Dynamically convert hstore keys into columns for an unknown set of keys

Question

I have a database that stores a bunch of custom fields using hstore. In order to merge it into another database that doesn't support hstore, I'd like to split the keys into extra columns.

Users can add new custom fields and so I can't rely on knowledge of the keys ahead of time. Which makes the answer at "Attributes from an hstore-column as separate columns in a view?" not applicable for my problem.

Where a record doesn't have a key present in other records, it should get the same column with a null value.

How do I do this?

Answer 1

This can be done, very efficiently, too. Not in a single statement, though, since SQL demands to know the return type at call time. So you need two steps. The solution involves a number of advanced techniques ...

Assuming the same table as @Denver in his answer:

CREATE TABLE hstore_test (
  id serial PRIMARY KEY
, hstore_col hstore
);

Solution 1: Simple SELECT

After I wrote the crosstab solution below it struck me that a simple "brute force" solution is probably faster. Basically, the query @Denver already posted, built dynamically:

Step 1a: Generate query

SELECT format(
     'SELECT id, h->%s
      FROM  (SELECT id, hstore_col AS h FROM hstore_test) t;'
    , string_agg(quote_literal(key) || ' AS ' || quote_ident(key), ', h->')
   ) AS sql 
FROM  (
   SELECT DISTINCT key
   FROM   hstore_test, skeys(hstore_col) key
   ORDER  BY 1
   ) sub;

The subquery (SELECT id, hstore_col AS h FROM hstore_test) is just to get in the column alias h for your hstore column.

Step 1b: Execute query

This generates a query of the form:

SELECT id, h->'key1' AS key1, h->'key2' AS key2, h->'key3' AS key3
FROM  (SELECT id, hstore_col AS h FROM hstore_test) t;

Result:

 id | key1  | key2  | key3
----+-------+-------+-------
  1 | val11 | val12 | val13
  2 | val21 | val22 |
  3 |       |       |        -- for a row where hstore_col IS NULL

Solution 2: crosstab()

For lots of keys this may perform better. Probably not. You'll have to test. Result is the same as for solution 1.

You need the additional extension tablefunc which provides the crosstab() function. Read this first if you are not familiar:

• PostgreSQL Crosstab Query

Step 2a: Generate query

SELECT format(
   $s$SELECT * FROM crosstab(
     $$SELECT h.id, kv.*
       FROM   hstore_test h, each(hstore_col) kv
       ORDER  BY 1, 2$$
   , $$SELECT unnest(%L::text[])$$
   ) AS t(id int, %s text);
   $s$
 , array_agg(key)  -- escapes strings automatically
 , string_agg(quote_ident(key), ' text, ')  -- needs escaping!
   ) AS sql 
FROM  (
   SELECT DISTINCT key
   FROM   hstore_test, skeys(hstore_col) key
   ORDER  BY 1
   ) sub;

Note the nested levels of dollar-quoting.

I use this explicit form in the main query instead of the short CROSS JOIN in the auxiliary query to preserve rows with empty or NULL hstore values:

LEFT   JOIN LATERAL each(hstore_col) kv ON TRUE

• What is the difference between LATERAL and a subquery in PostgreSQL?

Related:

• Dynamic alternative to pivot with CASE and GROUP BY

Step 2b: Execute query

This generates a query of the form:

SELECT * FROM crosstab(
     $$SELECT h.id, kv.*
       FROM   hstore_test h
       LEFT   JOIN LATERAL each(hstore_col) kv ON TRUE
       ORDER  BY 1, 2$$
   , $$SELECT unnest('{key1,key2,key3}'::text[])$$
   ) AS t(id int, key1 text, key2 text, key3 text);

You may want to inspect it for plausibility before running the first time. This should deliver optimized performance.

Notes

• Both solutions work for any number of keys up to the physical limit of ~ 1600 columns in a Postgres table.

• Both also work for keys of any shape or form up to the maximum length for identifiers, which is 63 bytes per default.

• Besides the hstore function each() that was already mentioned by s.m., I also use the related function skeys() to identify keys.

• Be sure to quote column names correctly to avoid possible SQL injection attacks by way of maliciously formed key names. I take care of that with quote_literal() and quote_ident().

Answer 2

I realize I'm a bit late—and by now you sure have it figured out—but, seeing the comment you left on Denver Timothy's answer, I thought I would leave an answer for everybody else:

select (each(hstore_col)).key from hstore_test;

This will create a row for each key contained in hstore_col, so you won't need to know what the keys are beforehand.

Answer 3

You'll want to use the -> operator on the column (see here).

Records without the same key in other records will show as NULL.

create table hstore_test (id serial, hstore_col hstore);
insert into hstore_test (hstore_col) values ('key1=>val11, key2=>val12, key3=>val13'), ('key1=>val21, key2=>val22');
select hstore_col->'key1' as key1, hstore_col->'key2' as key2, hstore_col->'key3' as key3 from hstore_test;
┌───────┬───────┬───────┐
│ key1  │ key2  │ key3  │
├───────┼───────┼───────┤
│ val11 │ val12 │ val13 │
│ val21 │ val22 │ NULL  │
└───────┴───────┴───────┘
(2 rows)

Here is a similar answer.

Answer 4

Based on s.m.'s answer it seems you can combine EACH() with both .key and .value to generate columns automatically.

Setup test table like in Ian Timothy's answer:

create table hstore_test (id serial, hstore_col hstore);
insert into hstore_test (hstore_col) values ('key1=>val11, key2=>val12, key3=>val13'), ('key1=>val21, key2=>val22');

Query

SELECT id, (EACH(hstore_col)).key, (EACH(hstore_col)).value FROM hstore_test;

which returns

 id | key  | value 
----+------+-------
  1 | key1 | val11
  1 | key2 | val12
  1 | key3 | val13
  2 | key1 | val21
  2 | key2 | val22

I was a bit unsure whether two EACH() inside the same query are guaranteed to produce the same order. I found an example in the hstore documentation using the same mechanism so I guess it's explicitly supported.