i have two sql queries which give resultset with same attributes... i want to combine these two result sets...
my first query gives
order_id frequency
-------------------------
1 5
3 7
10 2
12 3
and second query gives
order_id frequency
-------------------------
1 3
10 2
12 8
what i finally want in result is
order_id frequency
-------------------------
1 5
3 7
10 2
12 3
1 3
10 2
12 8
here union will not work as if there is are two same tuples such as pair
10 2
it should appear twice.
please suggest some mysql query;
have you tried with UNION ALL?
Use
UNION ALL
to avoid removing dups.
You need Union All
Union must have an equal number of expressions in their target lists
Select order_id, frequency from Table_A
Union All
Select order_id, frequency from Table_B
Related
I have the following data,
id emp_id csa_taken
1 100 2
2 100 2
3 100 0
4 100 2
5 101 2
6 101 2
7 101 0
8 101 0
I expect a result with count where csa_taken=2 for individual employee.
expected result:
emp_id count_csa_taken
100 3
101 2
I have tried the following query with a failed attempt.
Select count(employee_id) From $employeeCSA where csa_taken=2
Please suggest as I am new to sql.
If I understand you correctly you like to count all employees with a cas_taken of two. As there are multiple entries for the csa_taken for one employee you need to group them.
E.g.:
SELECT COUNT(*) FROM $employeeCSA WHERE csa_taken = 2 GROUP_BY employee_id
Please note that COUNT(*) counts the rows (not the fields).
You also need group by. Try like:
Select count(employee_id),emp_id From $employeeCSA where csa_taken=2
group by emp_id
If i understand correctly, then you can try this:
SELECT emp_id,COUNT(emp_id) from dbo.Sample WHERE csa_token = 2 GROUP BY emp_id
Please help me with writing a query for the following condition. I have a table which I have listed below
ID Wt1 Wt1_Type Wt2 Wt2_Type Wt3 Wt3_Type Wt4 Wt4_Type
--------------------------------------------------------------
1 200 1 220 1 300 2 400 3
2 100 4 150 3 100 5 120 1
3 100 3 110 1 200 5 100 4
I want a query to sum all the the weights (wt1, wt2, wt3, wt4) grouped on the weight type (wt1_type, wt2_type, wt3_type, wt4_type).
The output should look like
Wt_type Total
1 650
2 300
3 650
4 200
5 300
Can someone please help me draft a mysql query to get this result ?
Thanks
You can try below - using union all and subquery
select Wt_Type,sum(Wt) as total from
(
select Wt1_Type as Wt_Type,Wt1 as Wt from tablename
union all
select Wt2_Type ,Wt2 from tablename
union all
select Wt3_Type ,Wt3 from tablename
union all
select Wt4_Type ,Wt4 from tablename
)A group by Wt_Type
Rather than giving the answer by #fa06, which should work for you, I am going to suggest using a better table design. Here is how you should be storing your data:
ID Type Wt
-------------
1 1 200
2 4 100
3 3 100
4 1 220
5 3 150
6 1 110
7 2 300
8 5 100
9 5 200
10 3 400
11 1 120
12 4 100
Note that there is a single column which stores the type and a single column for that type's weight. Now your expected output just requires a very simple query:
SELECT Type, SUM(Wt) AS Total
FROM yourTableUpdated
GROUP BY Type;
Databases are really good at performing operations across rows, much less so across columns.
Use this it should be work
select Wt_Type,sum(Wt) as total from ( select Wt1_Type as Wt_Type,Wt1 as Wt from tablename union all select Wt2_Type ,Wt2 from tablename union all select Wt3_Type ,Wt3 from tablename union all select Wt4_Type ,Wt4 from tablename )A group by Wt_Type
I have a database with a table called BOOKINGS containing the following values
main-id place-id start-date end-date
1 1 2018-8-1 2018-8-8
2 2 2018-6-6 2018-6-9
3 3 2018-5-5 2018-5-8
4 4 2018-4-4 2018-4-5
5 5 2018-3-3 2018-3-10
5 1 2018-1-1 2018-1-6
4 2 2018-2-1 2018-2-10
3 3 2018-3-1 2018-3-28
2 4 2018-4-1 2018-4-6
1 5 2018-5-1 2018-5-15
1 3 2018-6-1 2018-8-8
1 4 2018-7-1 2018-7-6
1 1 2018-8-1 2018-8-18
1 2 2018-9-1 2018-9-3
1 5 2018-10-1 2018-10-6
2 5 2018-11-1 2018-11-5
2 3 2018-12-1 2018-12-25
2 2 2018-2-2 2018-2-19
2 4 2018-4-4 2018-4-9
2 1 2018-5-5 2018-5-23
What I need to do is for each main-id I need to find the largest total number of days for every place-id. Basically, I need to determine where each main-id has spend the most time.
This information must then be put into a view, so unfortunately I can't use temporary tables.
