I've been stuck on how to write a particular query for the following question:
How many employees are in how many businesses?
My end result should look like this:
EmployeeId Count
BusinessId Count
1
23473423
2
56245764
3
834456
So there are 23473423 businesses that have 1 employee, 23473423 businesses that have 2 employees, etc.
I have a table with a list of items including EmployeeId and BusinessId. A BusinessId can connect to many EmployeeIds.
So far I have the following code to get me employees per business
Select BusinessId,
Count(EmployeeId) as EIdCount
From Table
Group by BusinessId
Which gets me me a list of BusinessIds and how many EmployeeIds are attached to it.
BusinessId
EIdCount
23
2
24
5
25
1
26
3
But now I need to figure out how to further group it to where the BusinessId's can be grouped by the Grouped Counted Employee Ids. I've looked at subqueries, having by, and group but I am still at a loss how to progress this without running into an error.
Thank you for your help in advance!
Not sure if this is what you want:
Select EIdCount, Count(BusinessId)
From (
Select BusinessId,
Count(EmployeeId) As EIdCount
From Table
Group by BusinessId
) A
Group By EIdCount
Just use a subquery:
select
EIdCount,count(BusinessId) as [BusinessId Count]
from
(
--your original query/start
Select BusinessId,
Count(EmployeeId) as EIdCount
From Table
Group by BusinessId
--your original query/end
)t
group by EIdCount
I am very new to SQL and I am wondering how to solve this issue. For example, my table looks as follows:
As you see in the table item_id 1 appears in both city_id 1 and 2, so does the item_id 4, but I want to get all the items where appears only in one city_id.
In this example, these would be item_id 2 (appearing only in city_id 2) and item_id 3 (appearing in city_id 1).
Use aggregation on item_id and count distinct values of city_id. The having clause can be used to filter on aggregates.
select item_id from mytable group by id having count(distinct city_id) = 1
You can use the following query:
SELECT item_id
FROM table_name
GROUP BY item_id
HAVING COUNT(DISTINCT city_id) = 1
In case you want to see the city_id to you can use this query:
SELECT item_id, MIN(city_id) AS city_id
FROM example
GROUP BY item_id
HAVING COUNT(DISTINCT city_id) = 1
Since there is only one city_id you can use MIN or MAX to get the id.
demo on dbfiddle.uk
You want all the id where they have only one distinct city:
SELECT item_id
FROM table
GROUP BY item_id
HAVING count(distinct city_id) = 1
It works by counting all the different values that city_id has for the same item_id. For those item ids where they repeat a lot, but the city_id is always the same the count of unique values in the city id is 1, and we can look for these using a HAVING clause. "Having" is like a where clause that runs after a GROUP BY operation is completed. It is the conceptual equivalent of this:
SELECT item_id
FROM
(
SELECT item_id, count(distinct city_id) as cdci
FROM table
GROUP BY item_id
) x
WHERE cdci = 1
If you want the city id too you can either get the MAX city (because in this case there is only one city so it's safe to do):
SELECT item_id, MAX(city_id) as city_id
FROM table
GROUP BY item_id
HAVING count(distinct city_id) = 1
or you could join this query back to the item table as a subquery:
SELECT t.*
(
SELECT item_id
FROM table
GROUP BY item_id
HAVING count(distinct city_id) = 1
) x
INNER JOIN
table t
ON x.item_id = t.item_id
This technique is the more general process for performing a group by that finds some particular set of rows, then bringing in the rest of the data from that row. You cant always stick every other column you want in a MAX because it will mix row data up, and you can't put the extra columns in your group by because that will subdivide what you're grouping on, giving the wrong results. Doing the group as a subquery and joining it back is a typical way to get all the row data when you have to group it to find which rows are interesting
In your case this form of query will bring all the duplicated rows (whereas the group by/max won't). If you don't want the duplicate rows you can make the top line SELECT DISTINCT t.* but don't make a habit of slapping distinct in to get rid of duplicated rows; if your tables don't have duplicates to start with but suddenly after you wrote a JOIN you got duplicated rows, google fornwhat a Cartesian product is in database queries and how to prevent it
You just need a group by on item id with having
Select item_id from table group by
item_id having count(distinct city_id)
=1
Also, if you want to have majority of same no of rows as input then
Select item_id, city, rank()
over(partition by item_id order by city)
rn
From table where rn=1;
I'm trying to query each distinct medical speciality (e.g. oncologist, pediatrician, etc.) in a table and then count the number of times a claim (claim_id) is linked to it, which I've done using this:
select distinct specialization, count(distinct claim_id) AS Claim_Totals
from table1
group by specialization
order by Claim_Totals DESC
However, I also want to include an additional column which lists the % that each speciality makes up in the table (based on the number of claim_id related to it). So for instance, if there were 100 total claims and "cardiologist" had 25 claim_id records related to it, "oncologist" had 15, "general surgeon" had 10, and so forth, I want the output to look like this:
specialization | Claims_Totals | PERCENTAGE
___________________________________________
cardiologist 25 25%
oncologist 15 15%
general surgeon 10 10%
Could do this? I'm not familiar with Barbaros's syntax. If that works its more concise and better.
