How do i count occurences that were joined SQL
0
I am joining two tables:
DECLARE @Temp TABLE (
id INT)
INSERT INTO @Temp
VALUES (5)
,(2)
,(3)
DECLARE @Temp2 TABLE (
member_id INT)
INSERT INTO @Temp2
VALUES (5)
,(1)
,(3)
How do i count the number of rows that can be LEFT joined and the ones that can't.
In this example: 5 & 3 from @Temp can be joined to @Temp2 and only 2 from @Temp can't be joined.
I would like my output to show the following:
+--------+------------+
| Joined | Not_Joined |
+--------+------------+
| 2 | 1 |
+--------+------------+
sql
sql-server asked Nov 19 '18 at 21:20
Roger SteinbergRoger Steinberg
30113
add a comment |
0
I am joining two tables:
DECLARE @Temp TABLE (
id INT)
INSERT INTO @Temp
VALUES (5)
,(2)
,(3)
DECLARE @Temp2 TABLE (
member_id INT)
INSERT INTO @Temp2
VALUES (5)
,(1)
,(3)
How do i count the number of rows that can be LEFT joined and the ones that can't.
In this example: 5 & 3 from @Temp can be joined to @Temp2 and only 2 from @Temp can't be joined.
I would like my output to show the following:
+--------+------------+
| Joined | Not_Joined |
+--------+------------+
| 2 | 1 |
+--------+------------+
sql
sql-server asked Nov 19 '18 at 21:20
Roger SteinbergRoger Steinberg
30113
You would use COUNT or conditional SUM.
– Sean Lange
Nov 19 '18 at 21:22
add a comment |
0
0
0
I am joining two tables:
DECLARE @Temp TABLE (
id INT)
INSERT INTO @Temp
VALUES (5)
,(2)
,(3)
DECLARE @Temp2 TABLE (
member_id INT)
INSERT INTO @Temp2
VALUES (5)
,(1)
,(3)
How do i count the number of rows that can be LEFT joined and the ones that can't.
In this example: 5 & 3 from @Temp can be joined to @Temp2 and only 2 from @Temp can't be joined.
I would like my output to show the following:
+--------+------------+
| Joined | Not_Joined |
+--------+------------+
| 2 | 1 |
+--------+------------+
sql
sql-server asked Nov 19 '18 at 21:20
Roger SteinbergRoger Steinberg
30113
I am joining two tables:
DECLARE @Temp TABLE (
id INT)
INSERT INTO @Temp
VALUES (5)
,(2)
,(3)
DECLARE @Temp2 TABLE (
member_id INT)
INSERT INTO @Temp2
VALUES (5)
,(1)
,(3)
How do i count the number of rows that can be LEFT joined and the ones that can't.
In this example: 5 & 3 from @Temp can be joined to @Temp2 and only 2 from @Temp can't be joined.
I would like my output to show the following:
+--------+------------+
| Joined | Not_Joined |
+--------+------------+
| 2 | 1 |
+--------+------------+
sql
sql-server sql
sql-server asked Nov 19 '18 at 21:20
Roger SteinbergRoger Steinberg
30113
asked Nov 19 '18 at 21:20
Roger SteinbergRoger Steinberg
30113
asked Nov 19 '18 at 21:20
Roger SteinbergRoger Steinberg
30113
asked Nov 19 '18 at 21:20
Roger SteinbergRoger Steinberg
30113
asked Nov 19 '18 at 21:20
Roger SteinbergRoger Steinberg
30113
30113
You would use COUNT or conditional SUM.
– Sean Lange
Nov 19 '18 at 21:22
add a comment |
You would use COUNT or conditional SUM.
– Sean Lange
Nov 19 '18 at 21:22
You would use COUNT or conditional SUM.
– Sean Lange
Nov 19 '18 at 21:22
You would use COUNT or conditional SUM.
– Sean Lange
Nov 19 '18 at 21:22
add a comment |
3 Answers
3
active
oldest
votes
2
You can do this in a single query using COUNT and SUM. This should produce the results you are looking for.
