Newtonsoft.Json 无法从 JSON 转换为 ICollection<Object>

Question

嗨，我有以下JSON，希望对ICollection<Location>

进行挑选

" u0001 [{" id "：1，" name "：" a-01 "，" address "：" aa "， " children "：[{" id "：2，" name "：" a-01-01 "，" address "：" bb "， "儿童"：[{" id "：5，" name "：" A-01-02-01 "，" address "： " cc "}]}，{" id "：3，" name "：" A-01-02 "，" address "：" dd "}]}，}， {" id "：4，" name "：" b-01 "，" address "：" ee "，" children "： [{{" id "：6，" name "：" b-01-01 "，" address "：" ff "}]}]}]"

可读的JSON

[
  {
    "Id": 1,
    "Name": "A-01",
    "Address": "aa",
    "Children": [
      {
        "Id": 2,
        "Name": "A-01-01",
        "Address": "bb",
        "Children": [
          {
            "Id": 5,
            "Name": "A-01-02-01",
            "Address": "cc"
          }
        ]
      },
      {
        "Id": 3,
        "Name": "A-01-02",
        "Address": "dd"
      }
    ]
  },
  {
    "Id": 4,
    "Name": "B-01",
    "Address": "ee",
    "Children": [
      {
        "Id": 6,
        "Name": "B-01-01",
        "Address": "ff"
      }
    ]
  }
]

位置对象

public class Location
{
    [JsonProperty("Id")]
    public int Id { get; set; }
    [JsonProperty("Name")]
    [Required]
    public string Name { get; set; }
    [JsonProperty("Address")]
    public string Address { get; set; }
    public ICollection<Asset> Assets { get; set; }
    public Location ParentLocation { get; set; }
    [JsonProperty("Children")]
    public virtual ICollection<Location> ChildrenLocation { get; set; }
}

我尝试测试清理字符串，但没有希望。

    var jsonString = @"[{""Id"":1,""Name"":""A-01"",""Address"":""aa"",""Children"":[{""Id"":2,""Name"":""A-01-01"",""Address"":""bb",""Children"":[{""Id"":5,""Name"":""A-01-02-01"",""Address"":""cc""}]},{""Id"":3,""Name"":""A-01-02"",""Address"":""dd""}]},{""Id"":4,""Name"":""B-01"",""Address"":""ee"",""Children"":[{""Id"":6,""Name"":""B-01-01"",""Address"":""ff""}]}]";
    var locations = JsonConvert.DeserializeObject<ICollection<Location>>(jsonString.Trim());

错误

Newtonsoft.Json.JsonReaderException: Unexpected character encountered while parsing value: . Path '', line 0, position 0.
   at Newtonsoft.Json.JsonTextReader.ParseValue()
   at Newtonsoft.Json.JsonTextReader.Read()
   at Newtonsoft.Json.JsonReader.ReadForType(JsonContract contract, Boolean hasConverter)
   at Newtonsoft.Json.Serialization.JsonSerializerInternalReader.Deserialize(JsonReader reader, Type objectType, Boolean checkAdditionalContent)
   at Newtonsoft.Json.JsonSerializer.DeserializeInternal(JsonReader reader, Type objectType)
   at Newtonsoft.Json.JsonConvert.DeserializeObject(String value, Type type, JsonSerializerSettings settings)
   at Newtonsoft.Json.JsonConvert.DeserializeObject[T](String value, JsonSerializerSettings settings)
   at Newtonsoft.Json.JsonConvert.DeserializeObject[T](String value)

Answer 1

错误很明显，字符串不是有效的JSON，问题是第一个字符u0001。

"可读" JSON字符串和内饰测试不包含该字符，因此他们甚至无法重现问题。

trim（）也无法独自工作。它的工作是删除空格，但u0001是SOH（起始人）角色。那是不是被认为是空格。

为了实际解决该问题，应修复产生此特征的代码，以便它不会发出。如果JSON行来自使用SOH用作记录分配器的流json文件，则应将其修复，以便它在JSON字符串中包含记录定界符。

如果不可能，则应使用String.trim（char []）超载删除SHO字符：

var locations = JsonConvert.DeserializeObject(jsonString.Trim('u0001'));

一旦删除了SOH字符，避免序列化就可以了：

var jsonString = "u0001[{"Id": 1, "Name": "A-01", "Address": "aa", "Children": [{"Id": 2, "Name": "A-01-01", "Address": "bb", "Children": [{"Id": 5, "Name": "A-01-02-01", "Address": "cc"}]}, {"Id": 3, "Name": "A-01-02", "Address": "dd"}]}, {"Id": 4, "Name": "B-01", "Address": "ee", "Children": [{"Id": 6, "Name": "B-01-01", "Address": "ff"}]}]";
var cleaned=jsonString.Trim('u0001');
var locations = (JArray)JsonConvert.DeserializeObject(cleaned);
Debug.Assert(locations.Count==2);
var address=(JValue)locations[1]["Children"][0]["Address"];
Debug.Assert(String.Equals(address.Value,"ff"));

Answer 2

dai所说，有" u0001"字符（请参阅描述）。我认为您从Somwehere复制它，而不需要它，因此您可以将其删除：

  var str =
                "u0001[{"Id": 1, "Name": "A-01", "Address": "aa", "Children": [{"Id": 2, "Name": "A-01-01", "Address": "bb", "Children": [{"Id": 5, "Name": "A-01-02-01", "Address": "cc"}]}, {"Id": 3, "Name": "A-01-02", "Address": "dd"}]}, {"Id": 4, "Name": "B-01", "Address": "ee", "Children": [{"Id": 6, "Name": "B-01-01", "Address": "ff"}]}]";
            var unescapeStr = str.Replace("u0001", String.Empty);
            var result = JsonConvert.DeserializeObject<ICollection<Location>>(unescapeStr);

字符" u0001"根本不应该在您的字符串中，不仅在开始时，因此您必须使用替换not trim

将其从所有字符串中删除

测试代码中的第二个问题。当您复制一些Scpecial字符的Visual Studio问您时，您是否想逃脱它，如果您说"是"，您可以逃脱两次字符

因此，尝试使用上面的示例测试字符串