select specialization, count(distinct claim_id) AS Claim_Totals, count(distinct claim_id)/total_claims
from table1
INNER JOIN ( SELECT COUNT(DISTINCT claim_id)*1.0000 total_claims AS total_claims
FROM table1 ) TMP
ON 1 = 1
group by specialization
order by Claim_Totals DESC
select specialization,
count(distinct claim_id) AS claim_by_spec,
count(distinct claim_id)/
( SELECT COUNT(DISTINCT claim_id)*1.0000
FROM table1 ) AS percentage_calc
from table1
group by specialization
order by Claim_Totals DESC
You can use sum(count(distinct)) over() to get the overall claims and use it in the denominator to get the percentage.
select specialization
,count(distinct claim_id) AS Claim_Totals
,round(100*count(distinct claim_id)/sum(count(distinct claim_id)) over(),3) as percentage
from table1
group by specialization
You can use
,concat_ws('',count(distinct claim_id),'%') as percentage
or
,concat(count(distinct claim_id),'%') as percentage
as added to the select list's tail
Btw, distinct before specialization in the select list is redundant, since already included in the group by list.
Because you are using count(distinct), window functions are less useful. You can try:
select t1.specialization,
count(distinct t1.claim_id) AS Claim_Totals,
count(distinct t1.claim_id) / tt1.num_claims
from table1 t1 cross join
(select count(distinct claim_id) as num_claims
from table1
) tt1
group by t1.specialization
order by Claim_Totals DESC
I was told to Find out which occupation has the greatest number of patients with conditionID=MC8
I dk how to do the greatest part.....
Here my code right now
SELECT occupation
FROM Patient
WHERE EXISTS
(SELECT PatientID FROM PatientMedcon
Where conditionID=’MC8’)
GROUP BY occupation
HAVNG count(occupation) = (Select max(occupation)
From Patient
You should approach these types of queries using regular joins and then add additional factors. The following gets the count of patients for each occupation with that condition:
SELECT occupation, COUNT(*)
FROM Patient p JOIN
PatentMedcon pm
ON p.PatientId = pm.PatientId and
pm.conditionId = 'MC8'
GROUP BY occupation
ORDER BY COUNT(*) DESC;
If you want the top row, that depends on the database. It might be select top 1, limit 1 at the end, fetch first 1 rows only at the end, or even something else.
I am validating a table which has a transaction level data of an eCommerce site and find the exact errors.
I want your help to find duplicate records in a 50 column table on SQL Server.
Suppose my data is:
OrderNo shoppername amountpayed city Item
1 Sam 10 A Iphone
1 Sam 10 A Iphone--->>Duplication to be detected
1 Sam 5 A Ipod
2 John 20 B Macbook
3 John 25 B Macbookair
4 Jack 5 A Ipod
Suppose I use the below query:
Select shoppername,count(*) as cnt
from dbo.sales
having count(*) > 1
group by shoppername
will return me
Sam 2
John 2
But I don't want to find duplicate just over 1 or 2 columns. I want to find the duplicate over all the columns together in my data. I want the result as:
1 Sam 10 A Iphone
with x as (select *,rn = row_number()
over(PARTITION BY OrderNo,item order by OrderNo)
from #temp1)
select * from x
where rn > 1
you can remove duplicates by replacing select statement by
delete x where rn > 1
SELECT OrderNo, shoppername, amountPayed, city, item, count(*) as cnt
FROM dbo.sales
GROUP BY OrderNo, shoppername, amountPayed, city, item
HAVING COUNT(*) > 1
SQL> SELECT JOB,COUNT(JOB) FROM EMP GROUP BY JOB;
JOB COUNT(JOB)
--------- ----------
ANALYST 2
CLERK 4
MANAGER 3
PRESIDENT 1
SALESMAN 4
Just add all fields to the query and remember to add them to Group By as well.
