I have column options with type jsonb , in format {"names": ["name1", "name2"]} which was created with
UPDATE table1 t1 SET options = (SELECT jsonb_build_object('names', names) FROM table2 t2 WHERE t2.id= t1.id)
and where names have type jsonb array.
SELECT jsonb_typeof(names) FROM table2 give array
Now I want to extract value of names as jsonb array. But query
SELECT jsonb_build_array(options->>'names') FROM table
gave me ["[\"name1\", \"name2\"]"], while I expect ["name1", "name2"]
How can I get value in right format?
2 Answers
The ->> operator will return the value of the field (in your case, a JSON array) as a properly escaped text. What you are looking for is the -> operator instead.
However, note that using the jsonb_build_array on that will return an array containing your original array, which is probably not what you want either; simply using options->'names' should get you what you want.
Actually, you don't need to use jsonb_build_array() function.
Use select options -> 'names' from table; This will fix your issue.
jsonb_build_array() is for generating the array from jsonb object. You are following wrong way. That's why you are getting string like this ["[\"name1\", \"name2\"]"].
Try to execute this sample SQL script:
select j->'names'
from (
select '{"names": ["name1", "name2"]}'::JSONB as j
) as a;
