I have two tables:
Table 'bookings':
id | date | hours
--------------------------
1 | 06/01/2016 | 2
1 | 06/02/2016 | 1
2 | 06/03/2016 | 2
3 | 06/03/2016 | 4
Table 'lookupCalendar':
date
-----
06/01/2016
06/02/2016
06/03/2016
I want to join them together so that I have a date for each booking so that the results look like this:
Table 'results':
id | date | hours
--------------------------
1 | 06/01/2016 | 2
1 | 06/02/2016 | 1
1 | 06/03/2016 | 0 <-- Added by query
2 | 06/01/2016 | 0 <-- Added by query
2 | 06/02/2016 | 0 <-- Added by query
2 | 06/03/2016 | 2
3 | 06/01/2016 | 0 <-- Added by query
3 | 06/02/2016 | 0 <-- Added by query
3 | 06/03/2016 | 4
I have tried doing a cross-apply, but that doesn't get me there, neither does a full join. The FULL JOIN just gives me nulls in the id column and the cross-apply gives me too much data.
Is there a query that can give me the results table above?
More Information
It might be beneficial to note that I am doing this so that I can calculate an average hours booked over a period of time, not just the number of records in the table.
Ideally, I'd be able to do
SELECT AVG(hours) AS my_average, id
FROM bookings
GROUP BY id
But since that would just give me a count of the records instead of the count of the days I want to cross apply it with the dates. Then I think I can just do the query above with the results table.
select i.id, c.date, coalesce(b.hours, 0) as hours
from lookupCalendar c
cross join (select distinct id from bookings) i
left join bookings b
on b.id = i.id
and b.date = c.date
order by i.id, c.date
Try this:
select c.date, b.id, isnull(b.hours, 0)
from lookupCalendar c
left join bookings b on b.date = c.date
LookupCalendar is your main table because you want the bookings against each date, irrespective of whether there was a booking on that date or not, so a left join is required.
I am not sure if you need to include b.id to solve your actual problem though. Wouldn't you just want to get the total number of hours booked against each date like this, to then calculate the average?:
select c.date, sum(isnull(b.hours, 0))
from lookupCalendar c
left join bookings b on b.date = c.date
group by c.date
You can try joining all the combinations of IDs and dates and left joining the data;
WITH Booking AS (SELECT *
FROM (VALUES
( 1 , '06/01/2016', 2 )
, ( 1 , '06/02/2016', 1 )
, ( 2 , '06/03/2016', 2 )
, ( 3 , '06/03/2016', 4 )
) x (id, date, hours)
)
, lookupid AS (
SELECT DISTINCT id FROM Booking
)
, lookupCalender AS (
SELECT DISTINCT date FROM Booking
)
SELECT ID.id, Cal.Date, ISNULL(B.Hours,0) AS hours
FROM lookupid id
INNER JOIN lookupCalender Cal
ON 1 = 1
LEFT JOIN Booking B
ON id.id = B.id
AND Cal.date = B.Date
ORDER BY ID.id, Cal.Date
Related
I want to get the number of days someone logs on in a month. Using this query:
select id,
to_char(date_on, 'MM-DD') as mon_dd
from
logs
group by
id, to_char(date_on, 'MM-DD')
I get a table that looks like this:
id | mon_dd
0 | 01-27
3 | 02-23
1 | 01-05
0 | 01-31
2 | 02-01
3 | 02-05
1 | 02-09
I want to get a result that groups the id by the number of days they appear in a month like this:
id | month | days_appeared
0 | jan | 2
0 | feb | 0
1 | jan | 1
1 | feb | 1
2 | jan | 0
2 | feb | 1
3 | jan | 0
3 | feb | 2
You can generate a cartesian product of the distinct months and users in the table, and then bring the table with a left join:
select
i.id,
d.date_month,
count(distinct trunc(l.date_on)) days_appeared
from (select distinct trunc(date_on, 'month') date_month from logs) d
cross join (select distinct id from logs) i
left join logs l
on l.date_on >= d.date_month
and l.date_on < add_months(d.date_month, 1)
and l.id = i.id
group by i.id, d.date_month
If you want to get all months, even those with zeros, then:
select l.id, m.mon, count(distinct trunc(date_on)) as num_days
from (select distinct id from logs) i cross join
(select distinct trunc(date_on, 'month') as mon) m left join
logs l
on l.id = i.id and trunc(date_on, 'month') = m.mon
group by l.id, m.mon;
Note: You might have more efficient sources of the months and ids than using select distinct on the logs table.
