从 JSON 反序列化 java 枚举
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Deserialize java enum from JSON
提问by chetan
We use Hymanson 1.9.1 to serialize and deserialize JSON request response strings to/from Java objects. Primitive Java types, collection types, and custom objects are (de)serialized without issues. However, I have a problem trying to deserialize JSON string into java enum. JSON string is serialized like so:
我们使用 Hymanson 1.9.1 来序列化和反序列化 Java 对象的 JSON 请求响应字符串。原始 Java 类型、集合类型和自定义对象被（反）序列化没有问题。但是，我在尝试将 JSON 字符串反序列化为 java 枚举时遇到问题。JSON 字符串序列化如下：
"wt":{"wt":100.5,"unit":{"LBS":3}}
Java type for wt is like so:
wt 的 Java 类型如下所示：
public class Weight {
protected double weight;
protected Unit unit;
}
I referred to this, this, and thison SO and came up with enum for weight units like so:
我在 SO 上提到了this、this和this，并提出了重量单位的枚举，如下所示：
public enum Unit {
KG("kg"),
GM("gm"),
LBS("lbs"),
OZ("oz");
private String value;
private WeightMeasurementUnit(String value) { this.value = value; }
@JsonValue
public String getValue() { return this.value; }
@JsonCreator
public static Unit create(String val) {
Unit[] units = Unit.values();
for (Unit unit : units) {
if (unit.getValue().equals(val)) {
return unit;
}
}
return LBS;
}
}
The problem is, when ever I try to deserialize above mentioned JSON I get this error saying: "Unrecognized field "LBS" (Class a.b.c.d.Weight), not marked as ignorable" Exception stacktrace is like so:
问题是，当我尝试反序列化上述 JSON 时，我收到此错误消息：“无法识别的字段“LBS”（类 abcdWeight），未标记为可忽略的异常堆栈跟踪如下所示：
Caused by: org.codehaus.Hymanson.map.exc.UnrecognizedPropertyException: Unrecognized field "LBS" (Class a.b.c.d.Weight), not marked as ignorable
at [Source: java.io.ByteArrayInputStream@20172017; line: 1, column: 464] (through reference chain: a.b.c.d.MyRequest["blah"]->a.b.c.d.AnotherType["wt"]->a.b.c.d.Weight["LBS"])
at org.codehaus.Hymanson.map.exc.UnrecognizedPropertyException.from(UnrecognizedPropertyException.java:53)
at org.codehaus.Hymanson.map.deser.StdDeserializationContext.unknownFieldException(StdDeserializationContext.java:267)
at org.codehaus.Hymanson.map.deser.std.StdDeserializer.reportUnknownProperty(StdDeserializer.java:673)
at org.codehaus.Hymanson.map.deser.std.StdDeserializer.handleUnknownProperty(StdDeserializer.java:659)
at org.codehaus.Hymanson.map.deser.BeanDeserializer.handleUnknownProperty(BeanDeserializer.java:1365)
...
...
My questions are: Is the serialized JSON string for enum seem correct ? What else should I include (or annotate) for the enum to be properly deserialized ?
我的问题是： enum 的序列化 JSON 字符串是否正确？为了使枚举正确反序列化，我还应该包括（或注释）什么？
采纳答案by Hari Menon
I am assuming that in the public enum Unitcode block, you mean Unitinstead of WeightMeasurementUnit.
我假设在public enum Unit代码块中，您的意思是Unit而不是WeightMeasurementUnit.
The Weightclass has only a weightand a unit, so if you pass {"wt":100.5,"unit":"lbs"}, it should work, because a unitis just a unit without value. So there is no way for the deserializer to parse {"LBS":3}as a Unit. What is the 3for?
这个Weight类只有 aweight和 a unit，所以如果你通过了{"wt":100.5,"unit":"lbs"}，它应该可以工作，因为 aunit只是一个没有价值的单位。因此，反序列化器无法解析{"LBS":3}为Unit. 是3为了什么？
Another problem is that your value is "lbs" whereas you are passing "LBS". So either you need to standardise or you need to use unit.getValue().equalsIgnoreCase(val)
另一个问题是您的值是“lbs”，而您传递的是“LBS”。所以要么你需要标准化，要么你需要使用unit.getValue().equalsIgnoreCase(val)
回答by luckyhandler
I would suggest you update your Hymanson version to 2.7.0-rc2 (and probably also before) and then configure the ObjectMapper as follows:
我建议您将 Hymanson 版本更新到 2.7.0-rc2（可能也是之前），然后按如下方式配置 ObjectMapper：
private ObjectMapper createObjectMapper() {
final ObjectMapper mapper = new ObjectMapper();
// enable toString method of enums to return the value to be mapped
mapper.enable(SerializationFeature.WRITE_ENUMS_USING_TO_STRING);
mapper.enable(DeserializationFeature.READ_ENUMS_USING_TO_STRING);
return mapper;
}
In your enum you just have to override the toString() method:
在您的枚举中，您只需覆盖 toString() 方法：
public enum Unit {
KG,
GM,
LBS,
OZ;
// UPDATE: implicitly already the default so override not needed in this case
@Override
public String toString() {
return name();
}
}
You don't need any annotations or custom deserializers. This would be the way to map a simple enum to a json and vice-versa.
您不需要任何注释或自定义反序列化器。这将是将简单枚举映射到 json 的方法，反之亦然。
If your enum should be mapped from a special string you have to add a value field and a Constructor which assigns this field and return the value in the toString method.
如果您的枚举应该从特殊字符串映射，您必须添加一个值字段和一个构造函数，该构造函数分配此字段并在 toString 方法中返回值。
public enum Unit {
KG("kilogram"),
GM("gram"),
LBS("blah"),
OZ("anything");
Unit(final String value) {
this.value = value;
}
@Override
public String toString() {
return value;
}
}
