I have the table
_______________
| from | to |
| 1 | 2 |
| 2 | 2 |
| 1 | 2 |
| 3 | 2 |
Where from and to are countries.
I need to get top 10 countries met in from and to.
The result would be
________________
|country| count |
| 2 | 5 |
| 1 | 2 |
| 3 | 1 |
I reached what I need in two queries, but I am almost sure there is possibility to manage it with one.
Thanks in advance.
You want to use a UNION ALL and then GROUP BY using the union of from and to:
SELECT country, COUNT(*) AS count FROM
(SELECT from AS country FROM table_name
UNION ALL
SELECT to AS country FROM table_name)
GROUP BY country
ORDER BY count DESC
LIMIT 10
Related
I'm a starter at SQL and I have the following tables, ORDER_PRODUCTS, listing the products of an order and EXCHANGE_PRODUCTS, listing products that will be exchanged.
Both have the same fields, and I need to make a selection counting the amount of products in both tables, distinguishing them by the order_id, does anyone knows how I can do this?
ORDER_PRODUCTS
+-----+------------+----------+---------+
| id | product_id | order_id | amount |
+-----+------------+----------+---------+
| 1 | 5 | 1 | 2 |
| 2 | 7 | 1 | 1 |
| 3 | 13 | 5 | 1 |
| 4 | 18 | 8 | 3 |
| 5 | 45 | 11 | 4 |
+-----+------------+----------+---------+
EXCHANGE_PRODUCTS
+-----+------------+----------+---------+
| id | product_id | order_id | amount |
+-----+------------+----------+---------+
| 1 | 5 | 1 | 1 |
| 2 | 7 | 1 | 2 |
| 3 | 13 | 5 | 1 |
| 4 | 3 | 8 | 2 |
| 5 | 2 | 11 | 1 |
+-----+------------+----------+---------+
You want to use union all to combine the tables and then aggregate them. I might recommend:
select order_id, sum(ordered) as ordered, sum(exchanged) as exchanged,
sum(exchanged + ordered) as total
from ((select order_id, amount as ordered, 0 as exchanged
from order_products
) union all
(select order_id, 0 as ordered, amount as exchanged
from exhange_products
)
) oe
group by order_id;
It is important to use union all rather than union, because union removes duplicates (which can result in bad numbers). Union also incurs overhead that is unnecessary.
And, by "count amount" I assume you really mean to take the sum.
I think this query should do what you Need:
select sum(amount), order_id from (
select amount,order_id from order_products
union
select amount,order_id from Exchange_products)
group by order_id
I am trying to build a leaderboard based on the points (money) a user has. The points are stored in multiple tables and some users may not have points in a given table.
Table: account
--------------------------
| uid | name | locker |
|-----|---------|--------|
| 1 | Bob | 15 |
| 2 | Dave | 2 |
| 3 | Jim | 5 |
--------------------------
Table: container
------------------------
| account_uid | money |
|-------------|--------|
| 1 | 4 |
| 3 | 1 |
| 3 | 2 |
| 3 | 4 |
------------------------
Table: vehicle
------------------------
| account_uid | money |
|-------------|--------|
| 2 | 2 |
| 2 | 1 |
| 3 | 2 |
------------------------
I would like to see the results ouput as -
Bob 19
Jim 14
Dave 5
Note that some tables do not have points for some people.
This code did not work for me. It seems to have duplicated the points somehow.
SELECT
act.name,
act.uid,
SUM(COALESCE(act.locker,0) + COALESCE(con.money,0) + COALESCE(veh.money,0)) AS total
FROM account as act
LEFT JOIN container as con
ON act.uid = con.account_uid
LEFT JOIN vehicle as veh
ON act.uid = veh.account_uid
Group By act.name
ORDER BY total DESC
How about calculating the totals one at a time and then combining them into one?
select account.uid, account.name, sum(tot.Money) as TotalMoney from
(
(select uid, sum(Money) as Money from container
group by uid)
union all
(select uid, sum(Money) as Money from vehicle
group by uid)
) tot
inner join account on
tot.uid = account.uid
group by account.uid
Lets's say i have a table sign_ins which has data like so: (the real table has 3.5 million rows)
+-----------+---------+------------------+
| school_id | user_id | date(created_at) |
+-----------+---------+------------------+
| 1 | 4 | 2009-04-20 |
| 1 | 4 | 2009-04-21 |
| 1 | 4 | 2009-05-06 |
| 1 | 5 | 2009-04-20 |
| 1 | 5 | 2009-06-26 |
| 1 | 5 | 2009-06-26 |
| 2 | 6 | 2009-04-21 |
| 2 | 6 | 2009-06-26 |
| 2 | 7 | 2009-04-20 |
| 2 | 7 | 2009-04-20 |
+-----------+---------+------------------+
created_at is a datetime field but i'm calling date() on it to get the day.
I have the concept of a "login_days" which is the number of distinct days on which a given user has a sign_in record. I want to order the schools by the number of login days, highest first, and return the number of login days.
So, looking at the data above, school 1 has two users (4 & 5). User 4 has three sign_ins, on 3 distinct days, so 3 "login_days". User 5 has three logins, but only 2 distinct days, so 2 "login_days". Therefore school 1 has 5 login days.
Looking at school 2, it has 3 login days: 2 from user 6 and 1 from user 7.