DECLARE @Temp TABLE (
id INT)
INSERT INTO @Temp
VALUES (5)
,(2)
,(3)
DECLARE @Temp2 TABLE (
member_id INT)
INSERT INTO @Temp2
VALUES (5)
,(1)
,(3)
select Joined = count(t2.Member_id)
, NotJoined = sum(case when t2.Member_id is null then 1 end)
from @Temp t
left join @Temp2 t2 on t2.member_id = t.id
answered Nov 19 '18 at 21:26
Sean LangeSean Lange
25.1k21835
YUP thats exactly what i was looking for thank you!! why is my question downgraded?
– Roger Steinberg
Nov 19 '18 at 21:33
I up voted this too.. my lonely answer needs points...
– T McKeown
Nov 19 '18 at 21:34
Not my downvote but my guess is because you didn't show that you tried something. Seems pretty harsh to me. I thought your question was fine. And great job posting data and desired results!!
– Sean Lange
Nov 19 '18 at 21:35
thanks @SeanLange, i have the article and sources you posted on my last post in my bookmarks ;)
– Roger Steinberg
Nov 19 '18 at 21:59
add a comment |
0
The count from @Temp that EXISTS in @Temp2:
SELECT COUNT(*) FROM @TEMP WHERE ID IN(SELECT MEMBER_ID FROM @TEMP2)
The count from @Temp2 not in @Temp:
SELECT COUNT(*) FROM @TEMP2 WHERE MEMBER_ID NOT IN(ID FROM @TEMP)
Now to create a single result set, there are many ways but here is a simple one:
SELECT
(SELECT COUNT(*) FROM @TEMP2 WHERE MEMBER_ID IN(ID FROM @TEMP)) AS [Joined],
(SELECT COUNT(*) FROM @TEMP WHERE ID NOT IN(SELECT MEMBER_ID FROM @TEMP2)) AS [NotJoined]
@Sean Lange's answer is more specific to the JOIN question, my answer simply counts what exists in the lists.
answered Nov 19 '18 at 21:23
T McKeownT McKeown
11.5k11727
thats equally good as the answer above. nice to know alternatives. thank you
– Roger Steinberg
Nov 19 '18 at 21:33
add a comment |
-1
Select count(*) as 'NOT Joined ',
(Select t1.count(*) from table1
t1)-count(*) as 'Joined'
from table1 where id NOT IN (Select member_id from table2);
Its basically how a left join works that is Common values of both the
tables plus the value of table 1 which doesnt exists in table 2.
answered Nov 19 '18 at 21:35
Himanshu AhujaHimanshu Ahuja
6661216
The OP put in the effort to provide sample data and a nice consumable format. This as posted has several syntax errors. The OP deserves better.
– Sean Lange
Nov 19 '18 at 21:38
Can anybody explain why i have got a downvote.
– Himanshu Ahuja
Nov 19 '18 at 21:39
Because your code doesn't work as posted. If you fix the code so it works it would be considerably better.
– Sean Lange
Nov 19 '18 at 21:40
@Sean Lange Can you please check it now
– Himanshu Ahuja
Nov 19 '18 at 21:50
All you did was format it. There are still syntax errors. And since the OP posted such nice consumable data it really should work with what they posted.
– Sean Lange
Nov 20 '18 at 14:04
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
You can do this in a single query using COUNT and SUM. This should produce the results you are looking for.
DECLARE @Temp TABLE (
id INT)
INSERT INTO @Temp
VALUES (5)
,(2)
,(3)
DECLARE @Temp2 TABLE (
member_id INT)
INSERT INTO @Temp2
VALUES (5)
,(1)
,(3)
select Joined = count(t2.Member_id)
, NotJoined = sum(case when t2.Member_id is null then 1 end)
from @Temp t
left join @Temp2 t2 on t2.member_id = t.id
answered Nov 19 '18 at 21:26
Sean LangeSean Lange
25.1k21835
YUP thats exactly what i was looking for thank you!! why is my question downgraded?
– Roger Steinberg
Nov 19 '18 at 21:33
I up voted this too.. my lonely answer needs points...
– T McKeown
Nov 19 '18 at 21:34
Not my downvote but my guess is because you didn't show that you tried something. Seems pretty harsh to me. I thought your question was fine. And great job posting data and desired results!!
