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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
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  • Actually, your present solution doesn't work. Consider the extended test case in this SQLfiddle. Commented Mar 26, 2013 at 0:04

4 Answers 4

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8

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

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7 Comments

that works! only I have to add the GROUP BY Clause because the result is multiple!
Almost works, but not quite. See the SQLfiddle in my comment above.
Seems expensive with cross-joins if the tables involved are large--no way around it?
@ako - run the EXPLAIN statement on your query EXPLAIN SELECT ...the rest of your query and see if any optimizations can be done. Also ensure that indexes are set properly on your table.
@ako - LEFT JOIN works fine as expected, the point is that; in the sample data the user posted all the rows in the first tables have matching rows in the other table table2, so there are no null values because of that. But if you remove (for example) the last row from the table2, you will get a null values , since the name three has no matching rows in the second table. But this is not what the OP wants, if you look at the desired result, he wants to select all the names from the table1 with all possible combination from the table2.
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3

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.

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0

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
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2 Comments

That produces the following result: 1;"one";1;1 2;"two";1;1 1;"one";2;1 3;"three";2;1 but i want even the null values !
same result even with JOIN, even with LEFT OUTER JOIN :( I'm using postgreSQL 9.1
-5

This is enough:

select * from table1 left join table2 on table1.id=table2.tbl1_fk ;

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