jdbcTemplate and raw JSON column - sql

Question

Suppose I query a table which contains raw json in a column, or create json myself with a subquery like this:
select
p.name,
(select json_build_array(json_build_object('num', a.num, 'city', a.city)) from address a where a.id = p.addr) as addr
from person p;
How can I instruct Spring's NamedParameterJdbcTemplate not to escape
the addr column but leave it alone?
So far it insists returning me something like this:
[{
name: "John",
addr: {
type: "json",
value: "{"num" : "123", "city" : "Luxembourg"}"
}
}]

Answer 1

This is the working solution.
Service:
#Transactional(readOnly = true)
public String getRawJson() {
String sql = "select json_agg(row_to_json(json)) from ( "
+ "select "
+ "p.id, "
+ "p.name, "
+ "(select array_to_json(array_agg(row_to_json(c))) from ( "
+ " ... some subselect ... "
+ ") c ) as subq "
+ "from person p "
+ "where type = :type "
+ ") json";
MapSqlParameterSource params = new MapSqlParameterSource("type", 1);
return jdbcTemplate.queryForObject(sql, params, String.class);
}
Controller:
#ResponseBody
#GetMapping(value = "/json", produces = MediaType.APPLICATION_JSON_VALUE)
public String getRawJson() {
return miscService.getRawJson();
}
The key is the combination of json_agg and row_to_json / array_to_json plus returning a String to avoid any type conversions.