12 |
The problem is here
I need only the latest usage of year and month
I tried a lot of ways but none of them is good enough. because sometimes this query gets me one ofnot latest records of usages.
SELECT * FROM cars AS c
LEFT JOIN
(select *
from usages
) u on (c.id = u.car_id)
order by u.gas desc
You can do this with a DISTINCT ON in the derived table:
SELECT *
FROM cars AS c
LEFT JOIN (
select distinct on (u.car_id) *
from usages u
order by u.car_id, u.year desc, u.month desc
) lu on c.id = lu.car_id
order by u.gas desc;
I think you need window function row_number. Here is the demo.
select
id,
model,
reseller_id
from
(
select
c.id,
model,
reseller_id,
row_number() over (partition by u.car_id order by u.id desc) as rn
from cars c
left join usages u
on c.id = u.car_id
) subq
where rn = 1
Related
This question already has answers here:
Join to only the "latest" record with t-sql
(7 answers)
Fetch the rows which have the Max value for a column for each distinct value of another column
(35 answers)
Closed 4 months ago.
I want to list all customer with the latest phone number and most recent customer type
the phone number and type of customers are changing periodically so I want the latest record only without getting old values based on the lastestupdate column
Customer:
+------------+--------------------+------------+
|latestUpdate| CustID | AddID | TypeID |
+------------+--------+-----------+-------------
| 2020-03-01 | 1 | 1 | 1 |
| 2020-04-07 | 2 | 2 | 2 |
| 2020-06-13 | 3 | 3 | 3 |
| 2020-03-29 | 4 | 4 | 4 |
| 2020-02-06 | 5 | 5 | 5 |
+------------+--------+------------+----------+
CustomerAddress:
+------------+--------+-----------+
|latestUpdate| AddID | Mobile |
+------------+--------+-----------+
| 2020-03-01 | 1 | 66666 |
| 2020-04-07 | 1 | 55555 |
| 2020-06-13 | 2 | 99999 |
| 2020-03-29 | 3 | 11111 |
| 2020-02-06 | 3 | 22222 |
+------------+--------+-----------+
CustomerType:
+------------+--------+-----------+
|latestUpdate| TypeId | TypeName |
+------------+--------+-----------+
| 2020-03-01 | 1 | First |
| 2020-04-07 | 1 | Second |
| 2020-06-13 | 3 | Third |
| 2020-03-29 | 4 | Fourth |
| 2020-02-06 | 5 | Fifth |
+------------+--------+-----------+
When I tried to join I am always getting duplicated customerID not only the latest record
I want to Display Customer.CustID and CustomerType.TypeName and CustomerAddress.Mobile
You need to make sub-queries for most recent customer type and latest phone number like this:
SELECT *
FROM (
SELECT latestUpdate, CustID, AddID, TypeID,
ROW_NUMBER() OVER (PARTITION BY CustID ORDER BY latestUpdate DESC) AS RowNumber
FROM Customer
) AS c
INNER JOIN (
SELECT latestUpdate, AddID, Mobile,
ROW_NUMBER() OVER (PARTITION BY AddId ORDER BU ltestUpdate DESC) AS RowNumber
FROM CustomerAddress
) AS t
ON c.AddId = t.AddId
INNER JOIN CustomerType ct
ON ct.TypeId = c.TypeId
WHERE c.RowNumber = 1
AND t.RowNumber = 1
A simpler way than using row_number would be using cross apply together with top 1 in an ordered subquery:
select c.CustId, p.Mobile
from Customer c
cross apply (
select top 1 Mobile
from CustomerAddress a
where c.CustId = a.AddId
order by a.latestUpdate
) p
You need to use some subqueries :
SELECT *
FROM Customer AS C
LETF OUTER JOIN (SELECT *, ROW_NUMBER() OVER(PARTITION BY CustID ORDER BY LastestUpdate DESC) AS N
FROM CustomerAddress) AS A
ON C.CustID = A.CustID AND N = 1
LETF OUTER JOIN (SELECT *, ROW_NUMBER() OVER(PARTITION BY CustID ORDER BY LastestUpdate DESC) AS N
FROM CustomerType) AS T
ON C.CustID = T.CustID AND N = 1
If you have had used Temporal table which is an ISO SQL Standard feature for data history of table, you will always have the lastest rows inside the main table, old rows stays into history table and can be queried with a time point or date interval restriction.
