What query will count the number of rows, but distinct by three parameters?
Example:
Id Name Address
==============================
1 MyName MyAddress
2 MySecondName Address2
Something like:
select count(distinct id,name,address) from mytable
To get a count of the number of unique combinations of id, name and address:
SELECT Count(*)
FROM (
SELECT DISTINCT
id
, name
, address
FROM your_table
) As distinctified
Get all distinct id, name and address columns and count the resulting rows.
SELECT COUNT(*) FROM mytable GROUP BY id, name, address
Another (probably not production-ready or recommended) method I just came up with is to concat the values to a string and count this string distinctively:
SELECT count(DISTINCT concat(id, name, address)) FROM mytable;
You can also do something like:
SELECT COUNT(DISTINCT id + name + address) FROM mytable
Having to return the count of a unique Bill of Materials (BOM) where each BOM have multiple positions, I dd something like this:
select t_item, t_pono, count(distinct ltrim(rtrim(t_item)) + cast(t_pono as varchar(3))) as [BOM Pono Count]
from BOMMaster
where t_pono = 1
group by t_item, t_pono
Given t_pono is a smallint datatype and t_item is a varchar(16) datatype
Related
I need to show the ID (which is unique in every case) and the name, which is sometimes different. In my code I only want to show the names IF they are unique.
I tried with both distinct and count(*)=1, nothing solves my problem.
SELECT DISTINCT id, name
FROM person
GROUP BY id, name
HAVING count(name) = 1;
The result is still showing the names multiple times
By "unique", I assume you mean names that only appear once. That is not what "distinct" means in SQL; the use of distinct is to remove duplicates (either for counting or in a result set).
If so:
SELECT MAX(id), name
FROM person
GROUP BY name
HAVING COUNT(*) = 1;
If your DBMS supports it, you can use a window function:
SELECT id, name
FROM (
SELECT id, name, COUNT(*) OVER(PARTITION BY name) AS NameCount -- get count of each name
FROM person
) src
WHERE NameCount = 1
If not, you can do:
SELECT id, name
FROM person
WHERE name IN (
SELECT name
FROM person
GROUP BY name
HAVING COUNT(*) = 1 -- Only get names that occur once
)
Im new to DB2 , and tried based on some similar posts, I have a table where I need to find the count of IDs based on where status=P and
the count of(primary=1) more than once.
so my result should be 2 here - (9876,3456)
Tried:
SELECT id, COUNT(isprimary) Counts
FROM table
GROUP BY id
HAVING COUNT(isprimary)=1;
Try the query below:
select ID as IDs,Count(isPrimary) as isPrimary
From Table
where Status = 'p'
Group by ID
Having Count(isPrimary) >1
You are close, I think all you need to do is to add a where clause like:
SELECT id, COUNT(*) as Counted
FROM table
WHERE PrimaryFlag = 1
AND[status] = 'P'
GROUP BY id
EDIT: if you need to count only the distinct IDs, then try:
SELECT COUNT(t.ID) FROM
(
SELECT id, COUNT(*) as Counted
FROM table
WHERE PrimaryFlag = 1
AND[status] = 'P'
GROUP BY id
) as t
i have two columns - email id and customer id, where an email id can be associated with multiple customer ids. Now, I need to list only those email ids (along with their corresponding customer ids) which are having a count of more than 1 customer id. I tried using grouping sets, rollup and cube operators, however, am not getting the desired result.
Any help or pointers would be appreciated.
SELECT emailid
FROM
( SELECT emailid, count(custid)
FROM table
Group by emailid
Having count(custid) > 1
)
I think this will get you what you want, if I am understanding you question correctly
select emailid, customerid from tablename where emailid in
(
select emailid from tablename group by emailid having count(emailid) > 1
)
Sounds like you would need to use HAVING
e.g
SELECT email_id, COUNT(customer_id)
From sometable
GROUP BY email_id
HAVING COUNT(customer_id) > 1
HAVING allows you to filter following the grouping of a particular column.
WITH email_ids AS (
SELECT email_id, COUNT(customer_id) customer_count
FROM Table
GROUP BY email_id
HAVING count(customer_id) > 1
)
SELECT t.email_id, t.customer_id
FROM Table t
INNER JOIN email_ids ei
ON ei.email_id = t.email_id
If you need a comma separated list of all of their customer id's returned with the single email id, you could use GROUP_CONCAT for that.
This would find all email_id's with at least 1 customer_id, and give you a comma separated list of all customer_ids for that email_id:
SELECT email_id, GROUP_CONCAT(customer_id)
FROM your_table
GROUP BY email_id
HAVING count(customer_id) > 1;
Assuming email_id #1 was assigned to customer_ids 1, 2, & 3, your output would look like:
email_id | customer_id
1 | 1,2,3
I didn't realize you were using MS SQL, there's a thread here about simulating GROUP_CONCAT in MS SQL: Simulating group_concat MySQL function in Microsoft SQL Server 2005?
