Having two tables (table1, table2) with the same column names (generation, parent), the desired output would be the combination of all columns of both tables. Thereby the rows of table2 should join table1 so that the rows of table2 are matching those of table1 on generation column. The parent number should be ordered ascending for the entries in table1 as well as in table2. The number of rows of the query results should be equal of those of table1.
Given the following tables
table1:
| generation | parent |
|:----------:|:------:|
| 0 | 1 |
| 0 | 2 |
| 0 | 3 |
| 1 | 3 |
| 1 | 2 |
| 1 | 1 |
| 2 | 2 |
| 2 | 1 |
| 2 | 3 |
table2:
| generation | parent |
|:----------:|:------:|
| 1 | 3 |
| 1 | 1 |
| 1 | 3 |
| 2 | 1 |
| 2 | 2 |
| 2 | 3 |
The following queries are thought for creating and populating two sample tables as shown above:
create table table1(generation integer, parent integer);
insert into table1 (generation, parent) values(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(2,2),(2,1),(2,3);
create table table2(generation integer, parent integer);
insert into table2 (generation, parent) values(1,3),(1,1),(1,3),(2,1),(2,2),(2,3);
the imagined query should lead to the following desired result:
| table1_generation | table1_parent | table2_generation | table2_parent |
|:-----------------:|:-------------:|:-----------------:|:-------------:|
| 0 | 1 | | |
| 0 | 2 | | |
| 0 | 3 | | |
| 1 | 1 | 1 | 1 |
| 1 | 2 | 1 | 3 |
| 1 | 3 | 1 | 3 |
| 2 | 1 | 2 | 1 |
| 2 | 2 | 2 | 2 |
| 2 | 3 | 2 | 3 |
Current query looks as follows:
with
p as (
select
generation,
parent
from
table1
order by
generation,
parent
), o as(
select
generation,
parent
from
table2
order by
generation,
parent
)
select
p.generation as table1_generation,
p.parent as table1_parent,
o.generation as table2_generation,
o.parent as table2_parent
from
p
left join o on
o.generation=p.generation;
Which leads to the following result:
| table1_generation | table1_parent | table2_generation | table2_parent |
|:-----------------:|:-------------:|:-----------------:|:-------------:|
| 0 | 1 | | |
| 0 | 2 | | |
| 0 | 3 | | |
| 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 3 |
| 1 | 1 | 1 | 3 |
| 1 | 2 | 1 | 1 |
| 1 | 2 | 1 | 3 |
| 1 | 2 | 1 | 3 |
| 1 | 3 | 1 | 1 |
| 1 | 3 | 1 | 3 |
| 1 | 3 | 1 | 3 |
| 2 | 1 | 2 | 1 |
| 2 | 1 | 2 | 2 |
| 2 | 1 | 2 | 3 |
| 2 | 2 | 2 | 1 |
| 2 | 2 | 2 | 2 |
| 2 | 2 | 2 | 3 |
| 2 | 3 | 2 | 1 |
| 2 | 3 | 2 | 2 |
| 2 | 3 | 2 | 3 |
This link led to the conclusion, that any join command might not what is necessary here ... But union does only append rows... so for me it is absolutely unclear, how the desired result can be achieved o.O
Any help is highly appreciated. Thanks in advance!
The main misunderstanding on this question arose from the fact that you mentioned join, which is a very precisely mathematically defined concept based on the Cartesian product and can be applied to any two sets. So the current output is clear.
But as you wrote in the title, you want to put two tables side by side. You take advantage of the fact that they have the same number of rows (triples).
This select returns the output you want.
I made artificial join columns, row_number() OVER (order by generation, parent) as rnum, and moved the second table using the addition of three. I hope this helps you:
with
p as (
select
row_number() OVER (order by generation, parent) as rnum,
generation,
parent
from
table1
order by
generation,
parent
), o as(
select
row_number() OVER (order by generation, parent) as rnum,
generation,
parent
from
table2
order by
generation,
parent
)
select
p.generation as table1_generation,
p.parent as table1_parent,
o.generation as table2_generation,
o.parent as table2_parent
from
p
left join o on
o.rnum+3=p.rnum
order by 1,2,3,4;
Output:
table1_generation
table1_parent
table2_generation
table2_parent
0
1
(null)
(null)
0
2
(null)
(null)
0
3
(null)
(null)
1
1
1
1
1
2
1
3
1
3
1
3
2
1
2
1
2
2
2
2
2
3
2
3
I want to create a function that finds the the number of pno's with the sid (staff id) that have worked on it.
So for example if I wanted to find the sid's corresponding to pno = 1
select sid_worked_on(1)
count
-------
2
I would have 2 as sid 0 and 1 have worked on it.
This is the joined table from 3 different tables.
pno | a_sid | b_sid | c_sid
-----+--------+--------+--------
1 | 0 | 0 | 0
4 | 4 | 4 | 6
5 | 4 | 4 | 5
2 | 0 | 0 | 0
1 | 0 | 1 | 0
7 | 5 | 4 | 4
7 | 5 | 5 | 4
5 | 4 | 4 | 4
4 | 4 | 5 | 6
7 | 5 | 4 | 1
7 | 5 | 5 | 1
6 | 5 | 4 | 5
My only way of thinking how to do it would be to "flatten" the table into one column since there is no need for multiple columns and do distinct sid, but I haven't learnt how to do that yet.
pno | sid
-----+--------
1 | 0 |
4 | 4 |
5 | 4 |
2 | 0 |
1 | 0 |
7 | 5 |
7 | 5 |
5 | 4 |
4 | 4 |
7 | 5 |
7 | 5 |
6 | 5 |
--where the new table starts
1 | 0 |
4 | 4 |
5 | 4 |
2 | 0 |
1 | 1 |
7 | 4 |
...
