I've got a budget table:
user_id product_id budget created
-----------------------------------------------------------------
1 1 300 2011-12-01
2 1 400 2011-12-01
1 1 500 2011-12-03
2 2 400 2011-12-04
I've also got a manager_user table, joining a manager with the user
user_id manager_id product_id
------------------------------------
1 5 1
1 9 2
2 5 1
2 5 2
3 5 1
What I'd like to do is grab each of the user that's assigned to Manager #5, and also get their 'budgets'... but only the most recent one.
Right now my statement looks like this:
SELECT * FROM manager_user mu
LEFT JOIN budget b
ON b.user_id = mu.user_id AND b.product_id = mu.product_id
WHERE mu.manager_id = 5
GROUP BY mu.user_id, mu.product_id
ORDER BY b.created DESC;
The problem is it doesn't pull the most recent budget. Any suggestions? Thanks!
To accomplish your task you can do as follows:
select b1.user_id,
b1.budget
from budget b1 inner join (
select b.user_id,
b.product_id,
max(created) lastdate
from budget b
group by b.user_id, b.product_id ) q
on b1.user_id=q.user_id and
b1.product_id=q.product_id and
b1.created=q.lastdate
where b1.user_id in
(select user_id from manager_user where manager_id = 5);
I'm assuming here that your (user_id, product_id, created) combination is unique.
For what it's worth, here's the code that returned what I was looking for:
SELECT DISTINCT(b1.id),mu.user_id,mu.product_id,b1.budget,b1.created
FROM budget b1
INNER JOIN (
SELECT b.user_id, b.product_id, MAX(created) lastdate
FROM budget b
GROUP BY b.user_id, b.product_id) q
ON b1.user_id=q.user_id AND
b1.product_id=q.product_id AND
b1.created=q.lastdate
RIGHT JOIN manager_user mu
ON mu.user_id = b1.user_id AND
mu.product_id = b1.product_id
WHERE mu.manager_id = 5;
Thanks for the help Andrea!
Related
I have 2 tables with information: ID, persona_id, total_amount
The persona ID can repeat dozen of times. So i get all the one persons id total_amount with query:
select d.id as debt_id, p.name as persona, sum(d.total_amount) as total_amount
from debt d
join persons p on d.persona_id = p.id group by p.name
I want to get data from each table in one query and do aritmethic propertys with the total_amount column and return it as 1 tabel.
TABLE 1
id persons_id total_amount
1 2 50
2 3 100
3 2 200
4 5 300
5 1 500
TABLE 2
id persons_id total_amount
1 2 25
2 1 100
3 5 50
4 3 100
5 4 300
As a result i want to get the 2 tables comined with arithmetic operation (-, +, / , * ) of Total amount columns.Basicaly a change to get the ending result total amount in form i want for different cases.
What worked for me based on JohnHC answear was :
select c.id, c.persona_id, c.total_amount - d.total_amount as new_total
from ( select c.id , c.persona_id, sum(c.total_amount) as total_amount from credit c
join persons p on c.persona_id = p.id
group by p.name) c
inner join ( select d.id, d.persona_id, sum(d.total_amount) as total_amount from debt d
join persons p on d.persona_id = p.id
group by p.name) d
on c.persona_id = d.persona_id
group by c.id, c.persona_id
If you want the total, try:
select id, person_id, sum(total_amount)
from
(
select id, person_id, total_amount
from table1
union all
select id, person_id, total_amount
from table2
)
group by id, person_id
If you want to do other things, try:
select t1.id, t1.person_id, t1.total_amount [+ - / *] t2.total_amount as new_total
from table1 t1
inner join table2 t2
on t1.id = t2.person_id
group by t1.id, t1.person_id
I have 3 tables like following:
branch
id name
---------
1 abc
2 xyz
users
id branch_id name
-----------------
1 1 aa
2 1 bb
3 2 cc
4 1 dd
5 2 ee
sales
id user_id product price
1 1 xxxx 10
2 1 yyyy 20
3 2 zzzz 18
4 3 aaaa 12
5 2 bbbb 10
6 4 cccc 20
Now I want to get the total selling amount branch wise like:
branch_id total_price
---------------------
1 78
2 12
For that i write a sql query like:
SELECT SUM(s.price) , b.id
FROM sales s
JOIN branch b
GROUP BY id
HAVING s.user_id
IN (
SELECT id
FROM users
WHERE branch_id = b.id
)
But this does not provide the answer that I want. Please help me.
