無法在C＃中將Json字符串轉換為Json對象

Question

我有一個Json字符串，它是receiveCount({\\"url\\":\\"http://www.google.com\\",\\"count\\":75108})

我完整的方法是

public void GetPinCount(string url)
        {
            string QUrl = "https://api.pinterest.com/v1/urls/count.json?callback=receiveCount&url=" + url;
            System.Net.HttpWebRequest Request = (System.Net.HttpWebRequest)System.Net.WebRequest.Create(QUrl);
            Request.ContentType = "text/json";
            Request.Timeout = 10000;
            Request.Method = "GET";
            string content;
            using (WebResponse myResponse = Request.GetResponse())
            {
                using (System.IO.StreamReader sr = new System.IO.StreamReader(myResponse.GetResponseStream(), System.Text.Encoding.UTF8))
                {
                    content = sr.ReadToEnd();
                }
            };
           var json = JObject.Parse(content);
           var share_count = json["receiveCount"]["count"].ToString();
           Console.WriteLine("Share Count :" + share_count);
        }

當我嘗試訪問計數時，出現了類似的異常

Unexpected character encountered while parsing value: r. Path '', line 0, position 0.

請告訴我該怎么做。

Answer 1

您的字符串不是有效的JSON：

receiveCount(`{\"url\":\"http://www.google.com\",\"count\":75108}`)

有效的JSON部分是參數：

{"url":"http://www.google.com","count":75108}

您必須從字符串中提取有效的JSON部分以反序列化。

Answer 2

您正在調用錯誤的屬性。

你應該用

var share_count = json["count"].ToString();

代替，

var share_count = json["receiveCount"]["count"].ToString();

用於響應的代碼：

var json = JObject.Parse (content);
var share_url = json["url"];
Console.WriteLine ("Share URL:" + share_url);
var share_count = json["count"];
Console.WriteLine ("Share Count:" + share_count);

編輯1：

// Gets only the JSON string we need
content = content.Replace ("receiveCount(", "");
content = content.Remove (content.Length - 1);

// Converts JSON string to Object
var json = JObject.Parse (content);
var share_url = json["url"];
Console.WriteLine ("Share URL:" + share_url);
var share_count = json["count"];
Console.WriteLine ("Share Count:" + share_count);