blocks|key|484183|text|在第一种情况下，000没有坏处，值仍然是相同的，并且将被反序列化为完全相同的值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|484184|在第二种情况下，JavascriptSerializer不会对您有所帮助。JavacriptSerializer不应该更改数据，因为它将数据序列化为众所周知的格式，它不提供成员级别的数据转换(但它提供自定义对象转换器)。您需要的是转换%2B序列化，这是一个分两个阶段的任务。|484185|两个建议：|484186|1)使用DataContractJsonSerializer：添加另一个舍入值的属性：|484187|public+class+Money
{
++++public+string+Currency+{+get;+set;+}

++++[IgnoreDataMember]
++++public+decimal+Amount+{+get;+set;+}

++++[DataMember(Name+=+"Amount")]
++++public+decimal+RoundedAmount+{+get{+return+Math.Round(Amount,+2);+}+}
}|code-block|syntax|javascript|484188|2)克隆取值对象：|484189|public+class+Money+
{
++++public+string+Currency+{+get;+set;+}

++++public+decimal+Amount+{+get;+set;+}

++++public+Money+CloneRounding()+{
+++++++var+obj+=+(Money)this.MemberwiseClone();
+++++++obj.Amount+=+Math.Round(obj.Amount,+2);
+++++++return+obj;
++++}
}

var+roundMoney+=+money.CloneRounding();|484190|我猜json.net也不能做到这一点，但我不能百分之百确定。|484191|entityMap|0|LINK|mutability|MUTABLE|url|http://james.newtonking.com/projects/json-net.aspx^0|8|3|0|11|J|0|0|4|Q|0|0|0|0|2|8|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1A|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|L|3|M|5|N|7|1E|8|@]|D|@]|E|$O|P]]|$1|Q|3|R|5|6|7|1F|8|@]|D|@]|E|$]]|$1|S|3|T|5|N|7|1G|8|@]|D|@]|E|$O|P]]|$1|U|3|V|5|6|7|1H|8|@]|D|@$9|1I|A|1J|1|1K]]|E|$]]|$1|W|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|X|$Y|$5|Z|10|11|E|$12|13]]]]

In the first case the <code>000</code> does no harm, the value still is the same and will be deserialized to the exact same value. 

In the second case the JavascriptSerializer will not help you. The <code>JavacriptSerializer</code> is not supposed to change the data, since it serializes it to a well-known format it does not provide data conversion at member level (but it provides custom Object converters). What you want is a conversion + serialization, this is a two-phases task. 

Two suggestions:

1) Use <code>DataContractJsonSerializer</code>: add another property that rounds the value:

<pre><code>public class Money
{
 public string Currency { get; set; }

 [IgnoreDataMember]
 public decimal Amount { get; set; }

 [DataMember(Name = "Amount")]
 public decimal RoundedAmount { get{ return Math.Round(Amount, 2); } }
}
</code></pre>

2) Clone the object rounding the values:

<pre><code>public class Money 
{
 public string Currency { get; set; }

 public decimal Amount { get; set; }

 public Money CloneRounding() {
 var obj = (Money)this.MemberwiseClone();
 obj.Amount = Math.Round(obj.Amount, 2);
 return obj;
 }
}

var roundMoney = money.CloneRounding();
</code></pre>

I guess <a href="http://james.newtonking.com/projects/json-net.aspx" rel="nofollow">json.net</a> cannot do this either, but I'm not 100% sure.