I think this should do the trick:
SELECT branch.id AS branch_id, SUM(s.price) AS total_price
FROM branch
JOIN users ON branch.id = users.branch_id
JOIN sales ON users.id = sales.user_id
GROUP BY branch.id;
Also you could use INNER JOIN instead of JOIN(Both are doing the same thing). With INNER JOIN it is possibly easier to read, especially your query contains other types of JOIN's like LEFT JOIN or RIGHT JOIN
Hope that helps!
You could use something like this:
SELECT u.branch_id, SUM(s.price) AS total_price
FROM sales AS s INNER JOIN users u ON s.user_id = user.id
GROUP BY u.branch_id
ORDER BY u.branch_id
Im trying to get a combination of ids from a table like this:
Table activation:
user_id product_id reseller_id range_id Name
-------------------------------------------------
1 1 5 2 Oscar
1 1 5 3 Luis
2 1 5 4 Julian
Table prices (compType_id = reseller_id):
product_id compType_id price range_id
------------------------------------------------
1 5 38.60 2
1 5 48.60 3
1 5 58.60 4
Table users:
user_id name
----------------
1 lloyd
2 Mark
I want to select the activation Name and the price of prices based on the user_id.
How can i do that?
I tried something like this:
SELECT a.name
(SELECT price FROM prices WHERE product_id = 1 AND range_id =
AND compType_id = 5) AS price
FROM activation AS a
LEFT JOIN users AS u ON u.user_id = a.u_id
WHERE u.user_id = 1
The price columns have the same value in each row.
name price
Oscar 21.30
Luis 21.30
How can i change it to show the correct price?
I solved it with something like this:
SELECT a.company, a.name, a.email, a.phone, a.ruc, a.code, a.active,
a.numlic, (SELECT p.price FROM prices AS p WHERE p.product_id =
a.product_id AND p.range_id = a.range_id AND p.compType_id = c.type_id)
AS price
FROM activation AS a
LEFT JOIN users AS u ON u.user_id = a.u_id
LEFT JOIN companies AS c ON a.reseller_id = c.id
WHERE u.user_id = 1
Given 2 tables, I want to generate top 3 highest amount from [Purchase] table.
Additional criteria is [Crocs] must be included in top 3 of the records.
I have following SQL, but it cannot generates the result as I wanted (Result A), please guide me on how to pull out the result in Result B. Thank you.
Table (Purchase):
Purchase_ID | StoreID | Amount
------------|---------|--------
1 | 21 | 22
2 | 23 | 13
3 | 25 | 6
4 | 26 | 23
5 | 28 | 18
Table (Store):
Store_ID | StoreName
---------|----------
21 | Adidas
22 | Nike
23 | Puma
24 | New Balance
25 | Crocs
26 | Converse
SQL:
SELECT IF(SUM(amount) IS NULL, 0, SUM(amount)) as totalAmount
FROM (
SELECT a.amount
FROM purchase a
INNER JOIN store b
ON a.store_id = b.storeid
GROUP BY a.amount
HAVING b.StoreName = 'Crocs'
ORDER BY a.amount DESC
LIMIT 3
) t
Result A: $6
Explanation A: Amount of Crocs is $6
Result B: $51
Explanation B: Total Amount of top 3 = $22 (Adidas) + 23 (Puma) + $6 (Crocs)
The answer from scaisEdge is almost right, but the first query could also return a row with crocs and the sorting is wrong (order by max(a.amount) limit 2 means that the lowest 2 results will be shown). Additionally you could wrap the query in another select query to sort the results
SELECT * FROM (
SELECT b.storename, max(a.amount) as maxAmount
FROM purchase a
INNER JOIN store b ON a.store_id = b.storeid
WHERE b.storename != 'crocks'
GROUP BY a.storename
ORDER BY max(a.amount) DESC
LIMIT 2
UNION
SELECT b.storename, a.amount as maxAmount
FROM purchase a
INNER JOIN store b
ON a.store_id = b.storeid
WHERE b.storename='crocks'
ORDER BY a.amount DESC
LIMIT 1
) ORDER BY maxAmount DESC
You could use an union
SELECT b.storename, max(a.amount)
FROM purchase a
INNER JOIN store b
ON a.store_id = b.storeid
GROUP BY a.storename
order by max(a.amount) limit 2
union
SELECT b.storename, a.amount
FROM purchase a
INNER JOIN store b
ON a.store_id = b.storeid
where b.storename='crocks'
try this one:
SELECT sum(amount)as sum_amount,a.store_id,storename,category from
(select amount,store_id from tbl_purchase) as a
inner JOIN
(select store_id,storename,category from tbl_store)as b on a.store_id = b.store_id where b.category = 'supermarket' GROUP BY category
As I am not good with MySQL query's so I wish someone help me for creating this kind of sql query.