So, i would want to get this back from the query:
+-----------+------------+
| school_id | login_days |
+-----------+------------+
| 1 | 5 |
| 2 | 4 |
+-----------+------------+
I can't quite figure out how to do the query. I started off with this (i have the id < 11 part in there just to get my example data instead of my entire table of 3.5 million rows):
mysql> select school_id from sign_ins where id < 11 group by school_id, user_id, date(created_at);
+-----------+
| school_id |
+-----------+
| 1 |
| 1 |
| 1 |
| 1 |
| 1 |
| 2 |
| 2 |
| 2 |
+-----------+
8 rows in set (0.00 sec)
I can see in here that there are 5 rows for school 1 and 3 for school 2, which looks like it's worked. But i need to group that further, and order by that grouped number, to get it like in my required results. It must be something simple, can someone show me what i'm missing?
thanks, Max
MySQL allows you to count the number of distinct values for multiple expressions. So, this is basically an aggregation query with the appropriate count:
select school_id, count(distinct user_id, date(created_at)) as NumLoginDays
from sign_ins
group by school_id;
I have this 2 tables and I need to return the moset used office. Note: 1 office can be used by more than 1 guys and the column ido from TableB is populate from TableA
Probaly is a query with group by and desc limit 1
TableA
| ido| office | guy |
---------------------
| 1 | office1| guy1|
| 2 | office2| guy2|
| 3 | office1| guy3|
| 4 | office1| guy4|
| 5 | office5| guy5|
| 6 | office2| guy6|
TableB
| idb| vizit | ido|
---------------------
| 1 | date | 4 |
| 2 | date | 2 |
| 3 | date | 5 |
| 4 | date | 6 |
| 5 | date | 1 |
| 6 | date | 6 |
Thanks!
You were correct in that GROUP BY, LIMIT and DESC are useful here; it leads to a fairly straight forward query;
SELECT TableA.office
FROM TableA
JOIN TableB
ON TableA.ido = TableB.ido
GROUP BY TableA.office
ORDER BY COUNT(*) DESC
LIMIT 1
What it does is basically create rows with all valid combinations, counting the number of generated rows per office. A plain descending sort by that count will give you the most frequently used office.
An SQLfiddle to test with.
I want to select the data as per condition:I have a table with physician_key and corresponding quality score for a given month. I want to select count of distinct physicians with quality score 1,2.
For a month, there could be more entries for a physician_key and accordingly the quality assigned(on scale 1-7). I want to select only the count of those physicians which have quality (1,2) and if the same physician has quality >2 in given month, I don't want to count that physician.I want the information by product and month
I created an example table, since you didn't provide one:
mysql> select * from sales_mkt_rep_qual;
+-------------------+---------+-------+-------------------+
| GEO_PHYSICIAN_KEY | product | month | SALES_REP_QUALITY |
+-------------------+---------+-------+-------------------+
| 1 | a | 8 | 1 |
| 1 | a | 8 | 2 |
| 1 | a | 8 | 3 |
| 2 | b | 8 | 2 |
| 2 | b | 8 | 1 |
| 2 | b | 9 | 2 |
| 1 | a | 9 | 2 |
| 2 | b | 9 | 3 |
| 3 | a | 9 | 2 |
+-------------------+---------+-------+-------------------+
The query from your comment indeed gives an error:
SELECT COUNT(DISTINCT GEO_PHYSICIAN_KEY) AS encount_1to2,
product,MONTH
FROM sales_mkt_rep_qual
WHERE MAX(SALES_REP_QUALITY) = 2 ;
ERROR 1111 (HY000): Invalid use of group function
If you change that to:
SELECT DISTINCT geo_physician_key AS encount_1to2, product, month
FROM sales_mkt_rep_qual
WHERE (geo_physician_key,month,product)
NOT IN (
SELECT geo_physician_key, month, product
FROM sales_mkt_rep_qual
WHERE sales_rep_quality >2 );
you see the detailed result:
+--------------+---------+-------+
| encount_1to2 | product | month |
+--------------+---------+-------+
| 2 | b | 8 |
| 1 | a | 9 |
| 3 | a | 9 |
+--------------+---------+-------+
No, you can introduce the counting:
SELECT COUNT(distinct geo_physician_key ) AS no_of_physicians,product, month
FROM sales_mkt_rep_qual
WHERE (geo_physician_key,month,product)
NOT IN (
SELECT geo_physician_key, month, product
FROM sales_mkt_rep_qual WHERE sales_rep_quality >2 )
GROUP BY month, product;
+------------------+---------+-------+
| no_of_physicians | product | month |
+------------------+---------+-------+
| 1 | b | 8 |
| 2 | a | 9 |
+------------------+---------+-------+
If that still isn't what you are looking for, give more specific table structure and data example.
Try this:
SELECT count(DISTINCT physician_key)
FROM my_table
WHERE month = desired_month
AND max(quality) = 2
GROUP BY month
Actually I want the data to be like the output below:
+--------------+---------+-------+
| encount_1to2 | product | MONTH |
+--------------+---------+-------+
| 2 | b | 8 |
+--------------+---------+-------+
and for the criteria SALES_REP_QUALITY <= 2, isn't there a possibility that while selecting the distinct geo physician key, it might select out of first 2 considering it matches the criteria? Thats the reason I have used Thanix approach of max function with group by product and month, so that the aggregate function is applied on every product within a month