I like the use of the WITH clause to build your query in a modular way.
with vals as ( -- all information needed is here
select id
, to_char(mon_dd, 'mon') as month
, to_char(mon_dd,'mm') as mm
from logs
),
months as ( -- the distinct months,
select distinct month, mm -- including the month numbers
from vals -- for ordering the main query
),
ids as ( -- the distinct ids
select distinct id
from vals)
select i.id, m.month, (select count(id) from vals -- for every combination of id
where month=m.month -- and month
and id = i.id) as count -- count the number of ids
from ids i cross join months m
order by i.id, m.mm;
Let me ask you something I've been thinking about for a while. Imagine that you have two tables with data:
MAIN TABLE (A)
| ID | Date |
|:-----------|------------:|
| 1 | 01-01-1990|
| 2 | 01-01-1991|
| 3 | 01-01-1992|
| 4 | 01-01-2000|
| 5 | 01-01-2001|
| 6 | 01-01-2003|
SECONDARY TABLE (B)
| ID | Date | TOTAL |
|:-----------|------------:|--------:|
| 1 | 01-01-1990| 1 |
| 2 | 01-01-1991| 2 |
| 3 | 01-01-1992| 1 |
| 4 | 01-01-2000| 5 |
| 5 | 01-01-2001| 8 |
| 6 | 01-01-2003| 7 |
and you want to select only ID with date greater than 31-12-1999 and get the following columns: ID, Date and Total. For that we have many options but my question would be which of the following would be better in terms of performance:
OPTION 1
With main as(
select id,
date
from A
where date > '31-12-1999'
)
select main.id,
main.date,
B.total
from main inner join B on main.id = b.id
OPTION 1
With main as(
select id,
date
from A
where date > '31-12-1999'
),
secondary as (
select id,
total
from B
where date > '31-12-1999'
)
select main.id,
main.date,
secondary.total
from main inner join secondary on main.id = b.id
Which of both queries would be better in terms of performance? Thanks in advance!
DATE FOR BOTH TABLES MEANS THE SAME
You don't need to use CTE you can directly join two tables -
select A.id,
A.date,
B.total
from A inner join B on A.id = b.id
where A.date > '31-12-1999'
You would need to test on your data. But there is really no need for CTEs:
select a.id a.date, b.total
from a inner join
b
on a.id = b.id
where a.date > '1999-12-31' and b.date > '1999-12-31';
As for your specific question, the two queries are not the same, because the first is filtering on only one date and the second is filtering on two dates. You should run the query that implements the logic that you intend.
I have a column (PL.UNITS) that I need to Total at the bottom of the results of a query, is it possible to sum PL.UNITS that is already summed?
Please see query below.
SELECT ID.DUEDATE AS [DUE DATE], CD.RENEWALDATE, CD.RENEWALSTATUS, CD.CONTRACTNUMBER, L.LOCNAME, L.LOCADDRESS1, L.LOCADDRESS2, L.LOCADDRESS3, L.LOCADDRESS4, L.POSTCODE, SUM(PL.UNITS) AS UNITS from CLIENTDETAILS CD
INNER JOIN LOCATIONS L ON CD.CLIENTNUMBER = L.CLIENTNUMBER
INNER JOIN ITEMDETAILS ID ON L.LOCNUMBER = ID.LOCNUMBER
INNER JOIN PLANT PL ON ID.CODE = PL.CODE
WHERE L.OWNER = 210 and L.STATUSLIVE = 1 and ID.DUEDATE > '01/01/2017'
GROUP BY ID.DUEDATE, CD.RENEWALDATE, CD.RENEWALSTATUS, CD.CONTRACTNUMBER, L.LOCNAME, L.LOCADDRESS1, L.LOCADDRESS2, L.LOCADDRESS3, L.LOCADDRESS4, L.POSTCODE
It's probably best to do this sort of thing in front end development. Nevertheless, here is an example (quick and dirty, but shows the idea) for sql-server:
SELECT COALESCE(a.id, 'total') AS id
, SUM(a.thing) AS thing_summed
FROM (
SELECT '1' id
, 1 thing
UNION
SELECT '2'
, 2 thing
UNION
SELECT '1'
, 3 thing
) AS a
GROUP BY ROLLUP(a.id)
Result:
+-------+--------------+
| id | thing_summed |
+-------+--------------+
| 1 | 4 |
| 2 | 2 |
| total | 6 |
+-------+--------------+
I can't figure out a query that will add and compare across tables. I have three tables:
house
id
-----
1
2
month
id | house_id | btus
---------------------
3 | 1 | 100
4 | 2 | 200
car
id | month_id | btu
--------------------------
5 | 3 | 10
6 | 4 | 20
7 | 3 | 15
I need a query that will return the house ids sorted by total btus for the month and car.
