How to get those entries which have more than 1 records?
If it doesn't make sense... let me explain:
From the below table I want to access the sum of the commission of all rows where type is joining and "they have more than 1 entry with same downmem_id".
I have this query but it doesn't consider more entries scenario...
$search = "SELECT sum(commission) as income FROM `$database`.`$memcom` where type='joining'";
Here's the table:
id mem_id commission downmem_id type time
2 1 3250 2 joining 2019-09-22 13:24:40
3 45 500 2 egbvegr new time
4 32 20 2 vnsjkdv other time
5 23 2222 2 vfdvfvf some other time
6 43 42 3 joining time
7 32 353 5 joining time
8 54 35 5 vsdvsdd time
Here's the expected result: it should be the sum of the id no 2, 7 only
ie. 3250+353=whatever.
It shouldn't include id no 6 because it has only 1 row with the same downmem_id.
Please help me to make this query.
Another approach is two levels of aggregation:
select sum(t.commission) income
from (select sum(case when type = 'joining' then commission end) as commission
from t
group by downmem_id
having count(*) > 1
) t;
The main advantage to this approach is that this more readily supports more complex conditions on the other members of each group -- such as at most one "joining" record or both "joining" records and no more than two "vnsjkdv" records.
Use EXISTS:
select sum(t.commission) income
from tablename t
where t.type = 'joining'
and exists (
select 1 from tablename
where id <> t.id and downmem_id = t.downmem_id
)
See the demo.
Results:
| income |
| ----- |
| 3603 |
You can use subquery that will find all downmem_id having more than one occurrence in the table.
SELECT Sum(commission) AS income
FROM tablename
WHERE type = 'joining'
AND downmem_id IN (SELECT downmem_id
FROM tablename t
GROUP BY downmem_id
HAVING Count(id) > 1);
DEMO
Related
I have 2 tables in MySQL(InnoDB). The first is an employee table. The other table is the expense table. For simplicity, the employee table contains just id and first_name. The expense table contains id, employee_id(foreign key), amount_spent, budget, and created_time. What I would like is a query that returns the percentage of their budget spent for the most recent X number of expense they've registered.
So given the employee table:
| id | first_name
-------------------
1 alice
2 bob
3 mike
4 sally
and the expense table:
| id | employee_id | amount_spent | budget | created_time
----------------------------------------------------------
1 1 10 100 10/18
2 1 50 100 10/19
3 1 0 40 10/20
4 2 5 20 10/22
5 2 10 70 10/23
6 2 75 100 10/24
7 3 50 50 10/25
The query for the last 3 trips would return
|employee_id| first_name | percentage_spent |
--------------------------------------------
1 alice .2500 <----------(60/240)
2 bob .4736 <----------(90/190)
3 mike 1.000 <----------(50/50)
The query for the last 2 trips would return
|employee_id| first_name | percentage_spent |
--------------------------------------------
1 alice .3571 <----------(50/140)
2 bob .5000 <----------(85/170)
3 mike 1.000 <----------(50/50)
It would be nice if the query, as noted above, did not return any employees who have not registered any expenses (sally). Thanks in advance!!
I'll advise you to convert datatype of created_time as DATETIME in order to get accurate results.
As of now, I've assumed that most recent id indicates most recent spents as it's what sample data suggests.
Following query should work (didn't tested though):
select t2.employee_id,t1.first_name,
sum(t2.amount_spent)/sum(t2.budget) as percentage_spent
from employee t1
inner join
(select temp.* from
(select e.*,#num := if(#type = employee_id, #num + 1, 1) as row_number,
#type := employee_id as dummy
from expense e
order by employee_id,id desc) temp where temp.row_number <= 3 //write value of **n** here.
) t2
on t1.id = t2.employee_id
group by t2.employee_id
;
Click here for DEMO
Feel free to ask doubt(s), if you've any.
Hope it helps!
If you are using mysql 8.0.2 and higher you might use window function for it.
SELECT employee_id, first_name, sliding_sum_spent/sliding_sum_budget
FROM
(
SELECT employee_id, first_name,
SUM(amount_spent) OVER (PARTITION BY employee_id
ORDER BY created_time
RANGE BETWEEN 3 PRECEDING AND 0 FOLLOWING) AS sliding_sum_spent,
SUM(budget) OVER (PARTITION BY employee_id
ORDER BY created_time
RANGE BETWEEN 3 PRECEDING AND 0 FOLLOWING) AS sliding_sum_budget,
COUNT(*) OVER (PARTITION BY employee_id
ORDER BY created_time DESC) rn
FROM expense
JOIN employee On expense.employee_id = employee.id
) t
WHERE t.rn = 1
As mentioned by Harshil, order of row according to the created_time may be a problem, therefore, it would be better to use date date type.
