Json value may consist of a string value. eg.:
postgres=# SELECT to_json(\'Some \"text\"\'::TEXT);
to_json
-----------------
\"Some \\\"text\\\"\"
There is no way in PostgreSQL to deconstruct a scalar JSON object. Thus, as you point out,
select length(to_json('Some "text"'::TEXT) ::TEXT);
is 15,
The trick is to convert the JSON into an array of one JSON element, then extract that element using ->>
.
select length( array_to_json(array[to_json('Some "text"'::TEXT)])->>0 );
will return 11.
An easy way of doing this:
SELECT ('[' || to_json('Some "text"'::TEXT) || ']')::json ->> 0;
Just convert the json string into a json list
->> works for me.
postgres version:
<postgres.version>11.6</postgres.version>
Query:
select object_details->'valuationDate' as asofJson, object_details->>'valuationDate' as asofText from MyJsonbTable;
Output:
asofJson asofText
"2020-06-26" 2020-06-26
"2020-06-25" 2020-06-25
"2020-06-25" 2020-06-25
"2020-06-25" 2020-06-25
In 9.4.4 using the #>> operator works for me:
select to_json('test'::text) #>> '{}';
To use with a table column:
select jsoncol #>> '{}' from mytable;
Mr. Curious was curious about this as well. In addition to the #>> '{}'
operator, in 9.6+ one can get the value of a jsonb string with the ->>
operator:
select to_jsonb('Some "text"'::TEXT)->>0;
?column?
-------------
Some "text"
(1 row)
If one has a json value, then the solution is to cast into jsonb first:
select to_json('Some "text"'::TEXT)::jsonb->>0;
?column?
-------------
Some "text"
(1 row)