Whats wrong with this SQL?
SELECT Id, (select SUM(VALUE) from SomeTable) AS SumValue, GETDATE()
FROM MyTable
WHERE SumValue > 0
You cannot use aliased columns in the SELECT clause in the same query, except in ORDER BY.
It needs to be subqueried
SELECT Id, SumValue, GETDATE()
FROM (
SELECT Id, (select SUM(VALUE) from TABLE) AS SumValue
FROM MyTable
) X
WHERE SumValue > 0
That is the general case. For your specific query, it doesn't make sense because the subquery is not correlated to the outer query, so either NO rows show, or ALL rows show (with the same SumValue). I will simply assume you have simplified the query a lot since a table name of "table" doesn't really work.
I would probably rewrite like this:
SELECT a.Id, b.SumValue, GETDATE() as [now]
FROM MyTable a
Join
(
select id, SUM(VALUE) as [SumValue]
from [TABLE]
Group by id
)b on a.Id = b.Id
WHERE b.SumValue > 0
This is assuming that the value you are totalling relates to the ID in your table?
right way is
SELECT Id, (select SUM(VALUE) from TABLE) AS SumValue, GETDATE()
FROM MyTable
WHERE (select SUM(VALUE) from TABLE) > 0
Related
Query
select * from table1
where having count(reference)>1
I want to select * the data which have duplicate data,any idea why my query is not working?
Below are my expect result..
You can make use of window function count to find number of rows per id and reference and then filter to get those which have count more than 1.
;with cte as (
select t.*, count(*) over (partition by id, reference) cnt
from table1 t
)
select * from cte where cnt > 1;
Demo
In the above solution, I have made an assumption that name and id has one to one correspondence (which is true as per your given data). If that's not the case, add name too in the partition by clause:
;with cte as (
select t.*, count(*) over (partition by name, id, reference) cnt
from table1 t
)
select * from cte where cnt > 1;
I might actually approach this by using a subquery with GROUP BY:
SELECT t1.*
FROM table1 t1
INNER JOIN
(
SELECT Name, ID, reference
FROM table1
GROUP BY Name, ID, reference
HAVING COUNT(*) > 1
) t2
ON t1.Name = t2.Name AND
t1.ID = t2.ID AND
t1.reference = t2.reference
Demo here:
Rextester
Try this ), first i get count by partition, after that i get row with count > 1
select No, Name, ID, Reference
from (select count(*) over (partition by name, ID, reference) cnt, table1.* from table1)
where cnt>1
The easy way (although maybe not the best for performance) would be:
select * from table1 where reference in (
select reference from table1 group by reference having count(*)>1
)
In a subselect you have the duplicated data, and in the outter select you have all the data for these references.
I have a table like this:
Now I wish to GROUP BY the Field1, which is not that hard. After this, I want to add a row number to each group. And finally.. this has to be done in Access which is slightly different of course. So this is the code I have:
SELECT A.*, (SELECT COUNT(*) FROM Tabel1 WHERE A.ID>=ID) AS RowNum
FROM Tabel1 AS A
ORDER BY A.ID;
So, this works well, but now I can't group it. How can I group it?
You can wrap your entire query in a subquery:
select
B.*
from
( SELECT
A.*,
(SELECT COUNT(*) FROM Tabel1 WHERE A.ID>=ID) AS RowNum
FROM Tabel1 AS A
) as B
ORDER BY B.ID;
From here, you can do joins as though the output of your query was a table. I have no idea what you actually want to group on, but here is an example:
select
B.Field1, count (*) as count, max (B.RowNum) as max_row
from
( SELECT
A.*,
(SELECT COUNT(*) FROM Tabel1 WHERE A.ID>=ID) AS RowNum
FROM Tabel1 AS A
) as B
group by
b.Field1
-- Edit 11/14/2016 --
I think I see now. In that case, first build a query to handle your grouping. This is just an example:
SELECT Field1, min (Id) as min_id, max (id) as max_id
FROM Table1
group by Field1
Name this query Table1_Summary for the purposes of our example.
Now, in your new query, you will refer to Table1_Summary in the exact same way you did with your example:
SELECT
t.*,
(select count (*) from Table1_Summary t2 where t.Field1 >= t2.Field1) as RowNum
FROM Table1_Summary t
You could theoretically do this in a single query, but for readability/maintainability sake, I'd recommend you keep them split. Here is an example of the output:
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
In my table I need to know if each ID has one and only one ID_name. How can I write such query?
I tried:
select ID, count(distinct ID_name) as count_name
from table
group by ID
having count_name > 1
But it takes forever to run.
Any thoughts?
select ID
from YourTable
group by
ID
having count(distinct ID_name) > 1
or
select *
from YourTable yt1
where exists
(
select *
from YourTable yt2
where yt1.ID = yt2.ID
and yt1.ID_Name <> yt2.ID_Name
)
Now, most ID columns are defined as primary key and are unique. So in a regular database you'd expect both queries to return an empty set.
select tt.ID,max(tt.myRank)
from
(
select
ip.ID,ip.ID_name,
ROW_Number() over (partition by ip.ID,ip.ID_nameorder by ip.ID) as myRank
from YourTable ip
) tt
group by tt.ID
This gives you every ID with it's total number of ID_Name
If you want only those ID's which have more than one name associated just add a where clause
e.g.
select tt.ID,max(tt.myRank)
from
(
select
ip.ID,ip.ID_name,
ROW_NUMBER() over (partition by ip.ID,ip.ID_nameorder by ip.ID) as myRank
from YourTable ip
) tt
**where tt.myRank > 1**
group by tt.ID
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