Java 使用 Jackson 反序列化枚举
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Deserializing an enum with Hymanson
提问by jwilner
I'm trying and failing to deserialize an enum with Hymanson 2.5.4, and I don't quite see my case out there. My input strings are camel case, and I want to simply map to standard Enum conventions.
我正在尝试使用 Hymanson 2.5.4 反序列化枚举，但未能成功，而且我不太清楚我的情况。我的输入字符串是驼峰式大小写，我想简单地映射到标准 Enum 约定。
@JsonFormat(shape = JsonFormat.Shape.STRING)
public enum Status {
READY("ready"),
NOT_READY("notReady"),
NOT_READY_AT_ALL("notReadyAtAll");
private static Map<String, Status> FORMAT_MAP = Stream
.of(Status.values())
.collect(toMap(s -> s.formatted, Function.<Status>identity()));
private final String formatted;
Status(String formatted) {
this.formatted = formatted;
}
@JsonCreator
public Status fromString(String string) {
Status status = FORMAT_MAP.get(string);
if (status == null) {
throw new IllegalArgumentException(string + " has no corresponding value");
}
return status;
}
}
I've also tried @JsonValueon a getter to no avail, which was an option I saw reported elsewhere. They all blow up with:
我也试过@JsonValue一个吸气剂但无济于事，这是我在别处看到的一个选项。他们都炸了：
com.fasterxml.Hymanson.databind.exc.InvalidFormatException: Can not construct instance of ...Status from String value 'ready': value not one of declared Enum instance names: ...
What am I doing wrong?
我究竟做错了什么？
采纳答案by Federico Peralta Schaffner
EDIT:Starting from Hymanson 2.6, you can use @JsonPropertyon each element of the enum to specify its serialization/deserialization value (see here):
编辑：从 Hymanson 2.6 开始，您可以@JsonProperty在枚举的每个元素上使用来指定其序列化/反序列化值（请参见此处）：
public enum Status {
@JsonProperty("ready")
READY,
@JsonProperty("notReady")
NOT_READY,
@JsonProperty("notReadyAtAll")
NOT_READY_AT_ALL;
}
(The rest of this answer is still valid for older versions of Hymanson)
（这个答案的其余部分仍然适用于旧版本的Hyman逊）
You should use @JsonCreatorto annotate a static method that receives a Stringargument. That's what Hymanson calls a factory method:
您应该使用@JsonCreator注释接收String参数的静态方法。这就是Hyman逊所说的工厂方法：
public enum Status {
READY("ready"),
NOT_READY("notReady"),
NOT_READY_AT_ALL("notReadyAtAll");
private static Map<String, Status> FORMAT_MAP = Stream
.of(Status.values())
.collect(Collectors.toMap(s -> s.formatted, Function.identity()));
private final String formatted;
Status(String formatted) {
this.formatted = formatted;
}
@JsonCreator // This is the factory method and must be static
public static Status fromString(String string) {
return Optional
.ofNullable(FORMAT_MAP.get(string))
.orElseThrow(() -> new IllegalArgumentException(string));
}
}
This is the test:
这是测试：
ObjectMapper mapper = new ObjectMapper();
Status s1 = mapper.readValue("\"ready\"", Status.class);
Status s2 = mapper.readValue("\"notReadyAtAll\"", Status.class);
System.out.println(s1); // READY
System.out.println(s2); // NOT_READY_AT_ALL
As the factory method expects a String, you have to use JSON valid syntax for strings, which is to have the value quoted.
由于工厂方法需要 a String，因此您必须对字符串使用 JSON 有效语法，即引用值。
回答by Konstantin Kulagin
This is probably a faster way to do it:
这可能是一种更快的方法：
public enum Status {
READY("ready"),
NOT_READY("notReady"),
NOT_READY_AT_ALL("notReadyAtAll");
private final String formatted;
Status(String formatted) {
this.formatted = formatted;
}
@Override
public String toString() {
return formatted;
}
}
public static void main(String[] args) throws IOException {
ObjectMapper mapper = new ObjectMapper();
ObjectReader reader = mapper.reader(Status.class);
Status status = reader.with(DeserializationFeature.READ_ENUMS_USING_TO_STRING).readValue("\"notReady\"");
System.out.println(status.name()); // NOT_READY
}
回答by Stefan
The solutions on this page work only for single field and @JsonFormat(shape = JsonFormat.Shape.NATURAL) (default format)
此页面上的解决方案仅适用于单个字段和 @JsonFormat(shape = JsonFormat.Shape.NATURAL)（默认格式）
this works for multiple fields and @JsonFormat(shape = JsonFormat.Shape.OBJECT)
这适用于多个字段和 @JsonFormat(shape = JsonFormat.Shape.OBJECT)
@JsonFormat(shape = JsonFormat.Shape.OBJECT)
public enum PinOperationMode {
INPUT("Input", "I"),
OUTPUT("Output", "O")
;
private final String mode;
private final String code;
PinOperationMode(String mode, String code) {
this.mode = mode;
this.code = code;
}
public String getMode() {
return mode;
}
public String getCode() {
return code;
}
@JsonCreator
static PinOperationMode findValue(@JsonProperty("mode") String mode, @JsonProperty("code") String code) {
return Arrays.stream(PinOperationMode.values()).filter(pt -> pt.mode.equals(mode) && pt.code.equals(code)).findFirst().get();
}
}
回答by Sileria
For whoever is searching for enums with integer json properties. Here is what worked for me:
对于正在搜索具有整数 json 属性的枚举的人。这是对我有用的：
enum class Status (private val code: Int) {
PAST(0),
LIVE(2),
UPCOMING(1);
companion object {
private val codes = Status.values().associateBy(Status::code)
@JvmStatic @JsonCreator fun from (value: Int) = codes[value]
}
}
回答by Arun Kumar
@JsonCreator
public static Status forValue(String name)
{
return EnumUtil.getEnumByNameIgnoreCase(Status.class, name);
}
Adding this static method would resolve your problem of deserializing
添加此静态方法将解决您的反序列化问题
回答by alhamwa
You could use @JsonCreatorannotation to resolve your problem. Take a look at https://www.baeldung.com/Hymanson-serialize-enums, there's clear enough explanation about enum and serialize-deserialize with Hymanson lib.
您可以使用@JsonCreator注释来解决您的问题。看看https://www.baeldung.com/Hymanson-serialize-enums，关于 enum 和 serialize-deserialize with Hymanson lib 有足够清楚的解释。