blocks|key|484255|text|我刚刚经历了同样的麻烦，因为我有一些小数被序列化为1.00，另一些被序列化为1.0000。这是我的更改：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|484256|创建一个可以将值四舍五入为4个小数的JsonTextWriter。然后，每个小数将四舍五入为4个小数:+1.0变成1.0000，1.0000000变成1.0000|484257|private+class+JsonTextWriterOptimized+:+JsonTextWriter
{
++++public+JsonTextWriterOptimized(TextWriter+textWriter)
++++++++:+base(textWriter)
++++{
++++}
++++public+override+void+WriteValue(decimal+value)
++++{
++++++++//+we+really+really+really+want+the+value+to+be+serialized+as+"0.0000"+not+"0.00"+or+"0.0000"!
++++++++value+=+Math.Round(value,+4);
++++++++//+divide+first+to+force+the+appearance+of+4+decimals
++++++++value+=+Math.Round((((value%2B0.00001M)/10000)*10000)-0.00001M,+4);+
++++++++base.WriteValue(value);
++++}
}|code-block|syntax|javascript|484258|使用自己的编写器，而不是标准的编写器：|484259|var+jsonSerializer+=+Newtonsoft.Json.JsonSerializer.Create();
var+sb+=+new+StringBuilder(256);
var+sw+=+new+StringWriter(sb,+CultureInfo.InvariantCulture);
using+(var+jsonWriter+=+new+JsonTextWriterOptimized(sw))
{
++++jsonWriter.Formatting+=+Formatting.None;
++++jsonSerializer.Serialize(jsonWriter,+instance);
}|484260|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I just went through the same trouble as I had some decimals being serialized with 1.00 and some with 1.0000.
This is my change:

Create a JsonTextWriter that can round the value to 4 decimals. Every decimal will then be rounded to 4 decimals: 1.0 becomes 1.0000 and 1.0000000 becomes also 1.0000

<pre><code>private class JsonTextWriterOptimized : JsonTextWriter
{
 public JsonTextWriterOptimized(TextWriter textWriter)
 : base(textWriter)
 {
 }
 public override void WriteValue(decimal value)
 {
 // we really really really want the value to be serialized as "0.0000" not "0.00" or "0.0000"!
 value = Math.Round(value, 4);
 // divide first to force the appearance of 4 decimals
 value = Math.Round((((value+0.00001M)/10000)*10000)-0.00001M, 4); 
 base.WriteValue(value);
 }
}
</code></pre>

Use your own writer instead of the standard one:

<pre><code>var jsonSerializer = Newtonsoft.Json.JsonSerializer.Create();
var sb = new StringBuilder(256);
var sw = new StringWriter(sb, CultureInfo.InvariantCulture);
using (var jsonWriter = new JsonTextWriterOptimized(sw))
{
 jsonWriter.Formatting = Formatting.None;
 jsonSerializer.Serialize(jsonWriter, instance);
}
</code></pre>

blocks|key|267403|text|为了便于将来参考，这可以通过创建自定义JsonConverter在Json.net中很好地实现|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|267404|public+class+DecimalFormatJsonConverter+:+JsonConverter
{
++++private+readonly+int+_numberOfDecimals;

++++public+DecimalFormatJsonConverter(int+numberOfDecimals)
++++{
++++++++_numberOfDecimals+=+numberOfDecimals;
++++}

++++public+override+void+WriteJson(JsonWriter+writer,+object+value,+JsonSerializer+serializer)
++++{
++++++++var+d+=+(decimal)+value;
++++++++var+rounded+=+Math.Round(d,+_numberOfDecimals);
++++++++writer.WriteValue((decimal)rounded);
++++}

++++public+override+object+ReadJson(JsonReader+reader,+Type+objectType,+object+existingValue,
++++++++JsonSerializer+serializer)
++++{
++++++++throw+new+NotImplementedException("Unnecessary+because+CanRead+is+false.+The+type+will+skip+the+converter.");
++++}

