I'm trying to get the following result in mysql with below table: I
have tried to use sum (case when) but it give me the result for an
individual name only ,I want in the report the the total cost for each
user.
note: the ID column is not important you can ignore it .
mysql> select * from calls_records;
+----+-------+------------+------+
| id | month | name | cost |
+----+-------+------------+------+
| 1 | 1 | osama | 40 |
| 2 | 1 | rahman | 40 |
| 3 | 1 | ahmed | 30 |
| 4 | 1 | ali albann | 10.5 |
| 5 | 2 | osama | 10 |
| 6 | 2 | ali albann | 30 |
| 7 | 2 | ahmed | 10 |
| 8 | 2 | rahman | 10 |
+----+-------+------------+------+
expected result
+-----------+---------------------------+
| name | total_cost_for_each_user |
+------------+---------------------------+
| ahmed | 50 |
| ali albann | 40.5 |
| osama | 50 |
| rahman | 50 |
+------------+---------------------------+
query
select name, sum(cost) as totalcost
from calls_records
group by name
;
output
+------------+-----------+
| name | totalcost |
+------------+-----------+
| ahmed | 40 |
| ali albann | 40.5 |
| osama | 50 |
| rahman | 50 |
+------------+-----------+
sqlfiddle
Related
I'm trying to come up with a stored procedure that takes multiple rows that are exactly identical, and combines them into one row while summing one column, which can then be run through more stored procedures based on the sum of that one column.
I've tried a GROUP BY statement, but that doesn't actually group the rows together, because if I run the table through another procedure it performs actions as if each row were not combined. Performing a SELECT * FROM mytable query shows that each row was not actually combined into one.
Is there any way to permanently combine multiple rows into one singular row?
To start, I've got a table like this:
+-------+-----+--------+---------+------+-----+-----------+
| RowID | pID | Name | Date | Code | QTY | Purchased |
+-------+-----+--------+---------+------+-----+-----------+
| 1 | 1 | bob | 9/29/20 | 123 | 1 | |
| 2 | 1 | bob | 8/10/20 | 456 | 1 | |
| 3 | 2 | rob | 9/15/20 | 123 | 1 | |
| 4 | 2 | rob | 9/15/20 | 123 | 1 | |
| 5 | 2 | rob | 9/15/20 | 123 | 1 | |
| 6 | 2 | rob | 9/15/20 | 123 | 1 | |
| 7 | 2 | rob | 9/15/20 | 123 | 1 | |
| 8 | 3 | john | 7/12/20 | 987 | 1 | |
| 9 | 3 | john | 7/12/20 | 987 | 1 | |
| 10 | 4 | george | 9/12/20 | 684 | 1 | |
| 11 | 5 | paul | 2/2/20 | 454 | 1 | |
| 12 | 6 | amy | 1/12/20 | 252 | 1 | |
| 13 | 7 | susan | 5/30/20 | 131 | 1 | |
| 14 | 7 | susan | 6/6/20 | 252 | 1 | |
| 15 | 7 | susan | 5/30/20 | 131 | 1 | |
+-------+-----+--------+---------+------+-----+-----------+
By the end, i'd like to have a table like this:
+-------+-----+--------+---------+------+-----+-----------+
| RowID | pID | Name | Date | Code | QTY | Purchased |
+-------+-----+--------+---------+------+-----+-----------+
| 1 | 1 | bob | 9/29/20 | 123 | 1 | |
| 2 | 1 | bob | 8/10/20 | 456 | 1 | |
| 3 | 2 | rob | 9/15/20 | 123 | 5 | |
| 4 | 3 | john | 7/12/20 | 987 | 2 | |
| 5 | 4 | george | 9/12/20 | 684 | 1 | |
| 6 | 5 | paul | 2/2/20 | 454 | 1 | |
| 7 | 6 | amy | 1/12/20 | 252 | 1 | |
| 8 | 7 | susan | 5/30/20 | 131 | 2 | |
| 9 | 7 | susan | 6/6/20 | 252 | 1 | |
+-------+-----+--------+---------+------+-----+-----------+
Where exactly identical rows are combined into one row, and the QTY field is summed, that I can then add purchases to, or make deductions from the quantity as a total. Using GROUP BY statements can achieve this, but when I go to alter the quantity or add purchases to each person, it treats it like the first table, as if nothing was actually grouped.
