SELECT f_name,l_name,teachers.first_name,teachers.t_id,p_id,paid_amount,family_id,date,sum(payments.paid_amount)
FROM payments
LEFT JOIN family ON family.id = payments.family_id
LEFT JOIN teachers ON family.teacher_id = teachers.t_id
How can I get the selected columns fully and the sum column separately?
because that sum function makes all the selected result one row
SELECT f_name,l_name,teachers.first_name,teachers.t_id,p_id,paid_amount,family_id,date
FROM payments
LEFT JOIN family ON family.id = payments.family_id
LEFT JOIN teachers ON family.teacher_id = teachers.t_id
This query is working fine without the sum column
You didn't tell the database, which column to use for aggregating the data. Don't know which database you are using, but some complain, that there is no GROUP BY statement in the SQL text.
Please try with the following query:
SELECT f_name,l_name,teachers.first_name,teachers.t_id,p_id,paid_amount,family_id,date,sum(payments.paid_amount)
FROM payments
LEFT JOIN family ON family.id = payments.family_id
LEFT JOIN teachers ON family.teacher_id = teachers.t_id
GROUP BY f_name,l_name,teachers.first_name,teachers.t_id,p_id,paid_amount,family_id,date
GROUP BY tells the database, which are the key columns in the aggregation.
If you want all the payments, use a subquery or join:
SELECT f_name, l_name, t.first_name, t.t_id, p.p_id, p.paid_amount, p.family_id, date,
(select sum(p.paid_amount) from payments) as all_paid
FROM payments p LEFT JOIN
family f
ON f.id = p.family_id LEFT JOIN
teachers t
ON f.teacher_id = tetchers.t_id;
SELECT f_name,l_name,t.first_name,t.t_id,p_id,paid_amount,family_id,date,sum(p.paid_amount)
FROM payments p,family f,teachers t where f.id = p.family_id and f.teacher_id = t.t_id
Group by f_name,l_name,teachers.first_name,teachers.t_id,p_id,paid_amount,family_id
You can add date column also in Group by expression based on your requirement. Example:
f_name,l_name,teachers.first_name,teachers.t_id,p_id,paid_amount,family_id,date
Related
Outputted Date
I have a seperate Members table which has all the members ID's and I want to list all those but get rid of the ones that are displayed in this list.
SELECT DISTINCT tbl_classregistration.ClassID, tbl_classregistration.MemberID
FROM tbl_member INNER JOIN (tbl_classes
INNER JOIN tbl_classregistration ON
tbl_classes.ClassID = tbl_classregistration.ClassID) ON
tbl_member.MemberID = tbl_classregistration.MemberID
GROUP BY tbl_classregistration.ClassID, tbl_classregistration.MemberID
HAVING (((tbl_classregistration.ClassID)=[Enter ClassID]));
Thats the SQL View
Use not in:
select memberid from members where memberid not in (SELECT DISTINCT tbl_classregistration.MemberID
FROM tbl_member INNER JOIN (tbl_classes INNER JOIN tbl_classregistration ON tbl_classes.ClassID = tbl_classregistration.ClassID) ON tbl_member.MemberID = tbl_classregistration.MemberID
GROUP BY tbl_classregistration.ClassID, tbl_classregistration.MemberID
HAVING (((tbl_classregistration.ClassID)=[Enter ClassID])))
I am using SQL Server to query these three tables that look like (there are some extra columns but not that relevant):
Customers -> Id, Name
Addresses -> Id, Street, StreetNo, CustomerId
Sales -> AddressId, Week, Total
And I would like to get the total sales per week and customer (showing at the same time the address details). I have come up with this query
SELECT a.Name, b.Street, b.StreetNo, c.Week, SUM (c.Total) as Total
FROM Customers a
INNER JOIN Addresses b ON a.Id = b.CustomerId
INNER JOIN Sales c ON b.Id = c.AddressId
GROUP BY a.Name, c.Week, b.Street, b.StreetNo
and even if my SQL skill are close to none it looks like it's doing its job. But now I would like to be able to show 0 whenever the one customer don't have sales for a particular week (weeks are just integers). And I wonder if somehow I should get distinct values of the weeks in the Sales table, and then loop through them (not sure how)
Any help?
