I'd like to convert single rows into multiple rows in PostgreSQL, where some of the columns are removed. Here's an example of the current output:
name | st | ot | dt |
-----|----|----|----|
Fred | 8 | 2 | 3 |
Jane | 8 | 1 | 0 |
Samm | 8 | 0 | 6 |
Alex | 8 | 0 | 0 |
Using the following query:
SELECT
name, st, ot, dt
FROM
times;
And here's what I want:
name | t | val |
-----|----|-----|
Fred | st | 8 |
Fred | ot | 2 |
Fred | dt | 3 |
Jane | st | 8 |
Jane | ot | 1 |
Samm | st | 8 |
Samm | dt | 6 |
Alex | st | 8 |
How can I modify the query to get the above desired output?
SELECT
times.name, x.t, x.val
FROM
times cross join lateral (values('st',st),('ot',ot),('dt',dt)) as x(t,val)
WHERE
x.val <> 0;
The core problem is the reverse of a pivot / crosstab operation. Sometimes called "unpivot".
Basically, Abelisto's query is the way to go in Postgres 9.3 or later. Related:
You may want to use LEFT JOIN LATERAL ... ON u.val <> 0
to include names without valid values in the result (and shorten the syntax a bit).
If you have more than a few value columns (or varying lists of columns) you may want to use a function to build and execute the query automatically:
CREATE OR REPLACE FUNCTION f_unpivot_columns(VARIADIC _cols text[])
RETURNS TABLE(name text, t text, val int) AS
$func$
BEGIN
RETURN QUERY EXECUTE (
SELECT
'SELECT t.name, u.t, u.val
FROM times t
LEFT JOIN LATERAL (VALUES '
|| string_agg(format('(%L, t.%I)', c, c), ', ')
|| ') u(t, val) ON (u.val <> 0)'
FROM unnest(_cols) c
);
END
$func$ LANGUAGE plpgsql;
Call:
SELECT * FROM f_unpivot_times_columns(VARIADIC '{st, ot, dt}');
Or:
SELECT * FROM f_unpivot_columns('ot', 'dt');
Columns names are provided as string literals and must be in correct (case-sensitive!) spelling with no extra double-quotes. See:
dbfiddle here
Related with more examples and explanation:
One way:
with times(name , st , ot , dt) as(
select 'Fred',8 , 2 , 3 union all
select 'Jane',8 , 1 , 0 union all
select 'Samm',8 , 0 , 6 union all
select 'Alex',8 , 0 , 0
)
select name, key as t, value::int from
(
select name, json_build_object('st' ,st , 'ot',ot, 'dt',dt) as j
from times
) t
join lateral json_each_text(j)
on true
where value <> '0'
-- order by name, case when key = 'st' then 0 when key = 'ot' then 1 when key = 'dt' then 2 end
来源:https://stackoverflow.com/questions/45201170/convert-one-row-into-multiple-rows-with-fewer-columns