++++public+override+bool+CanRead
++++{
++++++++get+{+return+false;+}
++++}

++++public+override+bool+CanConvert(Type+objectType)
++++{
++++++++return+objectType+==+typeof(decimal);
++++}
}|code-block|syntax|javascript|267405|如果您使用构造函数显式地在代码中创建序列化程序，这将很好地工作，但我认为用JsonConverterAttribute修饰相关属性会更好，在这种情况下，类必须有一个公共的、无参数的构造函数。我通过创建一个特定于我想要的格式的子类解决了这个问题。|267406|public+class+SomePropertyDecimalFormatConverter+:+DecimalFormatJsonConverter
{
++++public+SomePropertyDecimalFormatConverter()+:+base(3)
++++{
++++}
}

public+class+Poco+
{
++++[JsonConverter(typeof(SomePropertyDecimalFormatConverter))]
++++public+decimal+SomeProperty+{+get;set;+}
}|267407|自定义转换器是从Json.NET+documentation派生而来的。|267408|entityMap|0|LINK|mutability|MUTABLE|url|http://james.newtonking.com/json/help/html/T_Newtonsoft_Json_JsonConverter.htm|1|http://james.newtonking.com/json/help/html/T_Newtonsoft_Json_JsonConverterAttribute.htm|2|http://james.newtonking.com/json/help/index.html?topic=html/CustomJsonConverter.htm^0|J|D|J|D|0|0|0|11|M|11|M|1|0|0|8|M|2|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@$9|1C|A|1D|1|1E]]|E|$]]|$1|M|3|N|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1G|8|@]|D|@$9|1H|A|1I|1|1J]]|E|$]]|$1|Q|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]|10|$5|T|U|V|E|$W|11]]]]

For future reference, this can be achieved in Json.net pretty elegantly by creating a custom <a href="http://james.newtonking.com/json/help/html/T_Newtonsoft_Json_JsonConverter.htm"><code>JsonConverter</code></a>

<pre><code>public class DecimalFormatJsonConverter : JsonConverter
{
 private readonly int _numberOfDecimals;

 public DecimalFormatJsonConverter(int numberOfDecimals)
 {
 _numberOfDecimals = numberOfDecimals;
 }

 public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
 {
 var d = (decimal) value;
 var rounded = Math.Round(d, _numberOfDecimals);
 writer.WriteValue((decimal)rounded);
 }

 public override object ReadJson(JsonReader reader, Type objectType, object existingValue,
 JsonSerializer serializer)
 {
 throw new NotImplementedException("Unnecessary because CanRead is false. The type will skip the converter.");
 }

 public override bool CanRead
 {
 get { return false; }
 }

 public override bool CanConvert(Type objectType)
 {
 return objectType == typeof(decimal);
 }
}
</code></pre>

If you're creating serializers in code using constructor explicitly, this will work fine but I think it's nicer to decorate the relevant properties with <a href="http://james.newtonking.com/json/help/html/T_Newtonsoft_Json_JsonConverterAttribute.htm"><code>JsonConverterAttribute</code></a>, in which case the class must have a public, parameterless constructor. I solved this by creating a subclass which is specific to the format I want.

<pre><code>public class SomePropertyDecimalFormatConverter : DecimalFormatJsonConverter
{
 public SomePropertyDecimalFormatConverter() : base(3)
 {
 }
}

public class Poco 
{
 [JsonConverter(typeof(SomePropertyDecimalFormatConverter))]
 public decimal SomeProperty { get;set; }
}
</code></pre>

The custom converter has been derived from <a href="http://james.newtonking.com/json/help/index.html?topic=html/CustomJsonConverter.htm">Json.NET documentation</a>.

blocks|key|484380|text|到目前为止，我对实现这一目标的所有技术并不完全满意。JsonConverterAttribute似乎是最有前途的，但我不能忍受硬编码的参数和各种选项组合的大量转换器类。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|484381|因此，我提交了一个PR，它添加了向JsonConverter和JsonProperty传递各种参数的功能。它已经被上游接受了，我希望在下一个版本中(6.0.5之后的下一个版本)|offset|length|484382|然后你可以这样做：|484383|public+class+Measurements
{
++++[JsonProperty(ItemConverterType+=+typeof(RoundingJsonConverter))]
++++public+List<double>+Positions+{+get;+set;+}

++++[JsonProperty(ItemConverterType+=+typeof(RoundingJsonConverter),+ItemConverterParameters+=+new+object[]+{+0,+MidpointRounding.ToEven+})]
++++public+List<double>+Loads+{+get;+set;+}