This is it:
select * from (select *,RANK() OVER (
PARTITION BY b.AddID
ORDER BY b.latestUpdate DESC,
) as rank1
from
Customer a
left join
CustomerAddress b
on
a.AddID=b.AddID
left join
CustomerType c
on
v.TypeId =c.TypeId
) where rank1=1;
You should join the tables using the "APPLY" operator.
See: Link
Table data
+-----+----------------+--------+----------------+
| ID | Required_by | Name | Another_Field |
+-----+----------------+--------+----------------+
| 1 | 7 August | cat | X |
| 2 | 7 August | cat | Y |
| 3 | 10 August | cat | Z |
| 4 | 11 August | dog | A |
+-----+----------------+--------+----------------+
What I want to do is group by the name, then for each group choose one of the rows with the earliest required by date.
For this data set, I would like to end up with either rows 1 and 4, or rows 2 and 4.
Expected result:
+-----+----------------+--------+----------------+
| ID | Required_by | Name | Another_Field |
+-----+----------------+--------+----------------+
| 1 | 7 August | cat | X |
| 4 | 11 August | dog | A |
+-----+----------------+--------+----------------+
OR
+-----+----------------+--------+----------------+
| ID | Required_by | Name | Another_Field |
+-----+----------------+--------+----------------+
| 2 | 7 August | cat | Y |
| 4 | 11 August | dog | A |
+-----+----------------+--------+----------------+
I have something that returns 1,2 and 4 but I'm not sure how to only pick one from the first group to get the desired result. I'm joining the grouping with the data table so that I can get the ID and another_field back after the grouping.
SELECT d.id, d.name, d.required_by, d.another_field
FROM
(
SELECT min(required_by) as min_date, name
FROM data
GROUP BY name
) agg
INNER JOIN
data d
on d.required_by = agg.min_date AND d.name = agg.name
This is typically solved using window functions:
select d.id, d.name, d.required_by, d.another_field
from (
select id, name, required_by, another_field,
row_number() over (partition by name order by required_by) as rn
from data
) d
where d.rn = 1;
In Postgres using distinct on() is typically faster:
select distinct on (name) *
from data
order by name, required_by
Online example
SELECT [id]
,[date]
,[name]
FROM [test].[dbo].[data]
WHERE date IN (SELECT min(date) FROM data GROUP BY name)
enter image description here
I have a query that looks like this:
select id, extension, count(distinct(id)) from publicids group by id,extension;
This is what the results looks like:
id | extension | count
-------------+-------------------------+-------
18459154909 | 12333 | 1
18459154909 | 9891114 | 1
18459154919 | 43244 | 1
18459154919 | 8776232 | 1
18766145025 | 12311 | 1
18766145025 | 1122111 | 1
18766145201 | 12422 | 1
18766145201 | 14141 | 1
But what I really want is for the results to look like this:
id | extension | count
-------------+-------------------------+-------
18459154909 | 12333 | 2
18459154909 | 9891114 | 2
18459154919 | 43244 | 2
18459154919 | 8776232 | 2
18766145025 | 12311 | 2
18766145025 | 1122111 | 2
18766145201 | 12422 | 2
18766145201 | 14141 | 2
I'm trying to get the count field to show the total number of records that have the same id.
Any suggestions would be appreciated
I think you want to count distincts extentions, not ids.
