如何解析 SQL Server 2012 中的 json 数据？

Question

I am using SQL Server 2012.I have been assigned a task where one of my column ( JsonText ) of table Sample contains json data. I want to pass parse that data and insert into columns of another table ( Test ). I searched on net 'openjson' is supported in SQL Server 2016. How to do in SQL Server 2012?

Table1 : Sample

Id JsonText Active 

JsonText

webaddress?{'data':'{"PId": "XXXX","Status": "YES","Name":"XXX","Address":"XXXX","MobileNumber":"xxx"}'}

I am intrested only 'PID,Address,MobileNumber' columns not all.

Table Test structure like this

Id, PID, Address, MobileNumber

Answer 1

I created a function compatible with SQL 2012 to take care of this

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Author:      Isaac Adams
-- Create date: 7/12/2018
-- Description: Give the JSON string and the name of the column from which you want the value
-- =============================================
CREATE FUNCTION JSON_VALUE
(
    @JSON NVARCHAR(3000),
    @column NVARCHAR(3000)
)
RETURNS NVARCHAR(3000)
AS
BEGIN

DECLARE @value NVARCHAR(3000);
DECLARE @trimmedJSON NVARCHAR(3000);

DECLARE @start INT;
DECLARE @length INT;

SET @start = PATINDEX('%' + @column + '":"%',@JSON) + LEN(@column) + 3;
SET @trimmedJSON = SUBSTRING(@JSON, @start, LEN(@JSON));
SET @length = PATINDEX('%", "%', @trimmedJSON);
SET @value = SUBSTRING(@trimmedJSON, 0, @length);

RETURN @value
END
GO

Answer 2

Isaac your code is not working with not quoted values eg {"isAuthorized":"false","customerID":null}. I fixed this and your function should look like this.

ALTER FUNCTION [dbo].[JSON_VALUE]
(
    @JSON NVARCHAR(3000),
    @column NVARCHAR(3000)
)
RETURNS NVARCHAR(3000)
AS
BEGIN

DECLARE @value NVARCHAR(3000);
DECLARE @trimmedJSON NVARCHAR(3000);

DECLARE @start INT;
DECLARE @end INT;

set @start = PATINDEX('%' + @column + '":%',@JSON) + LEN(@column) + 2;
SET @trimmedJSON = SUBSTRING(@JSON, @start, LEN(@JSON));
Set @end = CHARINDEX(',',@trimmedJSON);
SET @value = REPLACE(SUBSTRING(@trimmedJSON, 0, @end),'"','');

RETURN @value
END

Answer 3

>>> at JSON_VALUE function, at PATINDEX('%", "%', @trimmedJSON);

remove space from '%", "%'

if your JSON value is like

'{"street":"street1","street2":"street232423"}'

Answer 4

You can use JSON_VALUE(ColumnName,'$.Path') for pairs Json in TSQL, for example:

select JSON_VALUE(webaddress,'$.data.PID') as 'PID',
       JSON_VALUE(webaddress,'$.data.Status') as 'Status',
       JSON_VALUE(webaddress,'$.data.Name') as 'Name'
from test