I'm having a mind lapse on what I believe is a relatively easy script. Hopefully I'm overthinking the logic.
What I'm trying to do is perform two counts on a distinct column which is right joined.
What I want is:
count(a.book_id) as count_of_books
count(b.book_ref_number) as count_of_losses
Expected Output
--------------------------------------------------------
| Book | count_of_books | count of losses|
--------------------------------------------------------
|Hunger Games | 76 | 31 |
--------------------------------------------------------
|Hop on Pop | 27 | 6 |
--------------------------------------------------------
|Pout Pout Fish | 138 | 43 |
--------------------------------------------------------
I have tried a couple different scripts. Here are the two scripts I've tried.
(select count(*) from Inventory_Table x ) Count1,
(select count(*) from Loss_table b ) Count2
from Inventory_Table x
right join Loss_table b on b.book_ref_number = x.book_id
where rownum < 20
select
a.book_name,
count(distinct a.book_id),
count(b.book_ref_number)
from Inventory_Table x
right join Loss_table b on trim(b.book_ref_number) = trim(a.book_id)
Results I get
--------------------------------------------------------
| Book | count_of_books | count of losses|
--------------------------------------------------------
|Moby Dick | 4376 | 2574 |
--------------------------------------------------------
I'm looking for guidance in my neglectful mistake. Thank you in advance
and rownum <20 doesn't make sense. you are limiting your result set with 20 records.
try this:
select * from (
select
a.mrch_Nr,
count(distinct a.fdr_trac_nr),
count(b.auth_id)
from DATASTORE_FD.DEB_CRD_AUTH_LOG_REC a
right join jordab26.ft b on trim(b.auth_id) = trim(a.fdr_trac_nr)
where a.auth_log_dt between '20200101' and '20200408'
group by a.mrch_nr
)
where rownum < 20
Try this, I'm not sure about rownum < 20. Also, make sure your add correct group by condition.
select sum(case book_id when null then 0 else 1 end ) count_of_books,
sum(case book_ref_number when null then 0 else 1 end ) count_of_losses
from Inventory_Table x
right join Loss_table b on b.book_ref_number = x.book_id
where rownum < 20
Is this what you want?
Select distinct bookname,
count(distinct
a.bookid)+sum(
case when a.bookid IS NULL
THEN 1 END) ,
count(distinct b.id) as lossid
From inventary_table a
Left Join
Loss_table b
On
a.bookid=b.book_ref_number
SELECT book_name,COUNT(book_id),COUNT(book_ref_id) FROM Inventory_Table right join Loss_table on book_ref_number = book_id GROUP BY book_name
But if you need all the books in Inventory and only matching books from Loss_table then it should be left join:
SELECT book_name,COUNT(book_id),COUNT(book_ref_id) FROM Inventory_Table leftjoin Loss_table on book_ref_number = book_id GROUP BY book_name
0
SELECT book_name,COUNT(book_id),COUNT(book_ref_id)
FROM Inventory_Table
right join Loss_table on book_ref_number = book_id GROUP BY book_name
Related
Here's what I'm trying to do. Let's say I have this table t:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
2 | 18 | 2012-05-19 | y
3 | 18 | 2012-08-09 | z
4 | 19 | 2009-06-01 | a
5 | 19 | 2011-04-03 | b
6 | 19 | 2011-10-25 | c
7 | 19 | 2012-08-09 | d
For each id, I want to select the row containing the minimum record_date. So I'd get:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
The only solutions I've seen to this problem assume that all record_date entries are distinct, but that is not this case in my data. Using a subquery and an inner join with two conditions would give me duplicate rows for some ids, which I don't want:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
5 | 19 | 2011-04-03 | b
4 | 19 | 2009-06-01 | a
How about something like:
SELECT mt.*
FROM MyTable mt INNER JOIN
(
SELECT id, MIN(record_date) AS MinDate
FROM MyTable
GROUP BY id
) t ON mt.id = t.id AND mt.record_date = t.MinDate
This gets the minimum date per ID, and then gets the values based on those values. The only time you would have duplicates is if there are duplicate minimum record_dates for the same ID.
I could get to your expected result just by doing this in mysql:
SELECT id, min(record_date), other_cols
FROM mytable
GROUP BY id
Does this work for you?
To get the cheapest product in each category, you use the MIN() function in a correlated subquery as follows:
SELECT categoryid,
productid,
productName,
unitprice
FROM products a WHERE unitprice = (
SELECT MIN(unitprice)
FROM products b
WHERE b.categoryid = a.categoryid)
The outer query scans all rows in the products table and returns the products that have unit prices match with the lowest price in each category returned by the correlated subquery.
I would like to add to some of the other answers here, if you don't need the first item but say the second number for example you can use rownumber in a subquery and base your result set off of that.
