我只是使用XmlWriter创建了一些XML,以便在HTTP响应中发回。如何创建JSON字符串?我猜你会使用stringbuilder来构建JSON字符串,然后将响应格式化为JSON?


当前回答

看看http://www.codeplex.com/json/上的json-net。aspx项目。为什么要重新发明轮子?

其他回答

编码使用

JSON数组的简单对象EncodeJsObjectArray()

public class dummyObject
{
    public string fake { get; set; }
    public int id { get; set; }

    public dummyObject()
    {
        fake = "dummy";
        id = 5;
    }

    public override string ToString()
    {
        StringBuilder sb = new StringBuilder();
        sb.Append('[');
        sb.Append(id);
        sb.Append(',');
        sb.Append(JSONEncoders.EncodeJsString(fake));
        sb.Append(']');

        return sb.ToString();
    }
}

dummyObject[] dummys = new dummyObject[2];
dummys[0] = new dummyObject();
dummys[1] = new dummyObject();

dummys[0].fake = "mike";
dummys[0].id = 29;

string result = JSONEncoders.EncodeJsObjectArray(dummys);

论点: [29,“迈克”[5],“笨蛋”]

漂亮的用法

Pretty print JSON数组PrettyPrintJson()字符串扩展方法

string input = "[14,4,[14,\"data\"],[[5,\"10.186.122.15\"],[6,\"10.186.122.16\"]]]";
string result = input.PrettyPrintJson();

结果是:

[
   14,
   4,
   [
      14,
      "data"
   ],
   [
      [
         5,
         "10.186.122.15"
      ],
      [
         6,
         "10.186.122.16"
      ]
   ]
]

看看http://www.codeplex.com/json/上的json-net。aspx项目。为什么要重新发明轮子?

如果你需要复杂的结果(嵌入),创建你自己的结构:

class templateRequest
{
    public String[] registration_ids;
    public Data data;
    public class Data
    {
        public String message;
        public String tickerText;
        public String contentTitle;
        public Data(String message, String tickerText, string contentTitle)
        {
            this.message = message;
            this.tickerText = tickerText;
            this.contentTitle = contentTitle;
        }                
    };
}

然后通过调用获取JSON字符串

List<String> ids = new List<string>() { "id1", "id2" };
templateRequest request = new templeteRequest();
request.registration_ids = ids.ToArray();
request.data = new templateRequest.Data("Your message", "Your ticker", "Your content");

string json = new JavaScriptSerializer().Serialize(request);

结果是这样的:

json = "{\"registration_ids\":[\"id1\",\"id2\"],\"data\":{\"message\":\"Your message\",\"tickerText\":\"Your ticket\",\"contentTitle\":\"Your content\"}}"

希望能有所帮助!

这个库非常适合用于c#中的JSON

http://james.newtonking.com/pages/json-net.aspx

我发现您根本不需要序列化器。如果将对象作为List返回。 让我举个例子。

在asmx中,我们使用传递的变量获取数据

// return data
[WebMethod(CacheDuration = 180)]
public List<latlon> GetData(int id) 
{
    var data = from p in db.property 
               where p.id == id 
               select new latlon
               {
                   lat = p.lat,
                   lon = p.lon

               };
    return data.ToList();
}

public class latlon
{
    public string lat { get; set; }
    public string lon { get; set; }
}

然后使用jquery访问服务,传递该变量。

// get latlon
function getlatlon(propertyid) {
var mydata;

$.ajax({
    url: "getData.asmx/GetLatLon",
    type: "POST",
    data: "{'id': '" + propertyid + "'}",
    async: false,
    contentType: "application/json;",
    dataType: "json",
    success: function (data, textStatus, jqXHR) { //
        mydata = data;
    },
    error: function (xmlHttpRequest, textStatus, errorThrown) {
        console.log(xmlHttpRequest.responseText);
        console.log(textStatus);
        console.log(errorThrown);
    }
});
return mydata;
}

// call the function with your data
latlondata = getlatlon(id);

我们得到了答案。

{"d":[{"__type":"MapData+latlon","lat":"40.7031420","lon":"-80.6047970}]}