So you have this table:
| RowID | pID | Name | Date | Code | QTY | Purchased |
+-------+-----+--------+---------+------+-----+-----------+
| 1 | 1 | bob | 9/29/20 | 123 | 1 | |
| 2 | 1 | bob | 8/10/20 | 456 | 1 | |
| 3 | 2 | rob | 9/15/20 | 123 | 1 | |
| 4 | 2 | rob | 9/15/20 | 123 | 1 | |
| 5 | 2 | rob | 9/15/20 | 123 | 1 | |
| 6 | 2 | rob | 9/15/20 | 123 | 1 | |
| 7 | 2 | rob | 9/15/20 | 123 | 1 | |
| 8 | 3 | john | 7/12/20 | 987 | 1 | |
| 9 | 3 | john | 7/12/20 | 987 | 1 | |
| 10 | 4 | george | 9/12/20 | 684 | 1 | |
| 11 | 5 | paul | 2/2/20 | 454 | 1 | |
| 12 | 6 | amy | 1/12/20 | 252 | 1 | |
| 13 | 7 | susan | 5/30/20 | 131 | 1 | |
| 14 | 7 | susan | 6/6/20 | 252 | 1 | |
| 15 | 7 | susan | 5/30/20 | 131 | 1 | |
The best way, as has been suggested, is to create a new table with the content of your query, then to rename the old table, and the new table to the original table's name, to check if everything is all right, and to drop the original table if yes.
CREATE TABLE indata_new AS
WITH grp AS (
SELECT
MIN(rowid) AS orowid
, pid
, name
, MAX(date) AS date
, code
, SUM(qty) AS qty
FROM indata
GROUP BY
pid
, name
, code
)
SELECT
ROW_NUMBER() OVER(ORDER BY orowid ASC) AS rowid
, *
FROM grp;
ALTER TABLE indata RENAME TO indata_old;
ALTER TABLE indata_new RENAME TO indata;
-- if "indata" now contains the data you want ...
SELECT * FROM indata;
-- out rowid | orowid | pid | name | date | code | qty
-- out -------+--------+-----+--------+------------+------+-----
-- out 1 | 1 | 1 | bob | 2020-09-29 | 123 | 1
-- out 2 | 2 | 1 | bob | 2020-08-10 | 456 | 1
-- out 3 | 3 | 2 | rob | 2020-09-15 | 123 | 5
-- out 4 | 8 | 3 | john | 2020-07-12 | 987 | 2
-- out 5 | 10 | 4 | george | 2020-09-12 | 684 | 1
-- out 6 | 11 | 5 | paul | 2020-02-02 | 454 | 1
-- out 7 | 12 | 6 | amy | 2020-01-12 | 252 | 1
-- out 8 | 13 | 7 | susan | 2020-05-30 | 131 | 2
-- out 9 | 14 | 7 | susan | 2020-06-06 | 252 | 1
-- you can ...
DROP TABLE indata_old;
I have a table like this and i want to output without duplication of the same user. If i use group by it shows only one record on the same column. iam also using left join for location and user name. A little help
+------+---------+----------+---------+
| user | work id | location | time |
+------+---------+----------+---------+
| 1 | 42 | 1 | 2hr |
| 1 | 42 | 1 | 10min |
| 1 | 42 | 1 | 30min |
| 2 | 42 | 1 | 4hr |
| 2 | 42 | 1 | 2.30min |
| 1 | 50 | 2 | 4min |
| 1 | 50 | 2 | 5min |
| 2 | 20 | 3 | 3hr |
| 1 | 20 | 3 | 6hr |
+------+---------+----------+---------+
Iam looking for this
+------+---------+----------+
| user | work id | location |
+------+---------+----------+
| 1 | 42 | 1 |
| 1 | 50 | 2 |
| 1 | 20 | 3 |
| 2 | 42 | 1 |
| 2 | 20 | 3 |
+------+---------+----------+
You simply need a distinct clause here -
SELECT DISTINCT user
,workid
,location
FROM YOUR_TABLE
ORDER BY user
,location
I have two tables orders and customers:
select * from orders;
+------+---------------------+-------------+--------+
| oid | date | customer_id | amount |
+------+---------------------+-------------+--------+
| 102 | 2009-10-08 00:00:00 | 4 | 300 |
| 100 | 2009-10-08 00:00:00 | 3 | 15000 |
| 101 | 2008-10-08 00:00:00 | 2 | 1300 |
| 105 | 2010-10-08 00:00:00 | 1 | 400 |
| 106 | 2014-12-23 00:00:00 | 3 | 300 |
+------+---------------------+-------------+--------+
select * from customers;
+------+--------+------+-----------+----------+
| id | name | age | address | salary |
+------+--------+------+-----------+----------+
| 1 | ramesh | 32 | Ahmedabad | 2000.00 |
| 2 | khilan | 25 | delhi | 1500.00 |
| 3 | muffy | 22 | bhopal | 8500.00 |
| 4 | suresh | 48 | mumbai | 24000.00 |
| 1 | ramesh | 32 | Ahmedabad | 300.00 |
| 5 | akil | 21 | madurai | 1000.00 |
| 6 | rajesh | 22 | delhi | 5000.00 |
+------+--------+------+-----------+----------+