– Sean Lange
Nov 19 '18 at 21:35
thanks @SeanLange, i have the article and sources you posted on my last post in my bookmarks ;)
– Roger Steinberg
Nov 19 '18 at 21:59
add a comment |
2
You can do this in a single query using COUNT and SUM. This should produce the results you are looking for.
DECLARE @Temp TABLE (
id INT)
INSERT INTO @Temp
VALUES (5)
,(2)
,(3)
DECLARE @Temp2 TABLE (
member_id INT)
INSERT INTO @Temp2
VALUES (5)
,(1)
,(3)
select Joined = count(t2.Member_id)
, NotJoined = sum(case when t2.Member_id is null then 1 end)
from @Temp t
left join @Temp2 t2 on t2.member_id = t.id
answered Nov 19 '18 at 21:26
Sean LangeSean Lange
25.1k21835
YUP thats exactly what i was looking for thank you!! why is my question downgraded?
– Roger Steinberg
Nov 19 '18 at 21:33
I up voted this too.. my lonely answer needs points...
– T McKeown
Nov 19 '18 at 21:34
Not my downvote but my guess is because you didn't show that you tried something. Seems pretty harsh to me. I thought your question was fine. And great job posting data and desired results!!
– Sean Lange
Nov 19 '18 at 21:35
thanks @SeanLange, i have the article and sources you posted on my last post in my bookmarks ;)
– Roger Steinberg
Nov 19 '18 at 21:59
add a comment |
2
2
2
You can do this in a single query using COUNT and SUM. This should produce the results you are looking for.
DECLARE @Temp TABLE (
id INT)
INSERT INTO @Temp
VALUES (5)
,(2)
,(3)
DECLARE @Temp2 TABLE (
member_id INT)
INSERT INTO @Temp2
VALUES (5)
,(1)
,(3)
select Joined = count(t2.Member_id)
, NotJoined = sum(case when t2.Member_id is null then 1 end)
from @Temp t
left join @Temp2 t2 on t2.member_id = t.id
answered Nov 19 '18 at 21:26
Sean LangeSean Lange
25.1k21835
You can do this in a single query using COUNT and SUM. This should produce the results you are looking for.
DECLARE @Temp TABLE (
id INT)
INSERT INTO @Temp
VALUES (5)
,(2)
,(3)
DECLARE @Temp2 TABLE (
member_id INT)
INSERT INTO @Temp2
VALUES (5)
,(1)
,(3)
select Joined = count(t2.Member_id)
, NotJoined = sum(case when t2.Member_id is null then 1 end)
from @Temp t
left join @Temp2 t2 on t2.member_id = t.id
answered Nov 19 '18 at 21:26
Sean LangeSean Lange
25.1k21835
answered Nov 19 '18 at 21:26
Sean LangeSean Lange
25.1k21835
answered Nov 19 '18 at 21:26
Sean LangeSean Lange
25.1k21835
answered Nov 19 '18 at 21:26
Sean LangeSean Lange
25.1k21835
25.1k21835
YUP thats exactly what i was looking for thank you!! why is my question downgraded?
– Roger Steinberg
Nov 19 '18 at 21:33
I up voted this too.. my lonely answer needs points...
– T McKeown
Nov 19 '18 at 21:34
Not my downvote but my guess is because you didn't show that you tried something. Seems pretty harsh to me. I thought your question was fine. And great job posting data and desired results!!
– Sean Lange
Nov 19 '18 at 21:35
thanks @SeanLange, i have the article and sources you posted on my last post in my bookmarks ;)
– Roger Steinberg
Nov 19 '18 at 21:59
add a comment |
YUP thats exactly what i was looking for thank you!! why is my question downgraded?
– Roger Steinberg
Nov 19 '18 at 21:33
I up voted this too.. my lonely answer needs points...
– T McKeown
Nov 19 '18 at 21:34
Not my downvote but my guess is because you didn't show that you tried something. Seems pretty harsh to me. I thought your question was fine. And great job posting data and desired results!!