++++[JsonConverter(typeof(RoundingJsonConverter),+4)]
++++public+double+Gain+{+get;+set;+}
}|code-block|syntax|javascript|484384|有关示例，请参阅CustomDoubleRounding()测试。|484385|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/JamesNK/Newtonsoft.Json/pull/377|1|https://github.com/JamesNK/Newtonsoft.Json/blob/12c33e70d25370f68e02f78317a4e71b47e3ae49/Src/Newtonsoft.Json.Tests/JsonConvertTest.cs#L1137^0|0|9|2|0|0|0|0|8|M|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@$D|10|E|11|1|12]]|A|$]]|$1|F|3|G|5|6|7|13|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|14|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|15|8|@]|9|@$D|16|E|17|1|18]]|A|$]]|$1|O|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|X]]]]

I wasn't completely satisfied with all of the techniques thus far to achieve this. JsonConverterAttribute seemed the most promising, but I couldn't live with hard-coded parameters and proliferation of converter classes for every combination of options. 

So, I submitted a <a href="https://github.com/JamesNK/Newtonsoft.Json/pull/377">PR</a> that adds the ability to pass various arguments to JsonConverter and JsonProperty. It's been accepted upstream and I expect will be in the next release (whatever's next after 6.0.5)

You can then do it like this:

<pre><code>public class Measurements
{
 [JsonProperty(ItemConverterType = typeof(RoundingJsonConverter))]
 public List&lt;double&gt; Positions { get; set; }

 [JsonProperty(ItemConverterType = typeof(RoundingJsonConverter), ItemConverterParameters = new object[] { 0, MidpointRounding.ToEven })]
 public List&lt;double&gt; Loads { get; set; }

 [JsonConverter(typeof(RoundingJsonConverter), 4)]
 public double Gain { get; set; }
}
</code></pre>

Refer to the <a href="https://github.com/JamesNK/Newtonsoft.Json/blob/12c33e70d25370f68e02f78317a4e71b47e3ae49/Src/Newtonsoft.Json.Tests/JsonConvertTest.cs#L1137">CustomDoubleRounding()</a> test for an example.

I have a class

<pre><code>public class Money
{
 public string Currency { get; set; }
 public decimal Amount { get; set; }
}
</code></pre>

and would like to serialize it to JSON. If I use the <code>JavaScriptSerializer</code> I get

<pre><code>{"Currency":"USD","Amount":100.31000}
</code></pre>

Because of the API I have to conform to needs JSON amounts with maximum two decimal places, I feel it should be possible to somehow alter the way the <code>JavaScriptSerializer</code> serializes a decimal field, but I can't find out how. There is the <a href="http://msdn.microsoft.com/en-us/library/system.web.script.serialization.simpletyperesolver.aspx">SimpleTypeResolver</a> you can pass in the constructor, but it only work on types as far as I can understand. The <a href="http://msdn.microsoft.com/en-us/library/system.web.script.serialization.javascriptconverter.aspx">JavaScriptConverter</a>, which you can add through RegisterConverters(...) seems to be made for <code>Dictionary</code>.

I would like to get

<pre><code>{"Currency":"USD","Amount":100.31}
</code></pre>

after I serialize. Also, changing to double is out of the question. And I probably need to do some rounding (100.311 should become 100.31). 

Does anyone know how to do this? Is there perhaps an alternative to the <code>JavaScriptSerializer</code> that lets you control the serializing in more detail?

Serializing a decimal to JSON, how to round off?

数据转换

JSON

我有一节课public class Money{    public string Currency { get; set; }    public decimal Amount { get; set; }}并希望将其序列化为JSON。如果我使用JavaScriptSerializer，我会得到{"Currency":...

问将小数序列化为JSON，如何四舍五入？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问将小数序列化为JSON，如何四舍五入？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问将小数序列化为JSON，如何四舍五入？
EN