SELECT * FROM
(
SELECT
ROW_NUM() OVER (PARTITION BY Id ORDER BY record_date, other_cols) as rownum,
*
FROM products P
) INNER
WHERE rownum = 2
This also allows you to order off multiple columns in the subquery which may help if two record_dates have identical values. You can also partition off of multiple columns if needed by delimiting them with a comma
This does it simply:
select t2.id,t2.record_date,t2.other_cols
from (select ROW_NUMBER() over(partition by id order by record_date)as rownum,id,record_date,other_cols from MyTable)t2
where t2.rownum = 1
If record_date has no duplicates within a group:
think of it as of filtering. Simpliy get (WHERE) one (MIN(record_date)) row from the current group:
SELECT * FROM t t1 WHERE record_date = (
select MIN(record_date)
from t t2 where t2.group_id = t1.group_id)
If there could be 2+ min record_date within a group:
filter out non-min rows (see above)
then (AND) pick only one from the 2+ min record_date rows, within the given group_id. E.g. pick the one with the min unique key:
AND key_id = (select MIN(key_id)
from t t3 where t3.record_date = t1.record_date
and t3.group_id = t1.group_id)
so
key_id | group_id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
8 | 19 | 2009-06-01 | e
will select key_ids: #1 and #4
SELECT p.* FROM tbl p
INNER JOIN(
SELECT t.id, MIN(record_date) AS MinDate
FROM tbl t
GROUP BY t.id
) t ON p.id = t.id AND p.record_date = t.MinDate
GROUP BY p.id
This code eliminates duplicate record_date in case there are same ids with same record_date.
If you want duplicates, remove the last line GROUP BY p.id.
This a old question, but this can useful for someone
In my case i can't using a sub query because i have a big query and i need using min() on my result, if i use sub query the db need reexecute my big query. i'm using Mysql
select t.*
from (select m.*, #g := 0
from MyTable m --here i have a big query
order by id, record_date) t
where (1 = case when #g = 0 or #g <> id then 1 else 0 end )
and (#g := id) IS NOT NULL
Basically I ordered the result and then put a variable in order to get only the first record in each group.
The below query takes the first date for each work order (in a table of showing all status changes):
SELECT
WORKORDERNUM,
MIN(DATE)
FROM
WORKORDERS
WHERE
DATE >= to_date('2015-01-01','YYYY-MM-DD')
GROUP BY
WORKORDERNUM
select
department,
min_salary,
(select s1.last_name from staff s1 where s1.salary=s3.min_salary ) lastname
from
(select department, min (salary) min_salary from staff s2 group by s2.department) s3
I've got the below two queries which I'm trying to figure out if I can merge into a single query. Both tables have the exact same three columns, but the organization names will differ between the two in many cases. The queries are like so:
SELECT OrganizationHierarchyUnitLevelThreeNm, COUNT(*) AS Complete
FROM #complete c
JOIN #wanted w ON w.WorkerKey = c.WorkerKey
GROUP BY OrganizationHierarchyUnitLevelThreeNm
ORDER BY 1;
SELECT OrganizationHierarchyUnitLevelThreeNm, COUNT(*) AS Wanted
FROM #wanted
GROUP BY OrganizationHierarchyUnitLevelThreeNm
ORDER BY 1;
So the first one might end up with something like this:
OrganizationHierarchyUnitLevelThreeNm | Complete
------------------------------------------------
Foo | 2
Bar | 17
and then the second would would be
OrganizationHierarchyUnitLevelThreeNm | Wanted
------------------------------------------------
Foo | 27
Baz | 132
So in the resulting query I'd want:
OrganizationHierarchyUnitLevelThreeNm | Wanted | Complete
---------------------------------------------------------
Foo | 27 | 2
Bar | 0 | 17
Baz | 132 | 0
Is that possible?
SELECT OrganizationHierarchyUnitLevelThreeNm,
Count(w.WorkerKey) as wanted,
COUNT(c.WorkerKey) AS Complete
FROM #wanted w
LEFT JOIN #complete c ON w.WorkerKey = c.WorkerKey
GROUP BY OrganizationHierarchyUnitLevelThreeNm
ORDER BY 1;
You can use union all:
select OrganizationHierarchyUnitLevelThreeNm, sum(Wanted), sum(Complete)
from ( (select OrganizationHierarchyUnitLevelThreeNm, 1 as Wanted , 0 as Complete
from #wanted w
) union all
(select OrganizationHierarchyUnitLevelThreeNm, 0 as Wanted , 1 as Complete
from #complete c
)
) t
group by OrganizationHierarchyUnitLevelThreeNm;
I have a SQL Server database full of the following (fictional) data in the following structure:
ID | PatientID | Exam | (NON DB COLUMN FOR REFERENCE)
------------------------------------
1 | 12345 | CT | OK
2 | 11234 | CT | OK(Same PID but Different Exam)
3 | 11234 | MRI | OK(Same PID but Different Exam)
4 | 11123 | CT | BAD(Same PID, Same Exam)
5 | 11123 | CT | BAD(Same PID, Same Exam)
6 | 11112 | CT | BAD(Conflicts With ID 8)
7 | 11112 | MRI | OK(SAME PID but different Exam)
8 | 11112 | CT | BAD(Conflicts With ID 6)
9 | 11123 | CT | BAD(Same PID, Same Exam)
10 | 11123 | CT | BAD(Same PID, Same Exam)
I am trying to write a query with will go through an identify everything that isn't bad as per my example above.