– Sean Lange
Nov 19 '18 at 21:35
thanks @SeanLange, i have the article and sources you posted on my last post in my bookmarks ;)
– Roger Steinberg
Nov 19 '18 at 21:59
YUP thats exactly what i was looking for thank you!! why is my question downgraded?
– Roger Steinberg
Nov 19 '18 at 21:33
YUP thats exactly what i was looking for thank you!! why is my question downgraded?
– Roger Steinberg
Nov 19 '18 at 21:33
I up voted this too.. my lonely answer needs points...
– T McKeown
Nov 19 '18 at 21:34
I up voted this too.. my lonely answer needs points...
– T McKeown
Nov 19 '18 at 21:34
Not my downvote but my guess is because you didn't show that you tried something. Seems pretty harsh to me. I thought your question was fine. And great job posting data and desired results!!
– Sean Lange
Nov 19 '18 at 21:35
Not my downvote but my guess is because you didn't show that you tried something. Seems pretty harsh to me. I thought your question was fine. And great job posting data and desired results!!
– Sean Lange
Nov 19 '18 at 21:35
thanks @SeanLange, i have the article and sources you posted on my last post in my bookmarks ;)
– Roger Steinberg
Nov 19 '18 at 21:59
thanks @SeanLange, i have the article and sources you posted on my last post in my bookmarks ;)
– Roger Steinberg
Nov 19 '18 at 21:59
add a comment |
0
The count from @Temp that EXISTS in @Temp2:
SELECT COUNT(*) FROM @TEMP WHERE ID IN(SELECT MEMBER_ID FROM @TEMP2)
The count from @Temp2 not in @Temp:
SELECT COUNT(*) FROM @TEMP2 WHERE MEMBER_ID NOT IN(ID FROM @TEMP)
Now to create a single result set, there are many ways but here is a simple one:
SELECT
(SELECT COUNT(*) FROM @TEMP2 WHERE MEMBER_ID IN(ID FROM @TEMP)) AS [Joined],
(SELECT COUNT(*) FROM @TEMP WHERE ID NOT IN(SELECT MEMBER_ID FROM @TEMP2)) AS [NotJoined]
@Sean Lange's answer is more specific to the JOIN question, my answer simply counts what exists in the lists.
answered Nov 19 '18 at 21:23
T McKeownT McKeown
11.5k11727
thats equally good as the answer above. nice to know alternatives. thank you
– Roger Steinberg
Nov 19 '18 at 21:33
add a comment |
0
The count from @Temp that EXISTS in @Temp2:
SELECT COUNT(*) FROM @TEMP WHERE ID IN(SELECT MEMBER_ID FROM @TEMP2)
The count from @Temp2 not in @Temp:
SELECT COUNT(*) FROM @TEMP2 WHERE MEMBER_ID NOT IN(ID FROM @TEMP)
Now to create a single result set, there are many ways but here is a simple one:
SELECT
(SELECT COUNT(*) FROM @TEMP2 WHERE MEMBER_ID IN(ID FROM @TEMP)) AS [Joined],
(SELECT COUNT(*) FROM @TEMP WHERE ID NOT IN(SELECT MEMBER_ID FROM @TEMP2)) AS [NotJoined]
@Sean Lange's answer is more specific to the JOIN question, my answer simply counts what exists in the lists.
answered Nov 19 '18 at 21:23
T McKeownT McKeown
11.5k11727
thats equally good as the answer above. nice to know alternatives. thank you
– Roger Steinberg
Nov 19 '18 at 21:33
add a comment |
0
0
0
The count from @Temp that EXISTS in @Temp2:
SELECT COUNT(*) FROM @TEMP WHERE ID IN(SELECT MEMBER_ID FROM @TEMP2)
The count from @Temp2 not in @Temp:
SELECT COUNT(*) FROM @TEMP2 WHERE MEMBER_ID NOT IN(ID FROM @TEMP)
Now to create a single result set, there are many ways but here is a simple one:
SELECT
(SELECT COUNT(*) FROM @TEMP2 WHERE MEMBER_ID IN(ID FROM @TEMP)) AS [Joined],
(SELECT COUNT(*) FROM @TEMP WHERE ID NOT IN(SELECT MEMBER_ID FROM @TEMP2)) AS [NotJoined]
@Sean Lange's answer is more specific to the JOIN question, my answer simply counts what exists in the lists.
