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比较SQL Server 2016中使用openjson的JSON字符串_Sql_Json_Sql Server_Sql Server 2016 - Fatal编程技术网

比较SQL Server 2016中使用openjson的JSON字符串

比较SQL Server 2016中使用openjson的JSON字符串,sql,json,sql-server,sql-server-2016,Sql,Json,Sql Server,Sql Server 2016,我有两种插入json字符串的场景。 我使用下面的代码检查json字符串是否已经存在,然后插入到表中 **Scenario - 1** declare @jsonStr nvarchar(max) = {"Manufacurer": "test", "Model": "A", "Color":"New Color","Thickness":"1 mm"} declare @SqlStr nvarchar(max), @counts int Select @d

我有两种插入json字符串的场景。 我使用下面的代码检查json字符串是否已经存在,然后插入到表中

**Scenario - 1**


 declare @jsonStr nvarchar(max) = {"Manufacurer": "test", "Model": "A", "Color":"New Color","Thickness":"1 mm"}
        declare @SqlStr nvarchar(max), @counts int

        Select @dynamicFilter = Coalesce(@dynamicFilter,'') +' and Json_Value(ItemOfProduct,''$."'+ [key] +'"'') = ''' +[value]+'''' from openjson(jsonStr)

        set @SqlStr = 'select @cnt = count(*) from tblTest where '+@dynamicFilter
        Exec sp_executesql @SqlStr, N'@cnt int output', @cnt = @counts output

        if(@counts < 1)
        Begin
            //insert the data into the table
        End




**Scenario - 2** 


set @jsonStr = {"Manufacurer": "test", "Model": "A", "Color":"New Color"}
    //Do the same above process of inserting

在第二种情况下,由于计数为1,我的数据不会被插入表中。但我希望场景1和场景2中的字符串都被视为不同的字符串,并且两个数据都被插入到数据库中

我认为最好的解决方案实际上是使用JSON_查询而不是JSON_值,因为您的键可能是动态的。一旦切换到JSON_查询,就不再需要执行动态SQL

CREATE TABLE #tblTest
(
    id INT IDENTITY(1, 1)
    ,ItemOfProduct NVARCHAR(MAX)
);


/*Scenario 1*/
DECLARE @jsonStr NVARCHAR(MAX) = N'{"Manufacturer": "test", "Model": "A", "Color":"New Color","Thickness":"1 mm"}';
DECLARE @counts INT;

SELECT @counts = COUNT(*)
FROM #tblTest
WHERE JSON_QUERY(ItemOfProduct, '$') = JSON_QUERY(@jsonStr, '$');

IF(@counts < 1)
BEGIN
    INSERT INTO #tblTest(ItemOfProduct)
    VALUES(@jsonStr);
END;


/*Scenario 2*/
SET @jsonStr =
    N'{"Manufacturer": "test", "Model": "A", "Color":"New Color"}';

SELECT @counts = COUNT(*)
FROM #tblTest
WHERE JSON_QUERY(ItemOfProduct, '$') = JSON_QUERY(@jsonStr, '$');

IF(@counts < 1)
BEGIN
    INSERT INTO #tblTest(ItemOfProduct)
    VALUES(@jsonStr);
END;

SELECT *
FROM #tblTest;

我认为最好的解决方案实际上是使用JSON_查询而不是JSON_值,因为您的键可能是动态的。一旦切换到JSON_查询,就不再需要执行动态SQL

CREATE TABLE #tblTest
(
    id INT IDENTITY(1, 1)
    ,ItemOfProduct NVARCHAR(MAX)
);


/*Scenario 1*/
DECLARE @jsonStr NVARCHAR(MAX) = N'{"Manufacturer": "test", "Model": "A", "Color":"New Color","Thickness":"1 mm"}';
DECLARE @counts INT;

SELECT @counts = COUNT(*)
FROM #tblTest
WHERE JSON_QUERY(ItemOfProduct, '$') = JSON_QUERY(@jsonStr, '$');

IF(@counts < 1)
BEGIN
    INSERT INTO #tblTest(ItemOfProduct)
    VALUES(@jsonStr);
END;


/*Scenario 2*/
SET @jsonStr =
    N'{"Manufacturer": "test", "Model": "A", "Color":"New Color"}';

SELECT @counts = COUNT(*)
FROM #tblTest
WHERE JSON_QUERY(ItemOfProduct, '$') = JSON_QUERY(@jsonStr, '$');

IF(@counts < 1)
BEGIN
    INSERT INTO #tblTest(ItemOfProduct)
    VALUES(@jsonStr);
END;

SELECT *
FROM #tblTest;

tblTest的结构是什么?桌子上有什么限制吗?您实际使用什么代码尝试将记录插入表中?您的代码似乎缺少帮助您调试它所需的一部分。tblTest的结构是什么?桌子上有什么限制吗?您实际使用什么代码尝试将记录插入表中?您的代码似乎缺少帮助您调试它所需的部分。