Overall, a patient (identified by PatientId) can have many rows, but may not have 2 or more rows with the same exam!
I have attempted various modifications of exams I found on here but still with no luck.
Thanks.
You seem to want to identify duplicates, ranking them as good or bad. Here is a method using window functions:
select t.id, t.patientid, t.exam,
(case when cnt > 1 then 'BAD' else 'OK' end)
from (select t.*, count(*) over (partition by patientid, exam) as cnt
from table t
) t;
use Count() over() :
select *,case when COUNT(*) over(partition by PatientID, Exam) > 1 then 'bad' else 'ok'
from yourtable
You can also use:
;WITH CTE_Patients
(ID, PatientID, Exam, RowNumber)
AS
(
SELECT ID, PatientID, Exam
ROW_NUMBER() OVER (PARTITION BY PatientID, Exam ORDER BY ID)
FROM YourTableName
)
SELECT TableB.ID, TableB.PatientID, TableB.Exam, [DuplicateOf] = TableA.ID
FROM CTE_Patients TableB
INNER JOIN CTE_Patients TableA
ON TableB.PatientID = TableA.PatientID
AND TableB.Exam = TableA.Exam
WHERE TableB.RowNumber > 1 -- Duplicate rows
AND TableA.RowNumber = 1 -- Unique rows
I have a sample here: SQL Server – Identifying unique and duplicate rows in a table, you can identify unique rows as well as duplicate rows
If you don't want to use a CTE or Count Over, you can also group the Source table, and select from there...(but I'd be surprised if #Gordon was too far off the mark with the original answer :) )
SELECT a.PatientID, a.Exam, CASE WHEN a.cnt > 1 THEN 'BAD' ELSE 'OK' END
FROM ( SELECT PatientID
,Exam
,COUNT(*) AS cnt
FROM tableName
GROUP BY Exam
,PatientID
) a
Select those patients that never have 2 or more exams of same type.
select * from patients t1
where not exists (select 1 from patients t2
where t1.PatientID = t2.PatientID
group by exam
having count(*) > 1)
Or, if you want all rows, like in your example:
select ID,
PatientID,
Exam,
case when exists (select 1 from patients t2
where t1.PatientID = t2.PatientID
group by exam
having count(*) > 1) then 'BAD' else 'OK' end
from patients
Firstly, I apologize for my English. I want get auctions with count of bids and buys. It should look like this:
id | name | bids | buys
-----------------------
1 | Foo | 4 | 1
2 | Bar | 0 | 0
I have tables like following:
auction:
id | name
---------
1 | Foo
2 | Bar
auction_bid:
id | auction_id
---------------
1 | 1
2 | 1
3 | 1
4 | 1
auction_buy:
id | auction_id
---------------
1 | 1
I can get numbers in two queries:
SELECT *, COUNT(abid.id) AS `bids` FROM `auction` `t` LEFT JOIN auction_bid abid ON (t.id = abid.auction) GROUP BY t.id
SELECT *, COUNT(abuy.id) AS `buys` FROM `auction` `t` LEFT JOIN auction_buy abuy ON (t.id = abuy.auction) GROUP BY t.id
But when i combined it into one:
SELECT *, COUNT(abid.id) AS `bids`, COUNT(abuy.id) AS `buys` FROM `auction` `t` LEFT JOIN auction_bid abid ON (t.id = abid.auction) LEFT JOIN auction_buy abuy ON (t.id = abuy.auction) GROUP BY t.id
It was returning wrong amount (bids as much as buys).
How to fix this and get counts in one query?
You'll need to count DISTINCT abuy and abid IDs to eliminate the duplicates;
SELECT t.id, t.name,
COUNT(DISTINCT abid.id) `bids`,
COUNT(DISTINCT abuy.id) `buys`
FROM `auction` `t`
LEFT JOIN auction_bid abid ON t.id = abid.auction_id
LEFT JOIN auction_buy abuy ON t.id = abuy.auction_id
GROUP BY t.id, t.name;
An SQLfiddle to test with.
