How to write a select query to get the index value from Json object - sql

Question

I have the below JSON object. I need to write a select query to get the index values of Object JSON array. Kind of getting the sequence value.
{
"Model": [
{
"ModelName": "Test Model",
"Object": [
{
"ID": 1,
"Name": "ABC",
},
{
"ID": 11,
"Name": "ABCD",
},
{
"ID": 15,
"Name": "ABCDE",
},
]
}]}
Expected Output:
Index_Value
1
2
3

Answer 1

If I understand the question correctly and you want to get the index of the items in the Object JSON array, you need to use OPENJSON() with default schema. The result is a table with columns key, value and type and in case of JSON array, the key column holds the index of each item in the array (0-based):
JSON:
DECLARE #json nvarchar(max) = N'{
"Model":[
{
"ModelName":"Test Model",
"Object":[
{
"ID":1,
"Name":"ABC"
},
{
"ID":11,
"Name":"ABCD"
},
{
"ID":15,
"Name":"ABCDE"
}
]
}
]
}'
Statement:
SELECT CONVERT(int, j2.[key]) + 1 AS item_id
FROM OPENJSON (#json, '$.Model') j1
CROSS APPLY OPENJSON(j1.[value], '$.Object') j2
But if you want to get the values of the ID keys in the Object JSON array, the statement is different:
SELECT j2.ID
FROM OPENJSON (#json, '$.Model') j1
CROSS APPLY OPENJSON(j1.[value], '$.Object') WITH (
ID int '$.ID'
) j2
Note, that you need two OPENJSON() calls, because the input JSON has nested array structure. Of course, if Model JSON array has always one item, you may simplify the statement using an appropriate path:
SELECT CONVERT(int, [key]) + 1 AS item_id
FROM OPENJSON (#json, '$.Model[0].Object')
Finally, to get index, ID and Name, you should use the following statement, which assumes, that $.Model JSON array has more than one item and defines ID and Name columns with the appropraite data types:
SELECT
CONVERT(int, j2.[key]) + 1 AS ItemID,
j3.ID, j3.Name
FROM OPENJSON (#json, '$.Model') j1
CROSS APPLY OPENJSON(j1.[value], '$.Object') j2
CROSS APPLY OPENJSON(j2.[value], '$') WITH (
ID int '$.ID',
Name varchar(50) '$.Name'
) j3

Answer 2

DECLARE #json nvarchar(max) = N'{
"Model":[
{
"ModelName":"Test Model",
"Object":[
{
"ID":1,
"Name":"ABC"
},
{
"ID":11,
"Name":"ABCD"
},
{
"ID":15,
"Name":"ABCDE"
}
]
}
]
}'
declare #i int=0;
SELECT
j2.ID, j2.Name
FROM OPENJSON (#json, '$.Model') j1
CROSS APPLY OPENJSON(j1.[value],concat('$.Object[',#i,']')) WITH (
ID i`enter code here`nt '$.ID', Name varchar(100) '$.Name'
) j2
Results:-
ID
Name
11
ABCD

Answer 3

you can select the key columns in select clause no need to mention in with of crossjoin.
SELECT
distinct t.id,
JSON_VALUE(AttsData.[value], '$.address') as address,
JSON_VALUE(AttsData.[value], '$.name') as name,
JSON_VALUE(AttsData.[value], '$.owner_name') as owner_name,
JSON_VALUE(AttsData.[value], '$.project') as project
,CONVERT(int, AttsData.[key]) index_id
FROM mytablewithjsonfeild t
CROSS APPLY OPENJSON (t."jsonfeild",N'$.parentkey') as AttsData
Above query, from the table I have cross joined the JSON field. and in select statement i have taken the specific keys.
and CONVERT(int, AttsData.[key]) to get the index of the elements