So for the above example it would return 2,1 as (200+20) > (100 + 10 + 15)
SELECT h.*
FROM house
ORDER BY
(
SELECT SUM(c.btu)
FROM month m
JOIN cars c
ON c.month_id = m.id
WHERE m.house_id = h.id
) +
(
SELECT SUM(m.btus)
FROM month m
WHERE m.house_id = h.id
)
DESC
, or this one (probably slightly more efficient):
SELECT h.*
FROM house
ORDER BY
(
SELECT SUM
(
btus +
(
SELECT SUM(btu)
FROM cars c
WHERE c.month_id = m.id
)
)
FROM month m
WHERE m.house_id = h.id
)
DESC
Probably not the most performant use of joins, but:
SELECT house.id
FROM house JOIN month JOIN car
WHERE house.id = month.house_id AND month.id = car.month_id
GROUP BY house.id
ORDER BY sum(car.btu) + sum(month.btus);
The joins and where clause will explode the tables (try it with SELECT * and skipping the group/order clauses), group by will flatten them to one row each, and the sum()s will do the relevant math.
I want to create view that combine data from two tables, sample data in each table is like below.
SELECT Command for TableA
SELECT [ID], [Date], [SUM]
FROM TableA
Result
ID | Date | SUM
1 | 1/1/2010 | 2
1 | 1/2/2010 | 4
3 | 1/3/2010 | 6
SELECT Command for TableB
SELECT [ID], [Date], [SUM]
FROM TableB
Result
ID | Date | SUM
1 | 1/1/2010 | 5
1 | 2/1/2010 | 3
1 | 31/1/2010 | 2
2 | 1/2/2010 | 20
I want output like below
ID | Date | SUMA | SUMB
1 | 1/1/2010 | 2 | 10
1 | 1/2/2010 | 4 | 0
2 | 1/2/2010 | 0 | 20
3 | 1/3/2010 | 6 | 0
How can I do that on SQL Server 2005?
Date information be vary, as modify in table.
Try this...
SELECT
ISNULL(TableA.ID, TableB.ID) ID,
ISNULL(TableA.Date, TableB.Date),
ISNULL(TableA.Sum,0) SUMA,
ISNULL(TableB.Sum, 0) SUMB
FROM
TableA FULL OUTER JOIN TableB
ON TableA.ID = TableB.ID AND TableA.Date = TableB.Date
ORDER BY
ID
A full outer join is what you need because you want to include results from both tables regardless of whether there is a match or not.
I usually union the two queries together and then group them like so:
SELECT ID, [Date], SUM(SUMA) As SUMA, SUM(SUMB) AS SUMB
FROM (
SELECT ID, [Date], SUMA, 0 AS SUMB
FROM TableA
UNION ALL
SELECT ID, [Date], 0 As SUMA, SUMB
FROM TableB
)
GROUP BY ID, [Date]
SELECT
ISNULL(a.ID, b.ID) AS ID,
ISNULL(a.Date, b.Date) AS Date,
ISNULL(a.SUM, 0) AS SUMA,
ISNULL(b.SUM, 0) AS SUMB,
FROM
TableA AS a
FULL JOIN
TableB AS b
ON a.ID = b.ID
AND a.Date = b.Date;
It's not obvious how you want to combine the two tables. I think this is what you're after, but can you confirm please?
TableA.Date is the most important field; if a given date occurs in TableA then it will be included in the view, but not if it only occurs in TableB.
If a date has records in TableA and TableB and the records have a matching ID, they are combined into one row in the view with SUMA being taken from TableA.Sum and SUMB being TableA.Sum * TableB.Sum (e.g. Date: 01/01/2010, ID: 1) (e.g. Date: 01/03/2010 ID: 3).
If a date has records in TableA and TableB with different IDs, the view include these records separately without multiplying the Sum values at all (e.g. Date 02/01/2010, ID: 1 and ID: 2)