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如何用javascriptserializer解析多嵌套的JSON数组

javascriptserializer json c#

Mist · 技术社区 · 8 年前

现在我的JSON看起来很简单,只有一个对象。假设JSON中有许多对象,那么如何解析它。

现在我的JSON看起来像

    var jsonText = @"{
        ""some_number"": 108.541, 
        ""date_time"": ""2011-04-13T15:34:09Z"", 
        ""serial_number"": ""SN1234"",
        ""more_data"": {
            ""field1"": 1.0,
            ""field2"": ""hello""
        }
    }";

我是这样分析的

    var jss = new JavaScriptSerializer();
    var dict = jss.Deserialize<dynamic>(jsonText);

    Console.WriteLine(dict["some_number"]); 
    Console.WriteLine(dict["more_data"]["field2"]);
    Console.ReadLine();

当每个顺序的JSON数组中都有嵌套的对象时,我如何用 JavaScriptSerializer ?

假设我的JSON看起来有点像下面这样复杂,那么我如何解析它呢 Javascriptserializer .

    var jsonText = @"data[
    {
        ""some_number"": 108.541, 
        ""date_time"": ""2011-04-13T15:34:09Z"", 
        ""serial_number"": ""SN1234"",
        ""more_data"": {
            ""field1"": 1.0,
            ""field2"": ""hello""
        },
        ""Hobbies"" : [""game1"",""game2"",""game3""]
    },
    {
        ""some_number"": 200.541, 
        ""date_time"": ""2012-04-11T15:34:09Z"", 
        ""serial_number"": ""SN3333"",
        ""more_data"": {
            ""field1"": 2.0,
            ""field2"": ""hello1""
        },
        ""Hobbies"" : [""game4"",""game5"",""game6""]
    }]";

    var jss = new JavaScriptSerializer();
    var dict = jss.Deserialize<dynamic>(jsonText);

    Console.WriteLine(dict["some_number"]); 
    Console.WriteLine(dict["more_data"]["field2"]);
    Console.ReadLine();

请告诉我如何用javascriptserializer解析上面的JSON。谢谢

编辑

根据您的指导更改JSON。现在告诉我它看起来对吗?

    var jsonText = @"{ ""data""[
    {
        ""some_number"": 108.541, 
        ""date_time"": ""2011-04-13T15:34:09Z"", 
        ""serial_number"": ""SN1234"",
        ""more_data"": {
            ""field1"": 1.0,
            ""field2"": ""hello""
        },
        ""Hobbies"" : [""game1"",""game2"",""game3""]
    },
    {
        ""some_number"": 200.541, 
        ""date_time"": ""2012-04-11T15:34:09Z"", 
        ""serial_number"": ""SN3333"",
        ""more_data"": {
            ""field1"": 2.0,
            ""field2"": ""hello1""
        },
        ""Hobbies"" : [""game4"",""game5"",""game6""]
    }]}";

2 回复 | 直到 8 年前

itdoesntwork 8 年前

您的jContext不是有效的json。你可能需要这样的东西:

var jsonText = @"{
                  ""data"":[
                      {......

使JSON有效,JavaScriptSerializer将工作。

编辑

如果我猜对了,你的目标应该是

var jsonText = @"{""data"":[
{
    ""some_number"": 108.541, 
    ""date_time"": ""2011-04-13T15:34:09Z"", 
    ""serial_number"": ""SN1234"",
    ""more_data"": {
        ""field1"": 1.0,
        ""field2"": ""hello""
    },
    ""Hobbies"" : [""game1"",""game2"",""game3""]
},
{
    ""some_number"": 200.541, 
    ""date_time"": ""2012-04-11T15:34:09Z"", 
    ""serial_number"": ""SN3333"",
    ""more_data"": {
        ""field1"": 2.0,
        ""field2"": ""hello1""
    },
    ""Hobbies"" : [""game4"",""game5"",""game6""]
}]}";

现在你可以像…

dict["data"]  //which is your array containing your two objects
dict["data"][0]["some_number"] //eg first object property
dict["data"][1]["more_data"]["field2"] //second object property...

使用监视面板调试对象。您应该创建一个模型来将对象投射到模型中。除非需要,否则应避免使用未知模型。

Mist 8 年前

我把问题分类。@它没有网络提示真的帮助我修复JSON错误。在这里,我分享我的代码来展示我是如何做到的。

请添加参考 System.Web.Extensions 并使用添加此命名空间 System.Web.Script.Serialization; 在顶部

    var jsonText = @"{ ""data"":[
    {
        ""some_number"": 108.541, 
        ""date_time"": ""2011-04-13T15:34:09Z"", 
        ""serial_number"": ""SN1234"",
        ""more_data"": {
            ""field1"": 1.0,
            ""field2"": ""hello""
        },
        ""Hobbies"" : [""game1"",""game2"",""game3""]
    },
    {
        ""some_number"": 200.541, 
        ""date_time"": ""2012-04-11T15:34:09Z"", 
        ""serial_number"": ""SN3333"",
        ""more_data"": {
            ""field1"": 2.0,
            ""field2"": ""hello1""
        },
        ""Hobbies"" : [""game4"",""game5"",""game6""]
    }]}";

    var jss = new JavaScriptSerializer();
    var dict = jss.Deserialize<dynamic>(jsonText);

    Console.WriteLine(dict["data"][0]["some_number"]);
    Console.WriteLine(dict["data"][0]["more_data"]["field2"]);

    Console.WriteLine(dict["data"][1]["some_number"]);
    Console.WriteLine(dict["data"][1]["more_data"]["field2"]);
    Console.WriteLine(dict["data"][1]["Hobbies"][1]);

    Console.ReadLine();