I having two MySQL tables which is describe bellow:
Table Name: rating
-------------------
property_id user_id area_rate_count safety_rate_count friendly_rate_count walkability_rate_count
4 28 1 1 1 2
5 38 2 3 4 1
5 40 2 2 3 1
6 40 2 3 1 4
10 43 2 2 3 1
Table Name: listing
-------------------
property_id title
4 Sample 1
5 Sample 2
6 Sample 3
10 Sample 4
11 Sample 5
12 Sample 6
Now first I want to sum each column and divide. (area_rate_count, safety_rate_count, friendly_rate_count, walkability_rate_count). For example In property_id:5 having two times so first calculate column sum and divide by 2.
After calculation we will get this output:
Table Name: rating (After Calculation)
--------------------------------------
property_id rate
4 5
5 9 (Divided by 2 because this property_id is two times in table)
6 10
10 8
And Finally I want join this result to my listing table and result looks something like this:
Table Name: listing
-------------------
property_id title rate
4 Sample 1 5
5 Sample 2 9 (Divided by 2 becouse property_id is two times in table)
6 Sample 3 10
10 Sample 4 8
11 Sample 5 0
12 Sample 6 0
Thanks.
I think you want the avg() aggregation function along with a join:
select l.property_id, l.title,
coalesce(avg(area_rate_count + safety_rate_count + friendly_rate_count + walkability_rate_count
), 0) as rate
from listing l left outer join
property_id p
on l.property_id = p.property_id
group by l.property_id, l.title ;
If I understood it right I think you need this:
select l.property_id, l.title, coalesce(r.ssum/if(r.ct=0,1,r.ct), 0) as rate
from listing l LEFT JOIN
(select property_id,
sum(area_rate_count+safety_rate_count
+friendly_rate_count+walkability_rate_count) ssum,
count(*) ct
from rating
group by property_id ) r
ON l.property_id = r.property_id
order by l.property_id
See it here on fiddle: http://sqlfiddle.com/#!2/589d6/5
Edit
As OP asked on the comments that he wants all columns from listing here is what he want:
select l.*, coalesce(r.ssum/if(r.ct=0,1,r.ct), 0) as rate
from listing l LEFT JOIN
(select property_id,
sum(area_rate_count+safety_rate_count
+friendly_rate_count+walkability_rate_count) ssum,
count(*) ct
from rating
group by property_id ) r
ON l.property_id = r.property_id
order by l.property_id
CREATE TEMPORARY TABLE IF NOT EXISTS
temp_table ( INDEX(col_2) )
ENGINE=MyISAM
AS (
SELECT
property_id,
AVG(area_rate_count) as area_rate_count,
AVG(safety_rate_count) as safety_rate_count,
AVG(friendly_rate_count) as friendly_rate_count,
AVG(walkability_rate_count) as walkability_rate_count
FROM rating
GROUP BY property_id
)
SELECT * FROM listing L
JOIN temp_table T
ON L.property_id = T.property_id
Use the below statement to get distinct property_id with its own rate
select property_id, sum(separaterating)/count(property_id) from (
select property_id,sum(area_rate_count , safety_rate_count , friendly_rate_count , walkability_rate_count) as separaterating from rating group by property_id AS temp ) group by
property_id
you can then join with the other table to get the final result as below
select * from ( select property_id, sum(separaterating)/count(property_id) from (
select property_id,sum(area_rate_count , safety_rate_count , friendly_rate_count , walkability_rate_count) as separaterating from rating group by property_id AS temp ) group by
property_id) AS A inner join listing AS B on A.property_id = B.property_id
try this:
select a.prop_id as property_id, l.title, a.allratings / b.numberofreviews as rate
from
(
select property_id as prop_id, SUM(coalesce(area_rate_count,0) + coalesce(safety_rate_count,0) + coalesce(friendly_rate_count,0) + coalesce(walkability_rate_count,0)) as allratings
from rating
group by property_id
) a inner join
(
select property_id, count(distinct user_id) as numberofreviews
from rating
group by property_id
) b on a.property_id = b.property_id
inner join listing l on a.property_id = l.property_id
Try This Query
select ls.property_id,ls.title,inr.rate from listing as ls
left join
(select r.property_id as pid,r.rate/r.cnt as rate from
(select property_id,user_id,(area_rate_count+safefty_rate_count+friendly_rate_count+walkability_rate_count) as rate,count(*) as cnt from rating group by property_id) as r) as inr on inr.pid=ls.property_id