是否可能:在类中有一个字段，但在Jackson库中序列化/反序列化期间为它取不同的名称?
例如，我有一个类“coordindiantes”。
class Coordinates{
int red;
}
对于JSON的反序列化，希望有这样的格式:
{
"red":12
}
但是当我序列化对象时，结果应该是这样的:
{
"r":12
}
我试图通过在getter和setter上应用@JsonProperty注释来实现这一点(具有不同的值):
class Coordiantes{
int red;
@JsonProperty("r")
public byte getRed() {
return red;
}
@JsonProperty("red")
public void setRed(byte red) {
this.red = red;
}
}
但我有个例外:
org.codehaus.jackson。map。exx . unrecognizedpropertyexception:无法识别的字段“red”
刚刚测试，这是有效的:
public class Coordinates {
byte red;
@JsonProperty("r")
public byte getR() {
return red;
}
@JsonProperty("red")
public void setRed(byte red) {
this.red = red;
}
}
其思想是方法名应该是不同的，因此jackson将其解析为不同的字段，而不是一个字段。
下面是测试代码:
Coordinates c = new Coordinates();
c.setRed((byte) 5);
ObjectMapper mapper = new ObjectMapper();
System.out.println("Serialization: " + mapper.writeValueAsString(c));
Coordinates r = mapper.readValue("{\"red\":25}",Coordinates.class);
System.out.println("Deserialization: " + r.getR());
结果:
Serialization: {"r":5}
Deserialization: 25
你可以写一个序列化类来实现:
public class Symbol
{
private String symbol;
private String name;
public String getSymbol() {
return symbol;
}
public void setSymbol(String symbol) {
this.symbol = symbol;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
public class SymbolJsonSerializer extends JsonSerializer<Symbol> {
@Override
public void serialize(Symbol symbol, JsonGenerator jgen, SerializerProvider serializers) throws IOException, JsonProcessingException {
jgen.writeStartObject();
jgen.writeStringField("symbol", symbol.getSymbol());
//Changed name to full_name as the field name of Json string
jgen.writeStringField("full_name", symbol.getName());
jgen.writeEndObject();
}
}
ObjectMapper mapper = new ObjectMapper();
SimpleModule module = new SimpleModule();
module.addSerializer(Symbol.class, new SymbolJsonSerializer());
mapper.registerModule(module);
//only convert non-null field, option...
mapper.setSerializationInclusion(Include.NON_NULL);
String jsonString = mapper.writeValueAsString(symbolList);
可以有普通的getter/setter对。您只需要在@JsonProperty中指定访问模式
下面是单元测试:
public class JsonPropertyTest {
private static class TestJackson {
private String color;
@JsonProperty(value = "device_color", access = JsonProperty.Access.READ_ONLY)
public String getColor() {
return color;
};
@JsonProperty(value = "color", access = JsonProperty.Access.WRITE_ONLY)
public void setColor(String color) {
this.color = color;
}
}
@Test
public void shouldParseWithAccessModeSpecified() throws Exception {
String colorJson = "{\"color\":\"red\"}";
ObjectMapper mapper = new ObjectMapper();
TestJackson colotObject = mapper.readValue(colorJson, TestJackson.class);
String ser = mapper.writeValueAsString(colotObject);
System.out.println("Serialized colotObject: " + ser);
}
}
我得到的输出如下:
Serialized colotObject: {"device_color":"red"}
刚刚测试，这是有效的:
public class Coordinates {
byte red;
@JsonProperty("r")
public byte getR() {
return red;
}
@JsonProperty("red")
public void setRed(byte red) {
this.red = red;
}
}
其思想是方法名应该是不同的，因此jackson将其解析为不同的字段，而不是一个字段。
下面是测试代码:
Coordinates c = new Coordinates();
c.setRed((byte) 5);
ObjectMapper mapper = new ObjectMapper();
System.out.println("Serialization: " + mapper.writeValueAsString(c));
Coordinates r = mapper.readValue("{\"red\":25}",Coordinates.class);
System.out.println("Deserialization: " + r.getR());
结果:
Serialization: {"r":5}
Deserialization: 25
对于Kotlin的人:
data class TestClassDTO(
@JsonProperty("user_name")
val username: String
)
你将成功地从RestControllers中的POST有效载荷处理{"user_name": "John"}
但是当您需要用相同的@JsonProperty名称进行序列化时，您可以使用这种反射方法
fun Any.forceSerialize(separator: String, sorted: Boolean = false): String {
var fieldNameToAnnotatedNameMap = this.javaClass.declaredFields.map { it.name }.associateWith { fieldName ->
val jsonFieldName =
this::class.primaryConstructor?.parameters?.first { it.name == fieldName }?.annotations?.firstOrNull { it is JsonProperty }
val serializedName = if (jsonFieldName != null) (jsonFieldName as JsonProperty).value else fieldName
serializedName
}
if (sorted)
fieldNameToAnnotatedNameMap = fieldNameToAnnotatedNameMap.toList().sortedBy { (_, value) -> value}.toMap()
return fieldNameToAnnotatedNameMap.entries.joinToString(separator) { e ->
val field = this::class.memberProperties.first { it.name == e.key }
"${e.value}=${field.javaGetter?.invoke(this)}"
}
}