The query that gets me the closest is
CREATE VIEW `MOSTTIME` (`main-id`,`place-id`,`total`) AS
SELECT `BOOKINGS`.`main-id`, `BOOKINGS`.`place-id`, SUM(DATEDIFF(`end-date`, `begin-date`)) AS `total`
FROM `BOOKINGS`
GROUP BY `BOOKINGS`.`main-id`,`RESERVATION`.`place-id`
Which yields:
main-id place-id total
1 1 24
1 2 18
1 5 5
2 1 2
2 2 20
2 4 9
3 1 68
3 2 24
3 3 30
4 1 5
4 2 10
4 4 1
5 1 19
5 2 4
5 5 7
What I need is then the max total for each distinct main-id:
main-id place-id total
1 1 24
2 2 20
3 1 68
4 2 10
5 1 19
I've dug through a large amount of similar posts that recommend things like self joins; however, due to the fact that I have to create the new field total using an aggregate function (SUM) and another function (DATEDIFF) rather than just querying an existing field, my attempts at implementing those solutions have been unsuccessful.
I am hoping that my query that got me close will only require a small modification to get the correct solution.
Having hyphen character - in column name (which is also minus operator) is a really bad idea. Do consider replacing it with underscore character _.
One possible way is to use Derived Tables. One Derived Table is used to determine the total on a group of main id and place id. Another Derived Table is used to get maximum value out of them based on main id. We can then join back to get only the row corresponding to the maximum value.
CREATE VIEW `MOSTTIME` (`main-id`,`place-id`,`total`) AS
SELECT b1.main_id, b1.place_id, b1.total
FROM
(
SELECT `main-id` AS main_id,
`place-id` AS place_id,
SUM(DATEDIFF(`end-date`, `begin-date`)) AS total
FROM BOOKINGS
GROUP BY main_id, place_id
) AS b1
JOIN
(
SELECT dt.main_id, MAX(dt.total) AS max_total
FROM
(
SELECT `main-id` AS main_id,
`place-id` AS place_id,
SUM(DATEDIFF(`end-date`, `begin-date`)) AS total
FROM BOOKINGS
GROUP BY main_id, place_id
) AS dt
GROUP BY dt.main_id
) AS b2
ON b1.main_id = b2.main_id AND
b1.total = b2.max_total
MySQL 8+ solution would be utilizing the Row_Number() functionality:
CREATE VIEW `MOSTTIME` (`main-id`,`place-id`,`total`) AS
SELECT b.main_id, b.place_id, b.total
FROM
(
SELECT dt.main_id,
dt.place_id,
dt.total
ROW_NUMBER() OVER (PARTITION BY dt.main_id
ORDER BY dt.total DESC) AS row_num
FROM
(
SELECT `main-id` AS main_id,
`place-id` AS place_id,
SUM(DATEDIFF(`end-date`, `begin-date`)) AS total
FROM BOOKINGS
GROUP BY main_id, place_id
) AS dt
GROUP BY dt.main_id
) AS b
WHERE b.row_num = 1
I would like to join two tables but to specify that the result should have unique values on one of the columns.
http://sqlfiddle.com/#!2/70ded/4
Instead of
URLID DOMAINID
13 5
9 3
10 3
11 4
12 4
6 2
7 2
8 2
1 1
2 1
3 1
4 1
5 1
I would like to get:
URLID DOMAINID
13 5
9 3
11 4
6 2
1 1
Is it possible doing it and by that creating a faster query?
Your query will not be faster, because you have to remove the duplicates. Here is a MySQL way:
select urlID, domainID
from (<your query>) as t
group by domainID
This uses a MySQL mis(feature) called hidden columns. In most databases, you would use:
select min(urlID), domainID
from (<your query>) as t
group by domainID
How about this? I just applied an aggregate to the urlid and added a group by:
SELECT min(urlid) urlid, domain_stack.domainid
FROM domain_stack
INNER JOIN url_stack
ON url_stack.domainid = domain_stack.domainid
group by domain_stack.domainid
order by domain_stack.domainid desc
See SQL Fiddle with Demo
SELECT urlid,domain_stack.domainid
FROM domain_stack
INNER JOIN url_stack
ON url_stack.domainid = domain_stack.domainid
GROUP BY domainid
I have the following query, in the top select statement (sum(l.app_ln_amnt)/count(l.app_ln_amnt)) works well but in the union I want to find the total of (sum(l.app_ln_amnt)/count(l.app_ln_amnt)) query from the top select statement However my solution seems to be off I need some help please
select
(sum(l.app_ln_amnt)/count(l.app_ln_amnt)),
from receipt_history l
UNION
select
SUM(sum(l.app_ln_amnt)/count(l.app_ln_amnt)),
from receipt_history l
THIS is what the initial table looks like
id app_ln_amnt
1 2
1 2
1 2
1 2
2 5
3 7
4 9
id app_ln_amnt
1 2
2 5
3 7
4 9
total 23
Now my table looks like the second one but the total is 29 and im trying to get it to be 23
You could use a subquery rather than a union
SELECT SUM(Value) FROM
(select
(sum(l.app_ln_amnt)/count(l.app_ln_amnt)) AS value
from receipt_history l ) t