Thanks
Use CROSS JOIN to generate the rows for all customers and weeks. Then use LEFT JOIN to bring in the data that is available:
SELECT c.Name, a.Street, a.StreetNo, w.Week,
COALESCE(SUM(s.Total), 0) as Total
FROM Customers c CROSS JOIN
(SELECT DISTINCT s.Week FROM sales s) w LEFT JOIN
Addresses a
ON c.CustomerId = a.CustomerId LEFT JOIN
Sales s
ON s.week = w.week AND s.AddressId = a.AddressId
GROUP BY c.Name, a.Street, a.StreetNo, w.Week;
Using table aliases is good, but the aliases should be abbreviations for the table names. So, a for Addresses not Customers.
You should generate a week numbers, rather than using DISTINCT. This is better in terms of performance and reliability. Then use a LEFT JOIN on the Sales table instead of an INNER JOIN:
SELECT a.Name
,b.Street
,b.StreetNo
,weeks.[Week]
,COALESCE(SUM(c.Total),0) as Total
FROM Customers a
INNER JOIN Addresses b ON a.Id = b.CustomerId
CROSS JOIN (
-- Generate a sequence of 52 integers (13 x 4)
SELECT ROW_NUMBER() OVER (ORDER BY a.x) AS [Week]
FROM (VALUES(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) a(x)
CROSS JOIN (SELECT x FROM (VALUES(1),(1),(1),(1)) b(x)) b
) weeks
LEFT JOIN Sales c ON b.Id = c.AddressId AND c.[Week] = weeek.[Week]
GROUP BY a.Name
,b.Street
,b.StreetNo
,weeks.[Week]
Please try the following...
SELECT Name,
Street,
StreetNo,
Week,
SUM( CASE
WHEN Total IS NULL THEN
0
ELSE
Total
END ) AS Total
FROM Customers a
JOIN Addresses b ON a.Id = b.CustomerId
RIGHT JOIN Sales c ON b.Id = c.AddressId
GROUP BY a.Name,
c.Week,
b.Street,
b.StreetNo;
I have modified your statement in three places. The first is I changed your join to Sales to a RIGHT JOIN. This will join as it would with an INNER JOIN, but it will also keep the records from the table on the right side of the JOIN that do not have a matching record or group of records on the left, placing NULL values in the resulting dataset's fields that would have come from the left of the JOIN. A LEFT JOIN works in the same way, but with any extra records in the table on the left being retained.
I have removed the word INNER from your surviving INNER JOIN. Where JOIN is not preceded by a join type, an INNER JOIN is performed. Both JOIN and INNER JOIN are considered correct, but the prevailing protocol seems to be to leave the INNER out, where the RDBMS allows it to be left out (which SQL-Server does). Which you go with is still entirely up to you - I have left it out here for illustrative purposes.
The third change is that I have added a CASE statement that tests to see if the Total field contains a NULL value, which it will if there were no sales for that Customer for that Week. If it does then SUM() would return a NULL, so the CASE statement returns a 0 instead. If Total does not contain a NULL value, then the SUM() of all values of Total for that grouping is performed.
Please note that I am assuming that Total will not have any NULL values other than from the RIGHT JOIN. Please advise me if this assumption is incorrect.
Please also note that I have assumed that either there will be no missing Weeks for a Customer in the Sales table or that you are not interested in listing them if there are. Again, please advise me if this assumption is incorrect.
If you have any questions or comments, then please feel free to post a Comment accordingly.
I wrote a query. this query sum fields from 2 different table. And grouped by main table id field. But second left outer join is not grouped and giving me different results.
