SQL Server中将查询结果转换为Json格式脚本

这篇文章主要介绍了SQL Server中将查询结果转换为Json格式脚本分享,本文直接给出实现代码,需要的朋友可以参考下

原文地址：http://www.jb51.net/article/61462.htm

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE PROCEDURE[dbo].[SerializeJSON](
@ParameterSQL AS VARCHAR(MAX)
)
AS
BEGIN
  
DECLARE @SQL NVARCHAR(MAX)
DECLARE @XMLString VARCHAR(MAX)
DECLARE @XML XML
DECLARE @Paramlist NVARCHAR(1000)
SET @Paramlist = N'@XML XML OUTPUT'
SET @SQL = 'WITH PrepareTable (XMLString)'
SET @SQL = @SQL + 'AS('
SET @SQL = @SQL + @ParameterSQL+ 'FOR XML RAW,TYPE,ELEMENTS'
SET @SQL = @SQL + ')'
SET @SQL = @SQL + 'SELECT @XML=[XMLString]FROM[PrepareTable]'
EXEC sp_executesql @SQL, @Paramlist, @XML=@XML OUTPUT
SET @XMLString=CAST(@XML AS VARCHAR(MAX))
  
DECLARE @JSON VARCHAR(MAX)
DECLARE @Row VARCHAR(MAX)
DECLARE @RowStart INT
DECLARE @RowEnd INT
DECLARE @FieldStart INT
DECLARE @FieldEnd INT
DECLARE @KEY VARCHAR(MAX)
DECLARE @Value VARCHAR(MAX)
  
DECLARE @StartRoot VARCHAR(100);SET @StartRoot=''
DECLARE @EndRoot VARCHAR(100);SET @EndRoot=''
DECLARE @StartField VARCHAR(100);SET @StartField='<'
DECLARE @EndField VARCHAR(100);SET @EndField='>'
  
SET @RowStart=CharIndex(@StartRoot,@XMLString,0)
SET @JSON=''
WHILE @RowStart>0
BEGIN
    SET @RowStart=@RowStart+Len(@StartRoot)
    SET @RowEnd=CharIndex(@EndRoot,@XMLString,@RowStart)
    SET @Row=SubString(@XMLString,@RowStart,@RowEnd-@RowStart)
    SET @JSON=@JSON+'{'
  
    -- for each row
    SET @FieldStart=CharIndex(@StartField,@Row,0)
    WHILE @FieldStart>0
    BEGIN
        -- parse node key
        SET @FieldStart=@FieldStart+Len(@StartField)
        SET @FieldEnd=CharIndex(@EndField,@Row,@FieldStart)
        SET @KEY=SubString(@Row,@FieldStart,@FieldEnd-@FieldStart)
        SET @JSON=@JSON+'"'+@KEY+'":'
        -- parse node value
        SET @FieldStart=@FieldEnd+1
        SET @FieldEnd=CharIndex('0SET @JSON=SubString(@JSON,0,LEN(@JSON))
    SET @JSON=@JSON+'},'
    --/ for each row
  
    SET @RowStart=CharIndex(@StartRoot,@XMLString,@RowEnd)
END
IF LEN(@JSON)>0SET @JSON=SubString(@JSON,0,LEN(@JSON))
SET @JSON='['+@JSON+']'
SELECT @JSON
  
END
GO

转载于:https://www.cnblogs.com/caiyt/p/7905152.html