this my query but result false where number row different , that's to say whenever tableA select 2 row and tableB select 3 result is false
select sum(tableA.value)+sum(tableB.value1) )
from tableA,tableB
where tableA.data between '2016-01-21' and '2016-03-09'
and tableB.date2 between '2016-01-21' and '2016-03-09'
You need to do the sums in subqueries before joining. A simple rule: never use commas in the from clause.
select coalesce(avalue, 0) + coalesce(bvalue, 0)
from (select sum(a.value) as avalue
from tableA a
where a.data between '2016-01-21' and '2016-03-09'
) a cross join
(select sum(b.value) as bvalue
from tableB b
where b.data between '2016-01-21' and '2016-03-09'
) b;
OK . So here's what my understanding is.
You are trying to sum up two columns from two different tables and get the sum of the summed up columns. isn't ?? Correct me if I am wrong.If this is the case then
A Simple Subquery Can Come To Your Rescue.
Select
(Select SUM(value) From tableA
where data between '2016-01-21' and '2016-03-09') +
(Select SUM(value1) From tableB
where date2 between '2016-01-21' and '2016-03-09') FinalValue
Related
I have two queries:
select * from tableA
and
select a,b from tableA
group by a,b
the first query returns 2101 rows
the second query returns 2100 rows
I want to know which row is in the first but not in the second. It should be simple with NOT IN, but I can't find the correct syntax as NOT IN should be in WHERE statement. but I don't have a WHERE statement in my case.
There are N ways to do that and one of the simplest should be to find the rows that have a count > 1 when grouped on a,b.
select a,b from tableA
group by a,b
having count(*) > 1
Here is a sample:
with tableA as
(
select * from (values
(1,1,1),
(1,1,1),
(1,2,1)
) as t(a,b,c)
)
select a, b from tableA
group by a, b
having count(*) > 1;
You can get duplicates this way:
select a,b from tableA
group by a,b having count(1) > 1
I'm having a hard time summing up a column on two tables. The scenario is something like this (refer to the image below)
Table 1 may have a lot of rows per Date. But Table 2 may only consists of two rows of data per Date. What I wanted to do is to sum up all Item/Price (Table1) according to their Date and ADD them with another SUM of Item/Price of Table2. The category of SUM is by Date.
I tried any joins statement (left, right or inner) but it does not produce the result that I am expecting to. My expected result is the Result table. But on my query, it produces a very high value.
Thanks.
Use a UNION clause like this:
WITH t(d, p) AS (
SELECT [Date], Price FROM Table1
UNION ALL
SELECT [Date], Price FROM Table2
)
SELECT d, SUM(p) FROM t GROUP BY d
You can do this with UNION ALL in either a subquery or a cte, cte shown here:
;WITH cte AS (SELECT [Date], Price
FROM Table1
UNION ALL
SELECT [Date], Price
FROM Table2
)
SELECT [Date], SUM(Price) AS Total_Price
FROM cte
GROUP BY [Date]
Demo: SQL Fiddle
Try This,
with cte (C_Date,C_Price)
as
(
SELECT date,SUM(price) FROM table_1
group by date
union
SELECT date,SUM(price) FROM table_2
group by date
)
select c_date,SUM(c_price) from cte
group by C_Date
Try this
Select t.date,P1+P2
from(
Select Date,sum(Price) P1
from table1 t
group by Date
) t
left join
(
Select Date,sum(Price) P2
from table t2
group by date
) t1 on t.date = t1.date
group by date
I'm having difficulty with what I figure should be an easy problem. I want to select all the columns in a table for which one particular column has duplicate values.
I've been trying to use aggregate functions, but that's constraining me as I want to just match on one column and display all values. Using aggregates seems to require that I 'group by' all columns I'm going to want to display.
If I understood you correctly, this should do:
SELECT *
FROM YourTable A
WHERE EXISTS(SELECT 1
FROM YourTable
WHERE Col1 = A.Col1
GROUP BY Col1
HAVING COUNT(*) > 1)
You can join on a derived table where you aggregate and determine "col" values which are duplicated:
SELECT a.*
FROM Table1 a
INNER JOIN
(
SELECT col
FROM Table1
GROUP BY col
HAVING COUNT(1) > 1
) b ON a.col = b.col
This query gives you a chance to ORDER BY cola in ascending or descending order and change Cola output.
Here's a Demo on SqlFiddle.
with cl
as
(
select *, ROW_NUMBER() OVER(partition by colb order by cola ) as rn
from tbl)
select *
from cl
where rn > 1
I don't understand what I'm doing wrong. I'm trying to get a weekly COUNT of every ID that meets criteria A OR criteria B.
select CREATE_WEEK, count ( A.PK )
from TABLE1 A
where ( A.PK not in (select distinct ( B.FK )
from TABLE2 B
where B.CREATE_TIMESTAMP > '01-Jan-2013')
or A.PK in (select A.PK
from ( select A.PK, A.CREATE_TIMESTAMP as A_CRT, min ( B.CREATE_TIMESTAMP ) as FIRST_B
from TABLE1 A, TABLE2 B
where A.PK = B.FK
and A.CREATE_TIMESTAMP > '01-Jan-2013'
and B.CREATE_TIMESTAMP > '01-Jan-2013'
group by A.PK, A.CREATE_TIMESTAMP)
where A_CRT < FIRST_B) )
and A.CREATE_TIMESTAMP > '01-Jan-2013'
and CREATE_WEEK >= 2
and THIS_WEEK - CREATE_WEEK >= 1
group by CREATE_WEEK
order by CREATE_WEEK asc
**Note: PK in table1 = FK in table2, so in the first subquery, I'm checking whether the PK from table1 exists as FK in table2. Week comes from TO_CHAR (TO_DATE (TRUNC (A.CREATE_TIMESTAMP, 'IW')), 'IW')
When I take out the OR and run the query on either subquery the results are returned in 1-2 seconds. But when I try to run the combined query, the results aren't returned after 20 minutes.
I know I can run them separately and then sum them in a spreadsheet, but I'd rather just get one number.
I'm trying to get a weekly COUNT of every ID that meets criteria A OR criteria B
However your code is:
ID NOT IN (subquery A) OR ID IN (subquery B)
The NOT is at odds with your requirement.
Assuming you ID's that meet both criteria, use:
ID in (
select ... -- this is subquery A
union
select ... -- this is subquery B)
What is the method in T-SQL to select the orginal values limited by a HAVING attribute. For example, if I have
A|B
10|1
11|2
10|3
How would I get all the values of B (Not An Average or some other summary stat), Grouped by A, having a Count (Occurrences of A) greater than or equal two 2?
Actually, you have several options to choose from
1. You could make a subquery out of your original having statement and join it back to your table
SELECT *
FROM YourTable yt
INNER JOIN (
SELECT A
FROM YourTable
GROUP BY
A
HAVING COUNT(*) >= 2
) cnt ON cnt.A = yt.A
2. another equivalent solution would be to use a WITH clause
;WITH cnt AS (
SELECT A
FROM YourTable
GROUP BY
A
HAVING COUNT(*) >= 2
)
SELECT *
FROM YourTable yt
INNER JOIN cnt ON cnt.A = yt.A
3. or you could use an IN statement
SELECT *
FROM YourTable yt
WHERE A IN (SELECT A FROM YourTable GROUP BY A HAVING COUNT(*) >= 2)
A self join will work:
select B
from table
join(
select A
from table
group by 1
having count(1)>1
)s
using(A);
You can use window function (no joins, only one table scan):
select * from (
select *, cnt=count(*) over(partiton by A) from table
) as a
where cnt >= 2