我想以这种格式将数据保存在本地json文件中:
[
{"Name":"sdafdsf","Password":"dsfads","FirstName":"fsdf","LastName":"dsfdafas"},
{"Name":"sddafdsf","Password":"dsfadds","FirstName":"fdsdf","LastName":"dsfdafasdfs"}
]
我在控制器中使用它来序列化为json格式:
public ActionResult Index(demo obj)
{
String json = Newtonsoft.Json.JsonConvert.SerializeObject(obj);
string path = Server.MapPath("~/app/");
//// Write that JSON to txt file,
//var read = System.IO.File.ReadAllText(path + "output.json");
System.IO.File.WriteAllText(path + "output.json", json);
return View();
}
这是我的模特:
public class demo
{
public string Name { get; set; }
public string Password { get; set; }
public string FirstName { get; set; }
public string LastName { get; set; }
}
但是我没有使用正确的JSON格式,而是在我的output.json文件中找到它:
{"Name":"sdafdsf","Password":"dsfads","FirstName":"fsdf","LastName":"dsfdafas"}{"Name":"adfsdsfsafdsafasdfsdfsadf","Password":"dfsaasdsa","FirstName":"safd","LastName":"dfsafads"}
如何以正确的格式保存数据?
答案 0 :(得分:2)
这是正确的格式,如果你的意思是你需要它像一个数组然后将对象添加到一个数组,然后序列化它:
public ActionResult Index(demo obj)
{
var array = new [] {obj};
String json = Newtonsoft.Json.JsonConvert.SerializeObject(array);
string path = Server.MapPath("~/app/");
//// Write that JSON to txt file,
//var read = System.IO.File.ReadAllText(path + "output.json");
System.IO.File.WriteAllText(path + "output.json", json);
return View();
}
或者:
var list = new List<demo>();
list.Add(obj);
String json = Newtonsoft.Json.JsonConvert.SerializeObject(list);
如果您想将数据始终保存在数组中,那么您始终需要:
public ActionResult Index(demo obj)
{
string path = Server.MapPath("~/app/");
var read = System.IO.File.ReadAllText(path + "output.json");
List<demo> lst = Newtonsoft.Json.JsonConvert.DeserializeObject<List<demo>>(read);
if (lst == null)
{
List<demo> _data = new List<demo>();
_data.Add(obj);
String json = Newtonsoft.Json.JsonConvert.SerializeObject(_data.ToArray());
System.IO.File.WriteAllText(path + "output.json", json);
}
else
{
lst.Add(obj);
String json = Newtonsoft.Json.JsonConvert.SerializeObject(lst);
System.IO.File.WriteAllText(path + "output.json", json);
}
return View();
}
当然,您可以通过单独的部分编写更清晰的代码。
答案 1 :(得分:0)
首先创建一个对象列表,然后尝试序列化列表。我相信你会得到所需的输出。好像你通过序列化单个对象使用了AppendAllText方法。