Reputation: 211
I have a column in a table that has Jsonstring text:
[
{
"PType":{"code":"9","name":"Hospitality"},
"PSubType":{"code":"901","name":"Hotel"},
"AType":{"code":"9","name":"Hospitality"},
"ASubType":{"code":"901","name":"Hotel"}
}
]
How can I divide that into multiple columns using sql server query?
Upvotes: 0
Views: 2045
Reputation: 67311
With SQL-Server 2016+ there is native JSON support:
DECLARE @json NVARCHAR(MAX)=
N'[
{
"PType":{"code":"9","name":"Hospitality"},
"PSubType":{"code":"901","name":"Hotel"},
"AType":{"code":"9","name":"Hospitality"},
"ASubType":{"code":"901","name":"Hotel"}
}
]';
SELECT A.[key]
,JSON_VALUE(A.value,'$.code') AS Code
,JSON_VALUE(A.value,'$.name') AS [Name]
FROM OPENJSON(JSON_QUERY(@json,'$[0]')) A;
The result
key Code Name
---------------------------------
PType 9 Hospitality
PSubType 901 Hotel
AType 9 Hospitality
ASubType 901 Hotel
With JSON_QUERY()
you can get the element within the array, OPENJSON
will find all objects within and return them as derived table.
JSON_VALUE
will read the internals into columns.
with a version below v2016 you should use another tool or think about a CLR function...
Upvotes: 2
Reputation: 12355
In SQL Server 2016+ you can use a combination of openjson
(more info here) and cross apply
:
declare @json nvarchar(max) = '[ { "PType":{"code":"9","name":"Hospitality"}, "PSubType":{"code":"901","name":"Hotel"}, "AType":{"code":"9","name":"Hospitality"}, "ASubType":{"code":"901","name":"Hotel"} } ]'
select
[PT].[code] as Ptype_Code
,[PT].[name] as Ptype_Name
,[PS].[code] as PSubType_Code
,[PS].[name] as PSubType_Name
,[AT].[code] as AType_Code
,[AT].[name] as AType_Name
,[AS].[code] as ASubType_Code
,[AS].[name] as ASubType_Name
from openjson (@json)
with
(
PType nvarchar(max) as json,
PSubType nvarchar(max) as json,
AType nvarchar(max) as json,
ASubType nvarchar(max) as json
) as lev1
cross apply openjson (lev1.PType)
with
(
code int,
name nvarchar(100)
) as PT
cross apply openjson (lev1.PSubType)
with
(
code int,
name nvarchar(100)
) as PS
cross apply openjson (lev1.AType)
with
(
code int,
name nvarchar(100)
) as [AT]
cross apply openjson (lev1.ASubType)
with
(
code int,
name nvarchar(100)
) as [AS]
Result:
Upvotes: 0