I have an example where in a table there is ID,NAme and M_if(managerID). I populated the table in the following manner
Id Name M_id
1 A 2
2 B NUll
3 C 1
4 D 3
5 E 2
Id is employee ID, Name and M_id is manager ID. In above example A's manager is 2(B), B doesn't have manager, C's manager is 1(A) and so on. I need to find out the names of the employees and their managers name. I have written the following query by doing permutations and combinations which gives me proper result but I am not able to comprehend how exactly the query(left join) is working. Please make me explain the concept.
SELECT (e.Name), ee.name FROM test.employee e
left join test.employee ee on ee.Id = e.M_id
order by e.Id;
result i get
A B
B
C A
D C
E B
Please explain me the joint
two instances are there for same table as :
e
Id Name M_id
1 A 2
2 B NUll
3 C 1
4 D 3
5 E 2
ee
Id Name M_id
1 A 2
2 B NUll
3 C 1
4 D 3
5 E 2
according to your join condition on ee.Id = e.M_id
simply first row of instance e will be selected because of left join and e.M_id will get compared to ee.Id and 2nd row will be selected from second instance of same table.
selection of data from both the table is as :
e.Id e.Name e.M_id | ee.Id ee.Name ee.M_id
1 A 2 | 2 B NUll
2 B NUll |
3 C 1 | 1 A 2
4 D 3 | 3 C 1
5 E 2 | 2 B NUll
that is why it is showing
A B
Related
I have a table below which stores Connections between 2 person
TABLE (CONNECTION)
ID | REQUEST_PERSON | REQUESTEE_PERSON
I would like to build a VIEW which gets the REQUEST_PERSON, REQUESTEE_PERSON and MUTUAL_CONNECTION_COUNT(other common connections count between them). Any help is appreciated
For Example if we have a table data as below
ID | REQUEST_PERSON | REQUESTEE_PERSON
1 A B
2 A C
3 B C
4 D B
5 D A
6 A E
7 B E
8 A F
9 C G
I need a VIEW display below
ID | REQUEST_PERSON | REQUESTEE_PERSON | MUTUAL_CONNECTION_COUNT
1 A B 3
2 A C 1
3 B C 1
4 D B 1
5 D A 1
6 A E 1
7 B E 1
8 A F 0
9 C G 0
This is rather tricky. Here is code that does what you want:
select c.*,
(select count(*)
from (select v.person2
from connections c2 cross apply
(values (c2.REQUESTEE_PERSON, c2.REQUEST_PERSON), (c2.REQUEST_PERSON, c2.REQUESTEE_PERSON)
) v(person1, person2)
where v.person1 IN (c.Request_Person, c.Requestee_Person)
group by v.person2
having count(*) = 2
) v
) in_common
from connections c
order by id;
Here is a SQL Fiddle.
The essence of the problem is finding people who are connected to both people in each row. Your connections are unidirectional, which makes the logic hard to express -- C could be either the first or second person in either connection.
Arrgh!
So, the innermost subquery adds reverse links to the graph. Then, it can focus on filtering by the first person -- who has to match the persons in the outer query. The second person is the one that might be in common.
The inner aggregation is just summarizing by the second person. It filters using having count(*) = 2 to indicate that both people in the outer query need to be connected to the second person in the inner query. The count(*) assumes that you have no duplicates.
Then, these are counted, which is the value you want.
Table:
A | B | C | D
1 q 123 23
2 w 22 32
3 e 23 21
New table:
A | B | C | D
1 q 123 C
1 q 23 D
2 w 22 C
2 w 32 D
3 e 23 C
3 e 21 D
I want to derive a new table/view from an existing table, where I want the records in the first table to be split by a column name.
C and D are months in the original table. In the new table I want the months to be as records.
The records in the original table for the months (123,23 for 1) should match the months column and be put into another column in the new table.
Please let me know if it is not clear.
Do a UNION ALL, with one select for the c's and one select for the d's.
select a, b, c, 'c' from tablename
union all
select a, b, d, 'd' from tablename
I have 2 tables like below:
Table 1
---------
1
2
3
Table 2
--------
A
B
C
How do i join to get an output like below:
Output
---------
1 A
1 B
1 C
2 A
2 B
2 C
3 A
3 B
3 C
Use Cross join:
SELECT *
FROM Table1
CROSS JOIN Table2
You can change the order by replacing * with table fields.
Read more about Cross Join.