answered Nov 19 '18 at 21:23
T McKeownT McKeown
11.5k11727
The count from @Temp that EXISTS in @Temp2:
SELECT COUNT(*) FROM @TEMP WHERE ID IN(SELECT MEMBER_ID FROM @TEMP2)
The count from @Temp2 not in @Temp:
SELECT COUNT(*) FROM @TEMP2 WHERE MEMBER_ID NOT IN(ID FROM @TEMP)
Now to create a single result set, there are many ways but here is a simple one:
SELECT
(SELECT COUNT(*) FROM @TEMP2 WHERE MEMBER_ID IN(ID FROM @TEMP)) AS [Joined],
(SELECT COUNT(*) FROM @TEMP WHERE ID NOT IN(SELECT MEMBER_ID FROM @TEMP2)) AS [NotJoined]
@Sean Lange's answer is more specific to the JOIN question, my answer simply counts what exists in the lists.
answered Nov 19 '18 at 21:23
T McKeownT McKeown
11.5k11727
edited Nov 19 '18 at 21:28
answered Nov 19 '18 at 21:23
T McKeownT McKeown
11.5k11727
answered Nov 19 '18 at 21:23
T McKeownT McKeown
11.5k11727
answered Nov 19 '18 at 21:23
T McKeownT McKeown
11.5k11727
11.5k11727
thats equally good as the answer above. nice to know alternatives. thank you
– Roger Steinberg
Nov 19 '18 at 21:33
add a comment |
thats equally good as the answer above. nice to know alternatives. thank you
– Roger Steinberg
Nov 19 '18 at 21:33
thats equally good as the answer above. nice to know alternatives. thank you
– Roger Steinberg
Nov 19 '18 at 21:33
thats equally good as the answer above. nice to know alternatives. thank you
– Roger Steinberg
Nov 19 '18 at 21:33
add a comment |
-1
Select count(*) as 'NOT Joined ',
(Select t1.count(*) from table1
t1)-count(*) as 'Joined'
from table1 where id NOT IN (Select member_id from table2);
Its basically how a left join works that is Common values of both the
tables plus the value of table 1 which doesnt exists in table 2.
answered Nov 19 '18 at 21:35
Himanshu AhujaHimanshu Ahuja
6661216
The OP put in the effort to provide sample data and a nice consumable format. This as posted has several syntax errors. The OP deserves better.
– Sean Lange
Nov 19 '18 at 21:38
Can anybody explain why i have got a downvote.
– Himanshu Ahuja
Nov 19 '18 at 21:39
Because your code doesn't work as posted. If you fix the code so it works it would be considerably better.
– Sean Lange
Nov 19 '18 at 21:40
@Sean Lange Can you please check it now
– Himanshu Ahuja
Nov 19 '18 at 21:50
All you did was format it. There are still syntax errors. And since the OP posted such nice consumable data it really should work with what they posted.
– Sean Lange
Nov 20 '18 at 14:04
add a comment |
-1
Select count(*) as 'NOT Joined ',
(Select t1.count(*) from table1
t1)-count(*) as 'Joined'
from table1 where id NOT IN (Select member_id from table2);
Its basically how a left join works that is Common values of both the
tables plus the value of table 1 which doesnt exists in table 2.
answered Nov 19 '18 at 21:35
Himanshu AhujaHimanshu Ahuja
6661216
The OP put in the effort to provide sample data and a nice consumable format. This as posted has several syntax errors. The OP deserves better.
– Sean Lange
Nov 19 '18 at 21:38
Can anybody explain why i have got a downvote.
– Himanshu Ahuja
Nov 19 '18 at 21:39
Because your code doesn't work as posted. If you fix the code so it works it would be considerably better.
– Sean Lange
Nov 19 '18 at 21:40
@Sean Lange Can you please check it now
– Himanshu Ahuja
Nov 19 '18 at 21:50
All you did was format it. There are still syntax errors. And since the OP posted such nice consumable data it really should work with what they posted.