What I'm trying to do is to do SUM(salary) from customers and subtract it with the SUM(amount) from orders table. I have tried with this query:
SELECT id ,NAME,SUM(salary),SUM(amount),SUM(salary)-SUM(amount)
FROM customers a LEFT JOIN orders b ON a.id=b.customer_id
GROUP BY NAME;
This will return the following result, which some of them return incorrect value:
+------+--------+-------------+-------------+-------------------------+
| id | name | SUM(salary) | SUM(amount) | SUM(salary)-SUM(amount) |
+------+--------+-------------+-------------+-------------------------+
| 5 | akil | 1000.00 | NULL | NULL |
| 2 | khilan | 1500.00 | 1300 | 200.00 |
| 3 | muffy | 17000.00 | 15300 | 1700.00 |
| 6 | rajesh | 5000.00 | NULL | NULL |
| 1 | ramesh | 2300.00 | 800 | 1500.00 |
| 4 | suresh | 24000.00 | 300 | 23700.00 |
+------+--------+-------------+-------------+-------------------------+
My expected output is as following:
+------+--------+-------------+-------------+-------------------------+
| id | name | SUM(salary) | SUM(amount) | SUM(salary)-SUM(amount) |
+------+--------+-------------+-------------+-------------------------+
| 5 | akil | 1000.00 | NULL | 1000 |
| 2 | khilan | 1500.00 | 1300 | 200.00 |
| 3 | muffy | 8500 | 15300 | -6800 |
| 6 | rajesh | 5000.00 | NULL | 5000 |
| 1 | ramesh | 2300.00 | 400 | 1900.00 |
| 4 | suresh | 24000.00 | 300 | 23700.00 |
+------+--------+-------------+-------------+-------------------------+
One way is to make calculation on orders into sub-query.
To cater for NULL value, you can use IFNULL(value,0).
SELECT id,NAME,SUM(salary),amt,SUM(salary)-IFNULL(amt,0)
FROM customers a LEFT JOIN
(SELECT customer_id, SUM(amount) amt FROM orders GROUP BY customer_id) b
ON a.id=b.customer_id
GROUP BY NAME;
I have the following data table from which I would like to sum the values of the field 'pts' for each 'pid' as follows:
The sum of the top 3 values per 'cont' plus the values of any other 'cont' per 'pid'. The results should be presented in DESC order by 'total'
+--------+-----+------+
| pid | pts | cont |
+--------+-----+------+
| 121693 | 40 | 1 |
| 121693 | 80 | 2 |
| 121693 | 120 | 1 |
| 121693 | 100 | 1 |
| 121693 | 500 | 1 |
| 121694 | 20 | 1 |
| 121694 | 0 | 2 |
| 121694 | 30 | 3 |
| 121695 | 0 | 1 |
| 121695 | 30 | 2 |
| 121695 | 0 | 1 |
+--------+-----+------+
In this example the query should return something like this
+--------+-------+
| pid | total |
+--------+-------+
| 121693 | 800 |
| 121694 | 50 |
| 121695 | 30 |
+--------+-------+
Is this possible?
Thanks in advance.
SELECT DISTINCT pid, SUM(Pts) AS Total
FROM your tablename
GROUP BY Pid
ORDER BY TOTAL
(Requires testing and minor fixes on small syntax)
I'm trying to make query to find the Customer who has ordered the greatest quantity of Products from the following table!
mysql> select * from ORDERS;
+---------+---------+------------+-----+
| CUSTNUM | PRODNUM | DATE | QTY |
+---------+---------+------------+-----+
| 125216 | 2323 | 2016-03-21 | 2 |
| 136101 | 2357 | 2016-03-21 | 5 |
| 136101 | 2357 | 2016-10-12 | 1 |
| 136101 | 2357 | 2016-11-25 | 5 |
| 136101 | 3737 | 2016-10-12 | 10 |
| 136101 | 9193 | 2016-11-25 | 5 |
| 182764 | 2357 | 2015-03-21 | 12 |
| 182764 | 2357 | 2016-05-12 | 10 |
| 212836 | 3737 | 2015-09-16 | 6 |
| 455566 | 4143 | 2016-02-09 | 10 |
| 455566 | 4143 | 2016-05-12 | 10 |
+---------+---------+------------+-----+
expected result
+-------------+------------------+
| CUSTNUM | quantity_ordered |
+-------------+------------------+
| 136101 | 26 |
+-------------+------------------+
Thanks in advance for help.
Use group by clause.
For more info,
Please refer some tutorials
Or read the official docs
SELECT CUSTNUM, SUM(QTY) s FROM ORDERS GROUP BY CUSTNUM
ORDER BY s DESC LIMIT 1
SQLfiddle