PostgreSQL unnest with empty array

This works on Postgres 10 also:

SELECT id, UNNEST(CASE WHEN "values" <> '{}' THEN "values" ELSE '{null}' END)

Since Postgres 10

A LATERAL subquery is the clean, versatile solution, and works since Postgres 9.3, where LATERAL was added:

SELECT t.id, v.value
FROM   tbl t
LEFT   JOIN LATERAL unnest(t.values) AS v(value) ON true;

Or, with minimal syntax:

SELECT id, value
FROM   tbl
LEFT   JOIN LATERAL unnest(values) value ON true;

fiddle

See:

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

Original answer

This works up to Postgres 9.6. The behavior of set-returning functions in the SELECT list was sanitized in Postgres 10, the short syntax below stops working. See:

• What is the expected behaviour for multiple set-returning functions in SELECT clause?

Flip the logic in the currently accepted answer by @a_horse:

SELECT id, CASE WHEN values <> '{}' THEN unnest(values) END AS value
FROM   tbl;

fiddle

This returns a row with NULL in value for an empty array as well as for a NULL array, because only an array with elements in it produces true in the test values <> '{}'.

Works for arrays of any type, since the literal '{}' is automatically coerced to a matching type.

Without explicit ELSE branch, CASE returns NULL, which is what we want anyway.

Arrays with a NULL elements will return rows regardless.

However, I found an anomaly which I addressed in a related question:

• NULL emements lost when casting result of unnest()

Turned out to be a bug. It was fixed after my report for Postgres 9.3+.

select id,
    unnest (
        "values"
        ||
        (array[null]::integer[])[1:(array_upper("values", 1) is null)::integer]
    )
from "table"

You will need to use self LEFT JOIN, like this (also on SQL Fiddle):

SELECT t.id, u.u
  FROM tab t
  LEFT JOIN (SELECT id, unnest(vals) u FROM tab) u
    USING (id);

Note, that for bigger tables query will be performing badly.