Postgres: How to convert a json string to text?

Json value may consist of a string value. eg.:

postgres=# SELECT to_json('Some "text"'::TEXT);
to_json
-----------------
"Some \"text\""

How can I extract that string as a postgres text value?

::TEXT doesn't work. It returns quoted json, not the original string:

postgres=# SELECT to_json('Some "text"'::TEXT)::TEXT;
to_json
-----------------
"Some \"text\""

Thanks.

P.S. I'm using PostgreSQL 9.3

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最佳答案

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.

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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;

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An easy way of doing this:

SELECT  ('[' || to_json('Some "text"'::TEXT) || ']')::json ->> 0;

Just convert the json string into a json list

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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)

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->> 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