OR you could do this:
SELECT Table1.*,Table2.*
FROM Table2,Table1
Result:
ID NAME
1 A
1 B
1 C
2 A
2 B
2 C
3 A
3 B
3 C
You want to do a CROSS JOIN and that will give you the Cartesian product of all the rows.
See http://en.m.wikipedia.org/wiki/Join_(SQL)
select * from table 1,table 2 .
For o/p like A 1
A 2
A 3
B 1
B 2
B 3
C 1
C 2
C 3
just
I would like to know what's the logic for multiple joins (for example below)
SELECT * FROM B returns 100 rows
SELECT B.* FROM B LEFT JOIN C ON B.ID = C.ID returns 120 rows
As I know using left join will returns any matching data from the left table which is B if data are found for both table. But how come when using left join, it returns more data than table B itself?
What am I do wrong or misunderstood here? Any guidance are very appreciated. Thanks in advance.
Let be table B:
id
----
1
2
3
Let be table C
id name
------------
1 John
2 Mary
2 Anne
3 Stef
Any id from b is matched with ids from c, then id=2 will be matched twice. So a left join on id will return 4 rows even if base table B has 3 rows.
Now look at a more evil example:
Table B
id
----
1
2
2
3
4
table C
id name
------------
1 John
2 Mary
2 Anne
3 Stef
Every id from b is matched with ids from c, then first id=2 will be matched twice and second id=2 will be matched twice so the result of
select b.id, c.name
from b left join c on (b.id = c.id)
will be
id name
------------
1 John
2 Mary
2 Mary
2 Anne
2 Anne
3 Stef
4 (null)
The id=4 is not matched but appears in the result because is a left join.
Look at the following example :
B = {1,2}
C = {(1,a),(1,b),(1,c),(1,d),(1,e)}
The result of B left join C will be :
1 | a
1 | b
1 | c
1 | d
1 | e
2 | null
The number of rows in the result is definitely larger than rows in B (2).
In general the number of rows in result of B left join C is bounded by B.size + C.size and not only by B.size as you think...
As per your query it do the join to B Table with C and B table is Left Table so it will display all the records of Left table in our case it is B and related from other Table in our Case it is C.
I have only one table available. I want to show the grade and the count of the number of times an employee has that grade recorded, but it must show a 0 for the grade if there are no records for that employee. I know how to do this using left join when two tables are present, but I only have 1 table.
How is this possible?
For example:
TABLE
empID | dept | grade
1 | 11 | a
2 | 11 | a
3 | 11 | b
1 | 22 | c
2 | 22 | f
3 | 22 | d
1 | 33 | a
2 | 33 | a
3 | 33 | a
If I run SELECT grade, count(grade) from table where empID = 1 Group by grade;, for example, it ends up printing out only grades the employee got and the count. Now I want to also print out the 0s for grades the employee did not have.
i think you're asking for this?
SQL> select e.grade, count(e2.empid)
2 from (select distinct grade from e) e
3 left outer join e e2
4 on e2.grade = e.grade
5 and e2.empid = 1
6 group by e.grade
7 order by grade;
G COUNT(E2.EMPID)
- ---------------
a 2
b 0
c 1
d 0
f 0
or as you have no rows with "e" grade then if you have a lookup table called grade:
SQL> select * from grade;
G
-
a
b
c
d
e
f
SQL> select e.grade, count(e2.empid)
2 from grade emp
3 left outer join emp e2
4 on e2.grade = e.grade
5 and e2.empid = 1
6 group by e.grade
7 order by grade;
G COUNT(E2.EMPID)
- ---------------
a 2
b 0
c 1
d 0
e 0
f 0
Let's say your query to select a value is:
select value from tbl;
You can ensure a 0 is returned if there are no rows in tbl t:
select nvl(t.value, 0) value
from dual d
left join tbl t on 1=1;
Sounds like you want the NVL function. With NVL, you can conditionally return an alternate value if the value is null. See the documentation.
So, if you had the following...
SELECT fooName, fooNumber FROM foo
and these were your results
fooName, fooNumber
Blah, 1
Asdf, null
Qwer, 3
poiu, null
you could rewrite the query like this...
SELECT fooName, NVL(fooNumber, 0) FROM foo
and your results would now be...
fooName, fooNumber
Blah, 1
Asdf, 0
Qwer, 3
poiu, 0
http://docs.oracle.com/cd/B19306_01/server.102/b14200/functions105.htm