Using JSON_VALUE + JSON_QUERY to create new JSON - json

Question

I was wondering if it would be possible to reshape JSON and return it as JSON. I have JSON which, in simplified form, looks like:
Name
Details
fieldId
fieldValue
Other
Id
Value
And would like to return:
Name
Details
fieldId
fieldValue
I can return Name and Details with JSON_VALUE and JSON_QUERY but would like it as one combined JSON field.
create table #test (
[id] int,
[json] varchar(max)
);
insert into #test (id, json) values (1, '{
"Name": "Test 1",
"Details": [
{
"fieldId": "100",
"fieldValue": "ABC"
}],
"Other": [
{
"Id": "1",
"Value": "ABC"
}]
}');
insert into #test (id, json) values (2, '{
"Name": "Test 2",
"Details": [
{
"fieldId": "101",
"fieldValue": "ABCD"
}],
"Other": [
{
"Id": "2",
"Value": "ABCD"
}]
}');
select id, JSON_VALUE(json, '$.Name'), JSON_QUERY(json, '$.Details')
from #test

Answer 1

As an additional option, you may parse the JSON with OPENJSON() and explicit schema (columns definitions) and then build the new JSON again:
SELECT
id,
newjson = (
SELECT Name, Details
FROM OPENJSON(json) WITH (
Name varchar(max) '$.Name',
Details nvarchar(max) '$.Details' AS JSON
)
FOR JSON PATH, WITHOUT_ARRAY_WRAPPER
)
FROM #test
And the same approach with JSON_VALUE() and JSON_QUERY():
SELECT
id,
newjson = (
SELECT JSON_VALUE(json, '$.Name') AS [Name], JSON_QUERY(json, '$.Details') AS [Details]
FOR JSON PATH, WITHOUT_ARRAY_WRAPPER
)
FROM #test

Answer 2

One solution is to use JSON_MODIFY to re-construct the JSON:
SELECT
id,
JSON_MODIFY(
JSON_MODIFY(
'{}',
'$.Name',
JSON_VALUE(json, '$.Name')
),
'$.Details',
JSON_QUERY(json, '$.Details')
) AS new_json
FROM #test
An alternate would be to delete the Other node using JSON_MODIFY but you have to know the name of node(s) to remove.