How to get SUM of a column with number of affected rows
I am trying to get the sum of a column and I will also like to return the number of affected columns.
Example:
orders table
id customer_id name amount
----------------------------------------
1 2 burger 5.00
2 2 pizza 6.00
3 2 grape 1.00
4 1 sandwich 4.00
Now I want to SUM the amount column for a particular customer(customer_id) and also return number of items(count rows affected)
I am doing this but it only gets the sum of the amount, I would also like to get number number affected rows (count) from this single query:
SELECT SUM(amount) AS amount
FROM orders
WHERE customer_id = 2
Just do a COUNT as well:
SELECT SUM(amount) AS amount, COUNT(*) AS cnt
FROM orders
WHERE customer_id = 2
If amount is a nullable field and you want to only count NOT NULL rows, then do a COUNT(amount) instead of COUNT(*).
SELECT SUM(amount) AS amount, COUNT(1) AS cnt
FROM orders
WHERE customer_id = 2
Related
We require to get row if product_id is in sequence.
Table (product)
product_id name
1 Parle
2 Coconut
3 Pizza
4 Colgate
5 Ice Cream
6 Nuts
8 Britania
9 Pepsi
Require Output
product_id name
1 Parle
2 Coconut
3 Pizza
4 Colgate
5 Ice Cream
6 Nuts
product_id - 8 and 9 not getting because it is not in sequence.
My Try
select distinct t1.*, t1.product_id, (select GROUP_CONCAT(t2.product_id) from product as t2) as p_id
from product t1
having FIND_IN_SET(t1.product_id+1, p_id)
Output
product_id name
1 Parle
2 Coconut
3 Pizza
4 Colgate
5 Ice Cream
In this try i am not getting product_id - 6 row.
Note : I want MySql query not in PHP.
Thank You !
One way i can think of is to user user defined variable to get the rank of row and then calculate the difference of product id and rank and select only those rows where difference = 0
select *
from(
select f.*,#row:= #row + 1 rank,
product_id - #row as diff
from product f,
(select #row:= 0) t
order by product_id
) t1
where diff = 0
Demo
Or if you want to pick the least sequence no. automatically you can write it as
select *
from(
select f.*,#row:= #row + 1 rank,
product_id - #row as diff
from product f,
(select #row:= (select min(product_id) from product_sale_flag) a) t
order by product_id
) t1
where diff = -1
Demo
Explanation
First query assign's minimum value of product_id to variable #row,then it assigns a rank to each row ordered by product_id in ascending order, once rank is assigned for every row then it calculates the difference between the original value of product_id and lastly using the resultant difference it checks where difference is 0 get those rows because they follow the sequence. Hope this makes sense
s.no Name type of vehicle allotted
1 ram bus
2 krish car
3 raj bus
4 albert car
5 johnson bike
6 raghu car
consider this is a table now i want to retrieve the type of vehicle which is there more number of times in the table and count of it. is there any particular method(query).
You have to use COUNT and GROUP BY.
Count will give you the number of rows grouped by type field.
SELECT COUNT(id) as count, type FROM table_name GROUP BY type
The output should be like this:
count | type
2 | bus
3 | car
1 | Bike
If you need just the highest one, append to the query:
ORDER BY count DESC LIMIT 1
I think you have to count and select the top 1 row.
Query
SELECT t.* FROM(
SELECT `type`, COUNT(`type`) `count`
FROM `your_table_name`
GROUP BY `type`
)t
ORDER BY `count` DESC LIMIT 1;
How can I select last (=max) value from column and count of all rows in single query?
ID ITEM_ID VALUE
1 1 100
2 1 101
3 2 201
4 3 333
5 2 222
6 1 111
I want to select last / max value for particular ITEM_ID and count of all rows with this ID.
For ITEM_ID = 1 thus:
VALUE COUNT
111 3
My query is like this:
SELECT (SELECT COUNT(*) FROM table) AS count, (SELECT value FROM table ORDER BY id DESC LIMIT 1) AS value FROM table WHERE item_id = 1 LIMIT 1
It works but looks ... weird. Is there any better (simpler / faster) solution? Thanks
You need to do a GROUP BY on column ITEM_ID while getting the MAX() and COUNT() like
select max(value) as `VALUE`,
count(*) as `COUNT`
from your_table
group by ITEM_ID;
I have table Orders:
O_Id OrderDate OrderPrice Customer
1 2008/11/12 1000 Hansen
2 2008/10/23 1600 Nilsen
3 2008/09/02 700 Hansen
4 2008/09/03 300 Hansen
5 2008/08/30 2000 Jensen
6 2008/10/04 100 Nilsen
and query:
SELECT COUNT(Customer) AS CustomerNilsen FROM Orders
WHERE Customer='Nilsen'
but is possible add to this results IDs results?
I would like receive
count: 2
and
ids: 2 and 6
GROUP_CONCAT may help:
SELECT
COUNT(Customer) AS CustomerNilsen,
GROUP_CONCAT(O_Id) as IDS
FROM
Orders
WHERE
Customer='Nilsen'
Better to just fetch all the IDs and then use the appropriate row count function:
SELECT `O_Id` FROM `Orders`
WHERE `Customer` = 'Nilsen'
I don't really understand your question, but from what I can tell you just want to SELECT the o_id column in your query:
SELECT COUNT(Customer) AS CustomerNilsen, O_Id AS OrderID FROM Orders WHERE Customer='Nilsen'
You might check this
SELECT Customer,
COUNT(Customer) AS CustomerNilsen
FROM Orders
WHERE Customer = 'Nilsen'
GROUP BY Customer
HAVING CustomerNilsen = 1
Try this
SELECT O1.O_Id AS OrderID,Count(O1.O_Id) AS OrderCount FROM Orders O1
INNER JOIN Orders O2 ON O1.Customer = O2.Customer
WHERE O1.Customer='Nilsen'
GROUP BY O1.Customer,O1.O_Id
I have an SQL database with a table like this:
ObjID Rating
123 1
333 0
123 0
123 1
567 1
An Ojbect ID can show up in the table multiple times with different ratings, I need a way to select a "distinct" ObjID check all of its ratings and compare that to all other "distinct" ObjIDs and compare those ratings. In the end I should have the 1st - 3rd ObjIDs with the over all highest ratings, in this example 123, 567 and 333 would be printed out.
So the rating of each object is the total sum of all ratings for that object in the table?
SELECT ObjID, SUM(Rating) AS TotalRating
FROM ObjectTable
GROUP BY ObjID
ORDER BY SUM(Rating) DESC
LIMIT 3