Select shoppername, a, b, amountpayed, item, count(*) as cnt
from dbo.sales
group by shoppername, a, b, amountpayed, item
having count(*) > 1
To get the list of multiple records use following command
select field1,field2,field3, count(*)
from table_name
group by field1,field2,field3
having count(*) > 1
Try this instead
SELECT MAX(shoppername), COUNT(*) AS cnt
FROM dbo.sales
GROUP BY CHECKSUM(*)
HAVING COUNT(*) > 1
Read about the CHECKSUM function first, as there can be duplicates.
Try this
with T1 AS
(
SELECT LASTNAME, COUNT(1) AS 'COUNT' FROM Employees GROUP BY LastName HAVING COUNT(1) > 1
)
SELECT E.*,T1.[COUNT] FROM Employees E INNER JOIN T1 ON T1.LastName = E.LastName
with x as (
select shoppername,count(shoppername)
from sales
having count(shoppername)>1
group by shoppername)
select t.* from x,win_gp_pin1510 t
where x.shoppername=t.shoppername
order by t.shoppername
First of all, I doubt that the result it not accurate? Seem like there are Three 'Sam' from the original table. But it is not critical to the question.
Then here we come for the question itself. Based on your table, the best way to show duplicate value is to use count(*) and Group by clause. The query would look like this
SELECT OrderNo, shoppername, amountPayed, city, item, count(*) as RepeatTimes FROM dbo.sales GROUP BY OrderNo, shoppername, amountPayed, city, item HAVING COUNT(*) > 1
The reason is that all columns together from your table uniquely identified each record, which means the records will be considered as duplicate only when all values from each column are exactly the same, also you want to show all fields for duplicate records, so the group by will not miss any column, otherwise yes because you can only select columns that participate in the 'group by' clause.
Now I would like to give you any example for With...Row_Number()Over(...), which is using table expression together with Row_Number function.
Suppose you have a nearly same table but with one extra column called Shipping Date, and the value may change even the rest are the same. Here it is:
OrderNo shoppername amountpayed city Item Shipping Date
1 Sam 10 A Iphone 2016-01-01
1 Sam 10 A Iphone 2016-02-02
1 Sam 5 A Ipod 2016-03-03
2 John 20 B Macbook 2016-04-04
3 John 25 B Macbookair 2016-05-05
4 Jack 5 A Ipod 2016-06-06
Notice that row# 2 is not a duplicate one if you still take all columns as a unit. But what if you want to treat them as duplicate as well in this case? You should use With...Row_Number()Over(...), and the query would look like this:
WITH TABLEEXPRESSION
AS
(SELECT *,ROW_NUMBER() OVER (PARTITION BY OrderNo, shoppername, amountPayed, city, item ORDER BY [Shipping Date] as Identifier) --if you consider the one with late shipping date as the duplicate
FROM dbo.sales)
SELECT * FROM TABLEEXPRESSION
WHERE Identifier !=1 --or use '>1'
The above query will give result together with Shipping Date, for example:
OrderNo shoppername amountpayed city Item Shipping Date Identifier
1 Sam 10 A Iphone 2016-02-02 2
Note this one is different from the one with 2016-01-01, and the reason why 2016-02-02 has been filtered out is PARTITION BY OrderNo, shoppername, amountPayed, city, item ORDER BY [Shipping Date] as Identifier, and Shipping Date is NOT one of the column that need to be took care of for duplicate records, which means the one with 2016-02-02 still could be a perfect result for your question.
Now summarize it little bit, using count(*) and Group by clause together is the best choice when you only want to show all columns from Group byclause as the result, otherwise you will miss the columns that do not participate in group by.
While For With...Row_Number()Over(...), it is suitable in every scenario that you want to find duplicate records, however, it is little bit complicated to write the query and little bit over engineered compared to the former one.
If your purpose is to delete duplicate records from table, you have to use the later WITH...ROW_NUMBER()OVER(...)...DELETE FROM...WHERE one.
Hope this helps!
You can use below methods to find the output
with Ctec AS
(
select *,Row_number() over(partition by name order by Name)Rnk
from Table_A
)
select Name from ctec
where rnk>1
select name from Table_A
group by name
having count(*)>1
Select *
from dbo.sales
group by shoppername
having(count(Item) > 1)
Select EventID,count() as cnt
from dbo.EventInstances
group by EventID
having count() > 1
The following is running code:
SELECT abnno, COUNT(abnno)
FROM tbl_Name
GROUP BY abnno
HAVING ( COUNT(abnno) > 1 )