Try this:
SELECT t.*,COUNT(abid.id) as bids,buys
FROM auction t LEFT JOIN
auction_bid abid ON t.id = abid.auction_id LEFT JOIN
(SELECT t.id, Count(abuy.id) as buys
FROM auction t LEFT JOIN
auction_buy abuy ON t.id = abuy.auction_id
GROUP BY t.id) Temp ON t.id=Temp.id
GROUP BY t.id
Result:
ID NAME BIDS BUYS
1 Foo 2 0
2 Bar 1 1
Result in SQL Fiddle.
Here's what I'm trying to do. Let's say I have this table t:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
2 | 18 | 2012-05-19 | y
3 | 18 | 2012-08-09 | z
4 | 19 | 2009-06-01 | a
5 | 19 | 2011-04-03 | b
6 | 19 | 2011-10-25 | c
7 | 19 | 2012-08-09 | d
For each id, I want to select the row containing the minimum record_date. So I'd get:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
The only solutions I've seen to this problem assume that all record_date entries are distinct, but that is not this case in my data. Using a subquery and an inner join with two conditions would give me duplicate rows for some ids, which I don't want:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
5 | 19 | 2011-04-03 | b
4 | 19 | 2009-06-01 | a
How about something like:
SELECT mt.*
FROM MyTable mt INNER JOIN
(
SELECT id, MIN(record_date) AS MinDate
FROM MyTable
GROUP BY id
) t ON mt.id = t.id AND mt.record_date = t.MinDate
This gets the minimum date per ID, and then gets the values based on those values. The only time you would have duplicates is if there are duplicate minimum record_dates for the same ID.
I could get to your expected result just by doing this in mysql:
SELECT id, min(record_date), other_cols
FROM mytable
GROUP BY id
Does this work for you?
To get the cheapest product in each category, you use the MIN() function in a correlated subquery as follows:
SELECT categoryid,
productid,
productName,
unitprice
FROM products a WHERE unitprice = (
SELECT MIN(unitprice)
FROM products b
WHERE b.categoryid = a.categoryid)
The outer query scans all rows in the products table and returns the products that have unit prices match with the lowest price in each category returned by the correlated subquery.
I would like to add to some of the other answers here, if you don't need the first item but say the second number for example you can use rownumber in a subquery and base your result set off of that.
SELECT * FROM
(
SELECT
ROW_NUM() OVER (PARTITION BY Id ORDER BY record_date, other_cols) as rownum,
*
FROM products P
) INNER
WHERE rownum = 2
This also allows you to order off multiple columns in the subquery which may help if two record_dates have identical values. You can also partition off of multiple columns if needed by delimiting them with a comma
This does it simply:
select t2.id,t2.record_date,t2.other_cols
from (select ROW_NUMBER() over(partition by id order by record_date)as rownum,id,record_date,other_cols from MyTable)t2
where t2.rownum = 1
If record_date has no duplicates within a group:
think of it as of filtering. Simpliy get (WHERE) one (MIN(record_date)) row from the current group:
SELECT * FROM t t1 WHERE record_date = (
select MIN(record_date)
from t t2 where t2.group_id = t1.group_id)
If there could be 2+ min record_date within a group:
filter out non-min rows (see above)
then (AND) pick only one from the 2+ min record_date rows, within the given group_id. E.g. pick the one with the min unique key:
AND key_id = (select MIN(key_id)
from t t3 where t3.record_date = t1.record_date
and t3.group_id = t1.group_id)
so
key_id | group_id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
8 | 19 | 2009-06-01 | e
will select key_ids: #1 and #4
SELECT p.* FROM tbl p
INNER JOIN(
SELECT t.id, MIN(record_date) AS MinDate
FROM tbl t
GROUP BY t.id
) t ON p.id = t.id AND p.record_date = t.MinDate
GROUP BY p.id
This code eliminates duplicate record_date in case there are same ids with same record_date.
If you want duplicates, remove the last line GROUP BY p.id.
This a old question, but this can useful for someone
In my case i can't using a sub query because i have a big query and i need using min() on my result, if i use sub query the db need reexecute my big query. i'm using Mysql
select t.*
from (select m.*, #g := 0
from MyTable m --here i have a big query
order by id, record_date) t
where (1 = case when #g = 0 or #g <> id then 1 else 0 end )
and (#g := id) IS NOT NULL
Basically I ordered the result and then put a variable in order to get only the first record in each group.
The below query takes the first date for each work order (in a table of showing all status changes):
SELECT
WORKORDERNUM,
MIN(DATE)
FROM
WORKORDERS
WHERE
DATE >= to_date('2015-01-01','YYYY-MM-DD')
GROUP BY
WORKORDERNUM
select
department,
min_salary,
(select s1.last_name from staff s1 where s1.salary=s3.min_salary ) lastname
from
(select department, min (salary) min_salary from staff s2 group by s2.department) s3