Having trouble wrapping my head around having an efficient "duplicate entries" select in a single query.
In the below example, duplicate StockNo can exist spanning multiple Date. I want to search StockNo for duplicate entries, and if at least 1 StockNo record is found within the Date current YEAR-MONTH, then I also need to select its partner that could exist in any other YEAR-MONTH. Is this possible?
Example Query:
SELECT * FROM `sales`
WHERE `StockNo` IN
(SELECT `StockNo` FROM `sales` GROUP BY `StockNo` HAVING COUNT(*) > 1)
AND `Date` LIKE '2016-11-%'
ORDER BY `StockNo`, `TransactionID`;
Example Data:
ID | StockNo | Date
1 | 1 | 2016-11-01
2 | 1 | 2016-11-10
3 | 2 | 2016-11-05
4 | 2 | 2016-10-29
5 | 3 | 2016-10-25
6 | 3 | 2016-10-15
With my example query and data, I have 3 pairs of duplicate entries. It's pretty obvious that I will only return 3 records (ID's 1, 2 & 3) due to AND Date LIKE '2016-11-%', however I need to return ID's 1, 2, 3, 4. I want to ignore ID's 5 & 6 because neither of them fall within the current month.
Hope that makes sense. Thanks for any help you can provide.
SELECT StockNo
FROM sales
GROUP BY StockNo
HAVING SUM(CASE WHEN DATE_FORMAT(Date, '%Y-%m') = '2016-11' THEN 1 ELSE 0 END) > 0
If you also want to retrieve the full records for those matching stock numbers in the above query, you can just add a join:
SELECT s1.*
FROM sales s1
INNER JOIN
(
SELECT StockNo
FROM sales
GROUP BY StockNo
HAVING SUM(CASE WHEN DATE_FORMAT(Date, '%Y-%m') = '2016-11' THEN 1 ELSE 0 END) > 0
) s2
ON s1.StockNo = s2.StockNo
Demo here:
SQLFiddle
Thank you very much Tim for pointing me in the right direction. Your answer was close but it still only returned records from the current month and in the end I used the following query:
SELECT s1.* FROM `sales` s1
INNER JOIN
(
SELECT * FROM `sales` GROUP BY `StockNo` HAVING COUNT(`StockNo`) > 1
AND SUM(CASE WHEN DATE_FORMAT(`Date`, '%Y-%m')='2016-11' THEN 1 ELSE 0 END) > 0
) s2
ON s1.StockNo=s2.StockNo
This one had been eluding me for some time.
I'm attempting to join two tables and also get a SUM and flailing badly. I need to get the total commission amounts for each affiliate where affiliate.approved=1 AND order.status=3.
//affiliate table
affiliate_id | firstname | lastname | approved |
1 joe shmoe 1
2 frank dimag 0
3 bob roosky 1
here's the order table
//order
affiliate_id | order_status_id | commission
1 3 0.20
1 0 0.30
2 3 0.10
3 3 0.25
1 3 0.25
2 3 0.15
2 0 0.20
and here's what I'd like the query to return:
affiliate_id | commission
1 0.45
3 0.25
Here is my attempt that doesn't work. It outputs just one line.
SELECT order.affiliate_id, SUM(order.commission) AS total, affiliate.firstname, affiliate.lastname FROM `order`, `affiliate` WHERE order.order_status_id=3 AND affiliate.approved=1 AND order.affiliate_id = affiliate.affiliate_id ORDER BY total;
thanks for any help.
You've missed GROUP BY, try this:
SELECT
`order`.affiliate_id,
SUM(`order`.commission) AS total,
affiliate.firstname,
affiliate.lastname
FROM `order`
JOIN `affiliate`
ON `order`.order_status_id = 3 AND affiliate.approved = 1 AND `order`.affiliate_id = affiliate.affiliate_id
GROUP BY `order`.affiliate_id
ORDER BY total;
Demo Here
You can try this Query for your solution :-
SELECT order.affiliate_id, SUM(order.commission) AS total,affiliate.firstname,
affiliate.lastname
FROM `order`, `affiliate`
WHERE order.order_status_id=3
AND affiliate.approved=1
AND order.affiliate_id = affiliate.affiliate_id
GROUP BY order.affiliate_id
ORDER BY total;
Here is the solution:
select affiliate.affiliate_id,sum(`order`.commission) as total from affiliate left join `order` on affiliate.affiliate_id=`order`.affiliate_id
where affiliate.approved=1 and `order`.order_status_id=3 group by affiliate.affiliate_id
In addition,"order" is a key word of SQL , I recommend you not to use it as a table/column name.
