PostgreSQL LEFT OUTER JOIN query syntax - sql

Question

Lets say I have a table1:
id name
-------------
1 "one"
2 "two"
3 "three"
And a table2 with a foreign key to the first:
id tbl1_fk option value
-------------------------------
1 1 1 1
2 2 1 1
3 1 2 1
4 3 2 1
Now I want to have as a query result:
table1.id | table1.name | option | value
-------------------------------------
1 "one" 1 1
2 "two" 1 1
3 "three"
1 "one" 2 1
2 "two"
3 "three" 2 1
How do I achieve that?
I already tried:
SELECT
table1.id,
table1.name,
table2.option,
table2.value
FROM table1 AS table1
LEFT outer JOIN table2 AS table2 ON table1.id = table2.tbl1fk
but the result seems to omit the null vales:
1 "one" 1 1
2 "two" 1 1
1 "one" 2 1
3 "three" 2 1
SOLVED: thanks to Mahmoud Gamal: (plus the GROUP BY)
Solved with this query
SELECT
t1.id,
t1.name,
t2.option,
t2.value
FROM
(
SELECT t1.id, t1.name, t2.option
FROM table1 AS t1
CROSS JOIN table2 AS t2
) AS t1
LEFT JOIN table2 AS t2 ON t1.id = t2.tbl1fk
AND t1.option = t2.option
group by t1.id, t1.name, t2.option, t2.value
ORDER BY t1.id, t1.name

Answer 1

You have to use CROSS JOIN to get every possible combination of name from the first table with the option from the second table. Then LEFT JOIN these combination with the second table. Something like:
SELECT
t1.id,
t1.name,
t2.option,
t2.value
FROM
(
SELECT t1.id, t1.name, t2.option
FROM table1 AS t1
CROSS JOIN table2 AS t2
) AS t1
LEFT JOIN table2 AS t2 ON t1.id = t2.tbl1_fk
AND t1.option = t2.option
SQL Fiddle Demo

Answer 2

Simple version: option = group
It's not specified in the Q, but it seems like option is supposed to define a group somehow. In this case, the query can simply be:
SELECT t1.id, t1.name, t2.option, t2.value
FROM (SELECT generate_series(1, max(option)) AS option FROM table2) o
CROSS JOIN table1 t1
LEFT JOIN table2 t2 ON t2.option = o.option AND t2.tbl1_fk = t1.id
ORDER BY o.option, t1.id;
Or, if options are not numbered in sequence, starting with 1:
...
FROM (SELECT DISTINCT option FROM table2) o
...
Returns:
id | name | option | value
----+-------+--------+-------
1 | one | 1 | 1
2 | two | 1 | 1
3 | three | |
1 | one | 2 | 1
2 | two | |
3 | three | 2 | 1
Faster and cleaner, avoiding the big CROSS JOIN and the big GROUP BY.
You get distinct rows with a group number (grp) per set.
Requires Postgres 8.4+.
More complex: group indicated by sequence of rows
WITH t2 AS (
SELECT *, count(step OR NULL) OVER (ORDER BY id) AS grp
FROM (
SELECT *, lag(tbl1_fk, 1, 2147483647) OVER (ORDER BY id) >= tbl1_fk AS step
FROM table2
) x
)
SELECT g.grp, t1.id, t1.name, t2.option, t2.value
FROM (SELECT generate_series(1, max(grp)) AS grp FROM t2) g
CROSS JOIN table1 t1
LEFT JOIN t2 ON t2.grp = g.grp AND t2.tbl1_fk = t1.id
ORDER BY g.grp, t1.id;
Result:
grp | id | name | option | value
-----+----+-------+--------+-------
1 | 1 | one | 1 | 1
1 | 2 | two | 1 | 1
1 | 3 | three | |
2 | 1 | one | 2 | 1
2 | 2 | two | |
2 | 3 | three | 2 | 1
-> SQLfiddle for both.
How?
Explaining the complex version ...
Every set is started with a tbl1_fk <= the last one. I check for this with the window function lag(). To cover the corner case of the first row (no preceding row) I provide the biggest possible integer 2147483647 the default for lag().
With count() as aggregate window function I add the running count to each row, effectively forming the group number grp.
I could get a single instance for every group with:
(SELECT DISTINCT grp FROM t2) g
But it's faster to just get the maximum and employ the nifty generate_series() for the reduced CROSS JOIN.
This CROSS JOIN produces exactly the rows we need without any surplus. Avoids the need for a later GROUP BY.
LEFT JOIN t2 to that, using grp in addition to tbl1_fk to make it distinct.
Sort any way you like - which is possible now with a group number.

Answer 3

try this
SELECT
table1.id, table1.name, table2.option, table2.value FROM table1 AS table11
JOIN table2 AS table2 ON table1.id = table2.tbl1_fk

Answer 4

This is enough:
select * from table1 left join table2 on table1.id=table2.tbl1_fk ;