– Sean Lange
Nov 20 '18 at 14:04
add a comment |
-1
-1
-1
Select count(*) as 'NOT Joined ',
(Select t1.count(*) from table1
t1)-count(*) as 'Joined'
from table1 where id NOT IN (Select member_id from table2);
Its basically how a left join works that is Common values of both the
tables plus the value of table 1 which doesnt exists in table 2.
answered Nov 19 '18 at 21:35
Himanshu AhujaHimanshu Ahuja
6661216
Select count(*) as 'NOT Joined ',
(Select t1.count(*) from table1
t1)-count(*) as 'Joined'
from table1 where id NOT IN (Select member_id from table2);
Its basically how a left join works that is Common values of both the
tables plus the value of table 1 which doesnt exists in table 2.
answered Nov 19 '18 at 21:35
Himanshu AhujaHimanshu Ahuja
6661216
edited Nov 19 '18 at 21:48
answered Nov 19 '18 at 21:35
Himanshu AhujaHimanshu Ahuja
6661216
answered Nov 19 '18 at 21:35
Himanshu AhujaHimanshu Ahuja
6661216
answered Nov 19 '18 at 21:35
Himanshu AhujaHimanshu Ahuja
6661216
6661216
The OP put in the effort to provide sample data and a nice consumable format. This as posted has several syntax errors. The OP deserves better.
– Sean Lange
Nov 19 '18 at 21:38
Can anybody explain why i have got a downvote.
– Himanshu Ahuja
Nov 19 '18 at 21:39
Because your code doesn't work as posted. If you fix the code so it works it would be considerably better.
– Sean Lange
Nov 19 '18 at 21:40
@Sean Lange Can you please check it now
– Himanshu Ahuja
Nov 19 '18 at 21:50
All you did was format it. There are still syntax errors. And since the OP posted such nice consumable data it really should work with what they posted.
– Sean Lange
Nov 20 '18 at 14:04
add a comment |
The OP put in the effort to provide sample data and a nice consumable format. This as posted has several syntax errors. The OP deserves better.
– Sean Lange
Nov 19 '18 at 21:38
Can anybody explain why i have got a downvote.
– Himanshu Ahuja
Nov 19 '18 at 21:39
Because your code doesn't work as posted. If you fix the code so it works it would be considerably better.
– Sean Lange
Nov 19 '18 at 21:40
@Sean Lange Can you please check it now
– Himanshu Ahuja
Nov 19 '18 at 21:50
All you did was format it. There are still syntax errors. And since the OP posted such nice consumable data it really should work with what they posted.
– Sean Lange
Nov 20 '18 at 14:04
The OP put in the effort to provide sample data and a nice consumable format. This as posted has several syntax errors. The OP deserves better.
– Sean Lange
Nov 19 '18 at 21:38
The OP put in the effort to provide sample data and a nice consumable format. This as posted has several syntax errors. The OP deserves better.
– Sean Lange
Nov 19 '18 at 21:38
Can anybody explain why i have got a downvote.
– Himanshu Ahuja
Nov 19 '18 at 21:39
Can anybody explain why i have got a downvote.
– Himanshu Ahuja
Nov 19 '18 at 21:39
Because your code doesn't work as posted. If you fix the code so it works it would be considerably better.
– Sean Lange
Nov 19 '18 at 21:40
Because your code doesn't work as posted. If you fix the code so it works it would be considerably better.
– Sean Lange
Nov 19 '18 at 21:40
@Sean Lange Can you please check it now
– Himanshu Ahuja
Nov 19 '18 at 21:50
@Sean Lange Can you please check it now
– Himanshu Ahuja
Nov 19 '18 at 21:50
All you did was format it. There are still syntax errors. And since the OP posted such nice consumable data it really should work with what they posted.
– Sean Lange
Nov 20 '18 at 14:04
All you did was format it. There are still syntax errors. And since the OP posted such nice consumable data it really should work with what they posted.
– Sean Lange
Nov 20 '18 at 14:04
add a comment |
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You would use COUNT or conditional SUM.
– Sean Lange
Nov 19 '18 at 21:22