First: Remove the implicit join syntax. It's confusing.
Second: You needed to group by affiliate_id. Using aggregate function without group by collapses your result set into a single row.
Here's the query using INNER JOIN:
SELECT
`order`.affiliate_id,
SUM(`order`.commission) AS total,
affiliate.firstname,
affiliate.lastname
FROM `order`
INNER JOIN`affiliate` ON `order`.affiliate_id = affiliate.affiliate_id
WHERE `order`.order_status_id = 3
AND affiliate.approved = 1
GROUP BY affiliate.affiliate_id
ORDER BY total;
WORKING DEMO
Caution: You have picked one of the reserved words of MySQL as table name (order). Be aware to enclose it with (`)backtick always .
Just a gentle reminder
Below is my database table, where I will have Check In and Check Out entry records for attending the conference room.
id registration_id roomno day type
1 101 1 2 In
2 103 1 2 In
3 101 1 2 Out
4 105 1 2 In
5 103 1 2 Out
6 101 1 2 In
7 103 1 2 In
8 101 1 2 Out
9 105 1 2 In
10 103 1 2 Out
Now, I want to select those records, which are still attending the conference. Condition is like their last record should be type = In. There can be multiple In/Out entries for each user during a day.
Please let me know the quickest possible MySQL query.
Thanks
Answer which I ended up using:
select * from `registrations_inouts` t
group by t.registration_id
having max(id) = max(case when type = 'In' then id end)
order by rand() limit 1;
Here is one method using not exists:
select *
from t
where t.type = 'In' and
not exists (select 1
from t t2
where t2.registration_id = t.registration_id and t2.type = 'Out'
);
Another method uses conditional aggregation:
select t.registration_id
from t
group by t.registration_id
having max(id) = max(case when type = 'In' then id end);
Note: both of these assume that the ids are assigned sequentially in time, so larger ids are later in time.
Please review my tables below... Is it possible to build a single query capable of
1) calculating the SUM of total_time for all vehicles that have class_id 1 (regardless of feature_id)(result would be 6:35)
2) calculating the SUM of total_time for all vehicles that have class_id 1 AND have feature_id 2(result would be 5:35 based on vehicle_id 22 and 24)
I'm able to get the results in two seperate queries, but I was hoping to retrieve them in one single query.... something like:
SELECT
SUM((CASE WHEN (VEHICLE_TABLE.class_id = 1) then LOG_TABLE.total_time else 0 end)) **AS TOTAL_ALL**,
...here goes statement for 2)... AS TOTAL_DIESEL...
FROM LOG_TABLE, VEHICLE_TABLE .....
WHERE VEHICLE_TABLE.vehicle_id = LOG_TABLE.vehicle_id ......
TABLE 1: LOG_TABLE (vehicle_id is NOT unique)
vehicle_id | total_time
--------------|--------------
22 2:00
22 0:30
23 1:00
24 2:20
24 0:45
TABLE 2: VEHICLE_TABLE (vehicle_id is unique)
vehicle_id | class_id
--------------|--------------
22 1
23 3
24 1
TABLE 3: VEHICLE_FEATURES_TABLE (vehicle_id is NOT unique but feature_id is unique per vehicle_id)
vehicle_id | feature_id
--------------|--------------
22 1
22 2
23 1
23 2
23 6
24 2
24 6
SELECT SUM(lt.total_time) AS TOTAL_ALL,
SUM(CASE WHEN (vft.feature_id IS NOT NULL) then LOG_TABLE.total_time else 0 end) AS FEATURE_TOTAL
FROM VEHICLE_TABLE vt
JOIN LOG_TABLE lt
ON vt.vehicle_id = lt.vehicle_id
LEFT JOIN VEHICLE_FEATURES_TABLE vft
ON vt.vehicle_id = vft.vehicle_id AND vft.feature_id = 2
WHERE vt.class_id = 1
It seems that there is not much point in putting both of them in one query unless you want the results together.
If so, just add a UNION between the 2 queries.
If you want to have both values in the same row try something like this:
SELECT (SELECT Sum(X)
FROM TBL
WHERE CLASS_ID = 1) AS CLS_id1,
(SELECT Sum(X)
FROM TBL
WHERE CLASS_ID = 1
AND FEATURE_ID = 2) AS CLS_id1_FTR_ID2