SELECT s.*,
f.firma_adi,
sum(sd.fiyat) AS konak,
sum(ss.fiyat) AS sponsor
FROM fuar_sozlesme1 s
INNER JOIN fuar_firma_2012 f
ON ( s.cari = f.cari )
LEFT OUTER JOIN fuar_sozlesme1_detay sd
ON ( sd.sozlesme_id = s.id )
LEFT OUTER JOIN fuar_sozlesme1_sponsor ss
ON ( ss.sozlesme_id = s.id )
GROUP BY s.id
ORDER BY s.id DESC
I know, it is really complicated but I'm stucking on this issue.
My question is: why second left outer join is not correctly sum of field . If I remove second left outer join or first, everything is normal.
The problem is that you have multiple dimensions on your data, and the number of rows is multiplying beyond what you expect. I would suggest that you run the query for one id, without the group by, to see what rows the join is producing.
One way to fix this is by using correlated subqueries:
select s.*, f.firma_adi,
(select SUM(sd.fiyat)
from fuar_sozlesme1_detay fd
where sd.sozlesme_id = s.id
) as konak,
(select SUM(ss.fiyat)
from fuar_sozlesme1_sponsor ss
where (ss.sozlesme_id = s.id)
) as sponsor
from fuar_sozlesme1 s inner join
fuar_firma_2012 f
on (s.cari = f.cari)
order by s.id DESC
By the way, you appear to by using MySQL (because your query is not parsable in any other dialect). You should tag your questions with the version of the database you are using.
select
picks.`fbid`,
picks.`time`,
categories.`name` as cname,
options.`name` as oname,
users.`name`
from
picks
left join categories
on (categories.`id` = picks.`cid`)
left join options
on (options.`id` = picks.oid)
left join users
on (users.fbid = picks.`fbid`)
order by
time desc
that query returns a result that like:
my question is.... I would like to modify the query to select only DISTINCT fbid's. (perhaps the first row only sorted by time)
can someone help with this?
select
p2.fbid,
p2.time,
c.`name` as cname,
o.`name` as oname,
u.`name`
from
( select p1.fbid,
min( p1.time ) FirstTimePerID
from picks p1
group by p1.fbid ) as FirstPerID
JOIN Picks p2
on FirstPerID.fbid = p2.fbid
AND FirstPerID.FirstTimePerID = p2.time
LEFT JOIN Categories c
on p2.cid = c.id
LEFT JOIN Options o
on p2.oid = o.id
LEFT JOIN Users u
on p2.fbid = u.fbid
order by
time desc
I don't know why you originally had LEFT JOINs, as it appears that all picks must be associated with a valid category, option and user... I would then remove the left, and change them to INNER joins instead.
The first inner query grabs for each fbid, the FIRST entry time which will result in a single entity for the FBID. From that, it re-joins to the picks table for the same ID and timeslot... then continues for the rest of the category, options, users join criteria of that single entry.
2 options, you could write a group by clause.
Or you could write a nested query joined back to itself to get pertinent info.
Nested aliased table:
SELECT
n.fBids
FROM
MyTable t
INNER JOIN
(SELECT DISTINCT fBids
FROM MyTable) n
ON n.ID = t.ID
Or group by option
SELECT fBId from MyTable
GROUP BY fBID
select picks.`fbid`, picks.`time`, categories.`name` as cname,
options.`name` as oname, users.`name` from picks left join categories
on (categories.`id` = picks.`cid`) left join options on (options.`id` = picks.oid)
left join users on (users.fbid = picks.`fbid`)
order by time desc GROUP BY picks.`fbid`
select
picks.fbid,
MIN(picks.time) as first_time,
MAX(picks.time) as last_time
from
picks
group by
picks.fbid
order by
MIN(picks.time) desc
However, if you want only distinct fbid's you cannot display cname and other columns at the same time.
I have 2 tables AP and INV where both have the columns [PROJECT] and [Value].
