JSONConvert.Deserilize 對 JSON 對象值集合的擴展方法

Question

我有一種情況，我將 Json 值作為字符串列表。

List<string> values = new List<string>()
{
    "{\"Id\":\"SC\",\"Value\":8563}",
    "{\"Id\":\"SC\",\"Value\":8563}",
    "{\"Id\":\"SC\",\"Value\":8563}"
};

如何反序列化為對象列表：

public class ClassA
{
   public string Id {get; set;}
   public int Value {get;set;}
}

var objectValues = JsonConvert.DeserializeObject<IEnumerable<ClassA>>(values);

當我傳遞值列表並且它期望一個字符串作為參數時，我無法反序列化； 我可以創建擴展方法還是有更簡單的反序列化方法？

Answer 1

嘗試這個：

            List<ClassA> deserialized = new List<ClassA>();
            List<string> values = new List<string>()
            {   
                "{\"Id\":\"SC\",\"Value\":8563}",
                "{\"Id\":\"SC\",\"Value\":8563}",
                "{\"Id\":\"SC\",\"Value\":8563}"
            };
            foreach (var item in values)
            {
                var objectValue = JsonConvert.DeserializeObject<ClassA>(item);
                deserialized.Add(objectValue);
            }

作為擴展：

 public static List<ClassA> ToClassA(this List<string> stringList)
    {
        List<ClassA> deserialized = new List<ClassA>();
        foreach (var item in stringList)
        {
            var objectValue = JsonConvert.DeserializeObject<ClassA>(item);
            deserialized.Add(objectValue);
        }
        return deserialized;
    }

通用擴展

public static List<T> ToList<T>(this List<string> stringList) where T : class
        {
            List<T> deserialized = new List<T>();
            foreach (var item in stringList)
            {
                var objectValue = JsonConvert.DeserializeObject<T>(item);
                deserialized.Add(objectValue);
            }
            return deserialized;
        }

利用：

var result = values.ToList<ClassA>();

Answer 2

您可以通過構建新的Json將values列表轉換為Array string ，如下面的代碼：

var objectValues = JsonConvert.DeserializeObject<IEnumerable<ClassA>>($"[{string.Join(",", values)}]");

我希望這會幫助你。

Answer 3

public static PropertyBuilder<T> HasJsonConversion<T>(this PropertyBuilder<T> propertyBuilder,
    string columnType = null, string columnName = "", JsonSerializerSettings settings = null)
{
    var converter = new ValueConverter<T, string>(
        v => JsonConvert.SerializeObject(v, settings),
        v => JsonConvert.DeserializeObject<T>(v, settings));

    var comparer = new ValueComparer<T>(
        (l, r) => JsonConvert.SerializeObject(l, settings) == JsonConvert.SerializeObject(r, settings),
        v => v == null ? 0 : JsonConvert.SerializeObject(v, settings).GetHashCode(),
        v => JsonConvert.DeserializeObject<T>(JsonConvert.SerializeObject(v, settings), settings));

    propertyBuilder.HasConversion(converter);
    if (columnType != null) propertyBuilder.HasColumnType(columnType);

    if (columnName == "")
        propertyBuilder.HasColumnName($"Json_{propertyBuilder.Metadata.Name}");
    else if (columnName != null)
        propertyBuilder.HasColumnName(columnName);

    propertyBuilder.Metadata.SetValueConverter(converter);
    propertyBuilder.Metadata.SetValueComparer(comparer);

    return propertyBuilder;
}