...
I also thought to create a table and going through each value, so
create table
for each row where pno = 1
check if a_sid in table
if not then add a_sid to table
check if b_sid in table
if not then add b_sid to table
check if c_sid in table
if not then add c_sid to table
Would there be a better way of doing this?
Use UNION
SELECT pno, a_sid from table
UNION
SELECT pno, b_sid from table
UNION
SELECT pno, c_sid from table
Depending on whether you want duplicated entries of the same pno with the same column on the right side, you can use UNION ALL instead of UNION.
You can create a view with this query.
I have table AnalysisForm
a_id| a_description | medical_card_id
-------------------------
1 | Analysis1 | 5
2 | Analysis2 | 3
3 | Analysis3 | 2
4 | Analysis4 | 1
and table DicomForm
d_id| d_description | medical_card_id
-------------------------
1 | DicomForm1 | 5
2 | DicomForm2 | 3
3 | DicomForm3 | 2
4 | DicomForm4 | 1
Now I want to get info by medical_card_id = 5 like this
form_id| form_description | medical_card_id
-------------------------
1 | DicomForm1 | 5
1 | Analysis1 | 5
How can I make it in Postgres?
I actually think that you want a union query here, rather than a join:
SELECT a_id AS form_id, a_description AS form_description, medical_card_id
FROM AnalysisForm
WHERE medical_card_id = 5
UNION ALL
SELECT d_id, d_description, medical_card_id
FROM DicomForm
WHERE medical_card_id = 5;
I have the following Postres code:
SELECT
a.assessmentid,
b.groupid
FROM wo_assessment a
LEFT JOIN wo_group_info b ON a.assessmentid = b.assessmentid
WHERE a.workorderid=2
ORDER BY a.assessmentid
Which returns the following results:
|-------------------|------------|
| assessmentid | groupid |
|-------------------|------------|
| 5 | 5 |
|-------------------|------------|
| 6 | 4 |
|-------------------|------------|
| 7 | 0 |
|-------------------|------------|
| 8 | 5 |
|-------------------|------------|
| 9 | 0 |
|-------------------|------------|
| 10 | 0 |
|-------------------|------------|
I would like to populate the 0 values in the groupid field with the next number above in that column, that isn't 0.
So for example I want my table to look like this:
|-------------------|------------|
| assessmentid | groupid |
|-------------------|------------|
| 5 | 5 |
|-------------------|------------|
| 6 | 4 |
|-------------------|------------|
| 7 | 4 |
|-------------------|------------|
| 8 | 5 |
|-------------------|------------|
| 9 | 5 |
|-------------------|------------|
| 10 | 5 |
|-------------------|------------|
Here is what worked for me:
SELECT q.assessmentid,
first_value(b.groupid ) over (partition by value_partition order by q.assessmentid) FROM (
SELECT a.assessmentid,
b.groupid ,
sum(case when b.groupid is null then 0 else 1 end) over (order by a.assessmentid) as value_partition
FROM wo_assessment as a
LEFT JOIN wo_group_info b ON a.assessmentid = b.assessmentid
ORDER BY a.assessmentid ) as q
LEFT JOIN wo_group_info b ON q.assessmentid = b.assessmentid
I have a table Members(id, name, parent_id), where parent_id is the parent of the member(it is also a member which can have its parent). For example
id | name | parent_id
----------------------
1 | John | NULL
2 | Smith| 1
3 | Andy | 1
4 | Joe | 2
5 | Rick | 2
6 | Craig| 5
7 | Greg | NULL
8 | Bob | 5
9 | Mike | 8
And I'd like to run statement select from members, and I want to have
id | name | parent_id | root_parent_id
--------------------------------------
1 | John | NULL | NULL
2 | Smith| 1 | 1
3 | Andy | 1 | 1
4 | Joe | 2 | 1
5 | Rick | 2 | 1
6 | Craig| 5 | 1
7 | Greg | NULL | NULL
8 | Bob | 7 | 7
9 | Mike | 8 | 7
I want to find the root_parent_id for all members as deeply as possible. Help me please
with recursive recursive_members as (
select *, id root_id, 1 depth
from members
union all
select r.id, r.name, r.parent_id, m.parent_id, r.depth+ 1
from recursive_members r
join members m on r.root_id = m.id
where m.parent_id notnull
)
select distinct on (id) *
from recursive_members
order by id, depth desc;
id | name | parent_id | root_id | depth
----+-------+-----------+---------+-------
1 | John | | 1 | 1
2 | Smith | 1 | 1 | 2
3 | Andy | 1 | 1 | 2
4 | Joe | 2 | 1 | 3
5 | Rick | 2 | 1 | 3
6 | Craig | 5 | 1 | 4
7 | Greg | | 7 | 1
8 | Bob | 5 | 1 | 4
9 | Mike | 8 | 1 | 5
(9 rows)
Read about recursive WITH queries.