I want a query to return something like this :
PROJECT | SUM_AP | SUM_INV
I came up with the code below but it's returning the wrong results ( sum is wrong ).
SELECT AP.[PROJECT],
SUM(AP.Value) AS SUM_AP,
SUM(INV.Value) AS SUM_INV
FROM AP INNER JOIN INV ON (AP.[PROJECT] =INV.[PROJECT])
WHERE AP.[PROJECT] = 'XXXXX'
GROUP BY AP.[PROJECT]
The results from your query are wrong because the values you are trying to summarize are being grouped, which causes duplicate values to be included in the SUM.
You could solve it with a couple of sub-selects:
SELECT
AP1.[PROJECT],
(SELECT SUM(AP2.Value) FROM AP AS AP2 WHERE AP2.PROJECT = AP1.PROJECT) AS SUM_AP,
(SELECT SUM(INV2.Value) FROM INV AS INV2 WHERE INV2.PROJECT = AP1.PROJECT) AS SUM_INV
FROM AP AS AP1
INNER JOIN INV AS INV1
ON (AP1.[PROJECT] =INV1.[PROJECT])
WHERE AP1.[PROJECT] = 'XXXXX'
GROUP BY AP1.[PROJECT]
If you have N rows in AP with a given project ID, and M rows in INV with that ID, then the join between the two tables on the project ID will have a total of N*M rows for that project, because the same row in AP will be repeated for every row in INV that has that project ID, and vice versa. Hence why your counts are most likely off (because it's counting the same row in a given table multiple times due to repetition from the join).
Instead, you might want to try doing a join between the results of two subqueries, one which groups the first table by project ID and does that its sum, and the second which groups the other table by project ID and does that sum - then joining once you only have 1 row with sum for each project ID.
If PROJECT is the parent table, you should select FROM the project table, and do a left outer join on the two child tables:
SELECT PROJECT.PROJECT_ID, SUM(AP.Value) AS SUM_AP, SUM(INV.Value) AS SUM_INV
FROM PROJECT
LEFT OUTER JOIN AP ON (AP.[PROJECT] = PROJECT.[PROJECT_ID])
LEFT OUTER JOIN INV ON (INV.[PROJECT] = PROJECT.[PROJECT_ID])
WHERE PROJECT.[PROJECT_ID] = 'XXXXX'
GROUP BY PROJECT.[PROJECT_ID]
You could separate the two sum calculations. One way I can think of is to move the inventory calculation to a subquery, like:
SELECT
AP.[PROJECT]
, SUM(AP.Value) AS SUM_AP
, SummedInv as SUM_INV
FROM AP
LEFT JOIN (
SELECT PROJECT, SUM(Value) AS SUM_INV
FROM INV
GROUP BY PROJECT
) SummedInv ON SummedInv.Project = AP.Project
GROUP BY AP.PROJECT, SummedInv.SUM_INV
Because the SummedInv subquery is grouped on project, it's safe to group on SummedInv.SUM_INV in the outer query as well.
how about this query :
select SUM(gpCutBody.actualQty) as cutQty , SUM(gpSewBody.quantity) as sewQty
from jobOrder
inner join gpCutHead on gpCutHead.joNum = jobOrder.joNum
inner join gpSewHead on gpSewHead.joNum = jobOrder.joNum
inner join gpCutBody on gpCutBody.gpCutID = gpCutHead.gpCutID
inner join gpSewBody on gpSewBody.gpSewID = gpSewHead.gpSewID
where jobOrder.joNum = '36'
here is the link to the ERD: http://dl.dropbox.com/u/18794525/AUG%207%20DUMP%20STAN.png
Try:
SELECT AP.[PROJECT] AS PROJECT, SUM(AP.[Value]) AS SUM_AP, SUM(INV.[Value]) AS SUM_INV
FROM AP, INV
WHERE AP.[PROJECT] = INV.[PROJECT]
AND AP.[PROJECT] = 'XXXXX'
GROUP BY AP.[PROJECT]