Question

我试图在现有的Java项目中开始使用Kotlin，但是Jackson并未在我的Kotlin对象中检测到@JsonProperty（而它与Java对象完美结合）。

有什么方法可以配置Jackson使其像Kotlin一样适用于Java？

这是Java类：

public class TestJavaObj {
    @JsonProperty
    private String includeMe;
    private String dontIncludeMe;

    public TestJavaObj(String includeMe, String dontIncludeMe) {
        this.includeMe = includeMe;
        this.dontIncludeMe = dontIncludeMe;
    }
}

和Kotlin类：

class TestKotlinObj(
    @JsonProperty val includeMe: String,
    val dontIncludeMe : String?
)

这是测试代码：

public static void main(String[] args) throws JsonProcessingException {
    ObjectMapper mapper = new ObjectMapper()
            .registerModule(new KotlinModule()) //jackson-module-kotlin
            .configure(MapperFeature.AUTO_DETECT_GETTERS, false)
            .configure(MapperFeature.AUTO_DETECT_CREATORS, false)
            .configure(MapperFeature.AUTO_DETECT_FIELDS, false)
            .configure(MapperFeature.AUTO_DETECT_IS_GETTERS, false)
            .configure(MapperFeature.AUTO_DETECT_SETTERS, false);

    TestJavaObj javaObj = new TestJavaObj("hello", "world");
    TestKotlinObj kotlinObj = new TestKotlinObj("hello", "world");

    System.out.println("Expected: " + mapper.writeValueAsString(javaObj));
    System.out.println("Got: " + mapper.writeValueAsString(kotlinObj));
}

这是输出：

Expected: {"includeMe":"hello"}
Got: {}

我的gradle文件中的版本号：

kotlin_version = '1.3.72'
...
classpath "org.jetbrains.kotlin:kotlin-gradle-plugin:$kotlin_version"
classpath "org.jetbrains.kotlin:kotlin-allopen:$kotlin_version" 
...
apply plugin: 'kotlin'
...
compile('com.fasterxml.jackson.core:jackson-annotations:2.10.3')
compile('com.fasterxml.jackson.module:jackson-module-kotlin:2.9.8')
compile "org.jetbrains.kotlin:kotlin-stdlib:$kotlin_version"

Answer 1

指定注释use-site target：

class TestKotlinObj(
        @get:JsonProperty val includeMe: String,
        val dontIncludeMe : String
)

结果：

Expected: {"includeMe":"hello"}
Got: {"includeMe":"hello"}

背景：

翻译为Java字节码的类仅带有构造函数参数：

public final class TestKotlinObj {
   @NotNull // no annotation here
   private final String includeMe;
   @NotNull
   private final String dontIncludeMe;

   @NotNull // nor here
   public final String getIncludeMe() {
      return this.includeMe;
   }

   @NotNull
   public final String getDontIncludeMe() {
      return this.dontIncludeMe;
   }
                     // but here
                     // vvvv 
   public TestKotlinObj(@JsonProperty @NotNull String includeMe, @NotNull String dontIncludeMe) {
      Intrinsics.checkParameterIsNotNull(includeMe, "includeMe");
      Intrinsics.checkParameterIsNotNull(dontIncludeMe, "dontIncludeMe");
      super();
      this.includeMe = includeMe;
      this.dontIncludeMe = dontIncludeMe;
   }
}

序列化对象时不考虑哪个。

查看相关问题： Kotlin data class Jackson @JsonProperty not honored

Answer 2

尝试将Kotlin类更改为此：

tag={id:post}

您当前拥有的Kotlin类可能与java中的该类等效（显然不起作用）：

tag.update({id:post})

杰克逊与科特林：如何仅序列化带注释的属性

2 个答案: