我正在使用. net JSON解析器,并想序列化我的配置文件,使其可读。所以不要:

{"blah":"v", "blah2":"v2"}

我想要更好的东西,比如:

{
    "blah":"v", 
    "blah2":"v2"
}

我的代码是这样的:

using System.Web.Script.Serialization; 

var ser = new JavaScriptSerializer();
configSz = ser.Serialize(config);
using (var f = (TextWriter)File.CreateText(configFn))
{
    f.WriteLine(configSz);
    f.Close();
}

当前回答

. net 5内置了在System.Text.Json命名空间下处理JSON解析、序列化、反序列化的类。下面是一个序列化器的例子,它将。net对象转换为JSON字符串,

using System.Text.Json;
using System.Text.Json.Serialization;

private string ConvertJsonString(object obj)
{
    JsonSerializerOptions options = new JsonSerializerOptions();
    options.WriteIndented = true; //Pretty print using indent, white space, new line, etc.
    options.NumberHandling = JsonNumberHandling.AllowNamedFloatingPointLiterals; //Allow NANs
    string jsonString = JsonSerializer.Serialize(obj, options);
    return jsonString;
}

其他回答

onlineer使用Newtonsoft.Json.Linq:

string prettyJson = JToken.Parse(uglyJsonString).ToString(Formatting.Indented);

如果你有一个JSON字符串,想要“美化”它,但不想将它序列化到一个已知的c#类型,那么下面的方法就可以了(使用JSON.NET):

using System;
using System.IO;
using Newtonsoft.Json;

class JsonUtil
{
    public static string JsonPrettify(string json)
    {
        using (var stringReader = new StringReader(json))
        using (var stringWriter = new StringWriter())
        {
            var jsonReader = new JsonTextReader(stringReader);
            var jsonWriter = new JsonTextWriter(stringWriter) { Formatting = Formatting.Indented };
            jsonWriter.WriteToken(jsonReader);
            return stringWriter.ToString();
        }
    }
}

Json的简短示例代码。网络图书馆

private static string FormatJson(string json)
{
    dynamic parsedJson = JsonConvert.DeserializeObject(json);
    return JsonConvert.SerializeObject(parsedJson, Formatting.Indented);
}

我有一些非常简单的方法。你可以输入任何需要转换成json格式的对象:

private static string GetJson<T> (T json)
{
    return JsonConvert.SerializeObject(json, Formatting.Indented);
}

你可以使用以下标准方法来获取格式化的Json

JsonReaderWriterFactory。CreateJsonWriter(流流,编码编码,bool ownsStream, bool缩进,字符串indentChars)

只设置"缩进==true"

试试这样的方法

    public readonly DataContractJsonSerializerSettings Settings = 
            new DataContractJsonSerializerSettings
            { UseSimpleDictionaryFormat = true };

    public void Keep<TValue>(TValue item, string path)
    {
        try
        {
            using (var stream = File.Open(path, FileMode.Create))
            {
                //var currentCulture = Thread.CurrentThread.CurrentCulture;
                //Thread.CurrentThread.CurrentCulture = CultureInfo.InvariantCulture;

                try
                {
                    using (var writer = JsonReaderWriterFactory.CreateJsonWriter(
                        stream, Encoding.UTF8, true, true, "  "))
                    {
                        var serializer = new DataContractJsonSerializer(type, Settings);
                        serializer.WriteObject(writer, item);
                        writer.Flush();
                    }
                }
                catch (Exception exception)
                {
                    Debug.WriteLine(exception.ToString());
                }
                finally
                {
                    //Thread.CurrentThread.CurrentCulture = currentCulture;
                }
            }
        }
        catch (Exception exception)
        {
            Debug.WriteLine(exception.ToString());
        }
    }

注意线条

    var currentCulture = Thread.CurrentThread.CurrentCulture;
    Thread.CurrentThread.CurrentCulture = CultureInfo.InvariantCulture;
    ....
    Thread.CurrentThread.CurrentCulture = currentCulture;

对于某些类型的xml序列化器,您应该使用InvariantCulture来避免在具有不同区域设置的计算机上反序列化期间出现异常。例如,double或DateTime的无效格式有时会导致错误。

在反序列化

    public TValue Revive<TValue>(string path, params object[] constructorArgs)
    {
        try
        {
            using (var stream = File.OpenRead(path))
            {
                //var currentCulture = Thread.CurrentThread.CurrentCulture;
                //Thread.CurrentThread.CurrentCulture = CultureInfo.InvariantCulture;

                try
                {
                    var serializer = new DataContractJsonSerializer(type, Settings);
                    var item = (TValue) serializer.ReadObject(stream);
                    if (Equals(item, null)) throw new Exception();
                    return item;
                }
                catch (Exception exception)
                {
                    Debug.WriteLine(exception.ToString());
                    return (TValue) Activator.CreateInstance(type, constructorArgs);
                }
                finally
                {
                    //Thread.CurrentThread.CurrentCulture = currentCulture;
                }
            }
        }
        catch
        {
            return (TValue) Activator.CreateInstance(typeof (TValue), constructorArgs);
        }
    }

谢谢!