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

onlineer使用Newtonsoft.Json.Linq:

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

其他回答

下面是一个使用微软System.Text.Json库的解决方案:

static string FormatJsonText(string jsonString)
{
    using var doc = JsonDocument.Parse(
        jsonString,
        new JsonDocumentOptions
        {
            AllowTrailingCommas = true
        }
    );
    MemoryStream memoryStream = new MemoryStream();
    using (
        var utf8JsonWriter = new Utf8JsonWriter(
            memoryStream,
            new JsonWriterOptions
            {
                Indented = true
            }
        )
    )
    {
        doc.WriteTo(utf8JsonWriter);
    }
    return new System.Text.UTF8Encoding()
        .GetString(memoryStream.ToArray());
}

首先我想在Duncan Smart的帖子下添加评论,但不幸的是,我还没有足够的声誉来留下评论。我在这里试试。

我只是想警告一下副作用。

JsonTextReader在内部将json解析为类型化的jtoken,然后将它们序列化回来。

例如,如果您的原始JSON是

{ "double":0.00002, "date":"\/Date(1198908717056)\/"}

美化之后你得到

{ 
    "double":2E-05,
    "date": "2007-12-29T06:11:57.056Z"
}

当然,这两个json字符串是等价的,并且将反序列化为结构上相等的对象,但如果您需要保留原始字符串值,则需要考虑这一点

使用System.Text.Json设置JsonSerializerOptions。writeindent = true:

JsonSerializerOptions options = new JsonSerializerOptions { WriteIndented = true };
string json = JsonSerializer.Serialize<Type>(object, options);

Net Core App

var js = JsonSerializer.Serialize(obj, new JsonSerializerOptions {
             WriteIndented = true
         });

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!