Run this query:
select id
, extension
(select count(*) from publicids p1 where p.id = p1.id ) distinct_id_count
from publicids p
group by id,extension;
This is more or less the same as Pastor's answer. Depending on what the optimizer does it might be faster with higher record count source tables.
select p.id, p.extension, p2.id_count
from publicids p
inner join (
select id, count(*) as id_count
from publicids group by id
) as p2 on p.id = p2.id
I have two tables:
First table "persons"
id | name |
---------------
1 | peter |
3 | martin |
5 | lucy |
Second table "meetings"
id | date | id_persons |
--------------------------------
1 | 2014-12-08 | 1 |
2 | 2013-05-10 | 2 |
3 | 2015-08-25 | 1 |
4 | 2016-10-18 | 1 |
5 | 2012-01-01 | 3 |
6 | 2016-09-28 | 5 |
I need somehow get only last date from "meeting" table for every person (or selected). And result table must be order by name. I thought, it could be like this, but WHERE clause in LEFT JOIN can't be used:
SELECT meetings.id, meetings.date, persons.name FROM persons
LEFT JOIN (SELECT meetings.date, meetings.id, meetings.id_persons FROM
meetings WHERE persons.id = meetings.id_persons ORDER BY
meetings.date DESC LIMIT 1) m ON m.id_persons = persons.id
WHERE persons.id < 6 ORDER BY persons.name
So I started with DISTINCT and it worked, but I think that it is not good idea:
SELECT * FROM
(SELECT DISTINCT ON (persons.id) persons.id, persons.name,
m.date, m.id FROM persons
LEFT JOIN (SELECT meetings.id, meetings.date, meetings.id_persons
FROM meetings ORDER BY meetings.date DESC) m
ON m.id_persons = persons.id
WHERE persons.id < 6 ORDER BY persons.id) p
ORDER BY p.name
Result what I need is:
name | date | id_meetings
-----------------------------------
lucy | 2016-09-28 | 6
martin | 2012-01-01 | 5
peter | 2016-10-18 | 4
Could you help me with better solution?
In Postgres, the easiest way is probably distinct on:
select distinct on (p.id) p.*, m.*
from persons p left join
meetings m
on m.id_persons = p.id
order by p.id, m.date desc;
Note: distinct on is specific to Postgres.
I have 2 tables with similar layout, involving INCOME and EXPENSES.
The id column is a customer ID.
I need a result of customer TOTAL AMOUNT, summing up income and expenses.
Table: Income
| id | amountIN|
+--------------+
| 1 | a |
| 2 | b |
| 3 | c |
| 4 | d |
Table: Expenses
| id | amountOUT|
+---------------+
| 1 | -x |
| 4 | -z |
My problem is that some customers only have expenses and others just income... so cannot know in advance id I need to do a LEFT or RIGHT JOIN.
In the example above an RIGHT JOIN could do the trick, but if the situation is inverted (more customers on the Expenses table) it doesn't work.
Expected Result
| id | TotalAmount|
+--------------+
| 1 | a - x |
| 2 | b |
| 3 | c |
| 4 | d - z |
Any help?
select id, SUM(Amount)
from
(
select id, amountin as Amount
from Income
union all
select id, amountout as Amount
from Expense
) a
group by id
I believe a full join will solve your problem.
I would approach this as a union. Do that in your subquery then sum on it.
For instance:
select id, sum(amt) from
(
select i.id, i.amountIN as amt from Income i
union all
select e.id, e.amountOUT as amt from Expenses e
)
group by id
You should really have another table like client :
Table: Client
| id |
+----+
| 1 |
| 2 |
| 3 |
| 4 |
So you could do something like that
SELECT Client.ID, COALESCE(Income.AmountIN, 0) - COALESCE(Expenses.AmountOUT, 0)
FROM Client c
LEFT JOIN Income i ON i.ID = c.ID
LEFT JOIN Expense e ON e.ID = c.ID
Will be less complicated and i'm sure it will come handy another time :)