SELECT t1.email, t1.customer
FROM table t1
INNER JOIN (
SELECT email, COUNT(customer)
FROM table
GROUP BY email
HAVING COUNT(customer)>1
) t2 on t1.email = t2.email
This should get you what your looking for.
Basically, as other ppl have stated, you can filter group by results with HAVING. But since you want the customerids afterwards, join the entire select back to your original table to get your results. Could probably be done prettier but this is easy to understand.
SELECT
email_id,
STUFF((SELECT ',' + CONVERT(VARCHAR,customer_id) FROM cust_email_table T1 WHERE T1.email_id = T2.email_id
FOR
XML PATH('')
),1,1,'') AS customer_ids
FROM
cust_email_table T2
GROUP BY email_id
HAVING COUNT(*) > 1
this would give you a single row per email id and comma seperated list of customer id's.
I have one column for Farmer Names and one column for Town Names in my table TRY.
I want to find Average_Number_Of_Farmers_In_Each_Town.
Select TownName ,AVG(num)
FROM(Select TownName,Count(*) as num From try Group by TownName) a
group by TownName;
But this query always returns int values. How can i get values in float too?
;WITH [TRY]([Farmer Name], [Town Name])
AS
(
SELECT N'Johny', N'Bucharest' UNION ALL
SELECT N'Miky', N'Bucharest' UNION ALL
SELECT N'Kinky', N'Ploiesti'
)
SELECT AVG(src.Cnt) AS Average
FROM
(
SELECT COUNT(*)*1.00 AS Cnt
FROM [TRY]
GROUP BY [TRY].[Town Name]
) src
Results:
Average
--------
1.500000
Without ... *1.00 the result will be (!) 1 (AVG(INT 2 , INT 1) -truncated-> INT 1, see section Return types).
Your query is always returning int logically because the average is not doing anything. Both the inner and the outer queries are grouping by town name -- so there is one value for each average, and that average is the count.
If you are looking for the overall average, then something like:
Select AVG(cast(cnt as float))
FROM (Select TownName, Count(*) as cnt
From try
Group by TownName
) t
You can also do this without the subquery as:
select cast(count(*) as float) /count(distinct TownName)
from try;
EDIT:
The assumption was that each farmer in the town has one row in try. Are you just trying to count the number of distinct farmers in each town? Assuming you have a field like FarmerName that identifies a given farmer, that would be:
select TownName, count(distinct FarmerName)
from try
group by TownName;
In SQL Server 2005 I have a table cm_production that lists all the code that's been put into production. The table has a ticket_number, program_type, program_name and push_number along with some other columns.
GOAL: Count all the DISTINCT program names by program type and push number.
What I have so far is:
DECLARE #push_number INT;
SET #push_number = [HERE_ADD_NUMBER];
SELECT DISTINCT COUNT(*) AS Count, program_type AS [Type]
FROM cm_production
WHERE push_number=#push_number
GROUP BY program_type
This gets me partway there, but it's counting all the program names, not the distinct ones (which I don't expect it to do in that query). I guess I just can't wrap my head around how to tell it to count only the distinct program names without selecting them. Or something.
Count all the DISTINCT program names by program type and push number
SELECT COUNT(DISTINCT program_name) AS Count,
program_type AS [Type]
FROM cm_production
WHERE push_number=#push_number
GROUP BY program_type
DISTINCT COUNT(*) will return a row for each unique count. What you want is COUNT(DISTINCT <expression>): evaluates expression for each row in a group and returns the number of unique, non-null values.
I needed to get the number of occurrences of each distinct value. The column contained Region info.
The simple SQL query I ended up with was:
SELECT Region, count(*)
FROM item
WHERE Region is not null
GROUP BY Region
Which would give me a list like, say:
Region, count
Denmark, 4
Sweden, 1
USA, 10
You have to create a derived table for the distinct columns and then query the count from that table:
SELECT COUNT(*)
FROM (SELECT DISTINCT column1,column2
FROM tablename
WHERE condition ) as dt
Here dt is a derived table.
SELECT COUNT(DISTINCT program_name) AS Count, program_type AS [Type]
FROM cm_production
WHERE push_number=#push_number
GROUP BY program_type
try this:
SELECT
COUNT(program_name) AS [Count],program_type AS [Type]
FROM (SELECT DISTINCT program_name,program_type
FROM cm_production
WHERE push_number=#push_number
) dt
GROUP BY program_type
You can try the following query.
SELECT column1,COUNT(*) AS Count
FROM tablename where createddate >= '2022-07-01'::date group by column1
This is a good example where you want to get count of Pincode which stored in the last of address field
SELECT DISTINCT
RIGHT (address, 6),
count(*) AS count
FROM
datafile
WHERE
address IS NOT NULL
GROUP BY
RIGHT (address, 6)