C# 如何将此json字符串转换为jobject
如何将此json字符串转换为JobjectC# 如何将此json字符串转换为jobject,c#,C#,如何将此json字符串转换为Jobject "{\"user\":{\"id\":373697,\"username\":\"luckygirlxxx123\",\"counter_rechecking\":0,\"user_id\":76131,\"fb_id\":\"100047460285611\"}}"; JObject concac = new JObject(); concac1["id"] = 373697; concac1["us
"{\"user\":{\"id\":373697,\"username\":\"luckygirlxxx123\",\"counter_rechecking\":0,\"user_id\":76131,\"fb_id\":\"100047460285611\"}}";
JObject concac = new JObject();
concac1["id"] = 373697;
concac1["username"] = "luckygirlxxx123";
concac1["counter_rechecking"] = 0;
concac1["user_id"] = 76131;
concac1["fb_id"] = "100047460285611";
我需要它像这样转换成Jobject,因为我需要它将数据发布到网站api示例:
"{\"user\":{\"id\":373697,\"username\":\"luckygirlxxx123\",\"counter_rechecking\":0,\"user_id\":76131,\"fb_id\":\"100047460285611\"}}";
JObject concac = new JObject();
concac1["id"] = 373697;
concac1["username"] = "luckygirlxxx123";
concac1["counter_rechecking"] = 0;
concac1["user_id"] = 76131;
concac1["fb_id"] = "100047460285611";
因为在用户中,我不知道如何用它创建jobject,这段代码可能会对您有所帮助
publicstaticvoidmain(字符串[]args)
{
字符串jsonData=“{\'user\':{\'id\':373697,\'username\':\'luckygirlxx123\',\'counter\'u revecking\':0,\'user\'u id\':76131,\'fb\'u id\':\'100047460285611\'}”;
var details=JObject.Parse(jsonData);
Console.WriteLine(string.Concat(详细信息[“用户”][“id”],“+详细信息[“用户”][“用户名]));
}
此代码可能会对您有所帮助
publicstaticvoidmain(字符串[]args)
{
字符串jsonData=“{\'user\':{\'id\':373697,\'username\':\'luckygirlxx123\',\'counter\'u revecking\':0,\'user\'u id\':76131,\'fb\'u id\':\'100047460285611\'}”;
var details=JObject.Parse(jsonData);
Console.WriteLine(string.Concat(详细信息[“用户”][“id”],“+详细信息[“用户”][“用户名]));
}
最简单的方法是将这些JSON值赋予对象的约定
public JObject(int id, string username, int user_id ....)
然后声明对象并向其传递值
JObject concac = new JObject(concac1["id"], concac1["username"], concac1["user_id"]..... );
第二种方法是最简单的方法是将这些JSON值赋予对象的约定
public JObject(int id, string username, int user_id ....)
然后声明对象并向其传递值
JObject concac = new JObject(concac1["id"], concac1["username"], concac1["user_id"]..... );
第二种方法是问题是alll ID、用户名等在用户中问题是alll ID、用户名等在用户中