我正在尝试序列化具有如下字段的 Java 实例。
public class Person{
private String firstName;
private String lastName;
public String getFirstName() {
return firstName;
}
public void setFirstName(String firstName) {
this.firstName = firstName;
}
public String getLastName() {
return lastName;
}
public void setLastName(String lastName) {
this.lastName = lastName;
}
}
如何使用与实际字段名称不同的名称来序列化它们?在Gson
这是可以通过使用@SerializedName("first-name")
注释如下。
@SerializedName("first-name")
private String firstName;
有没有类似上面的东西snakeyaml
。的依赖项详细信息snakeyaml
如下,
<dependency>
<groupId>org.yaml</groupId>
<artifactId>snakeyaml</artifactId>
<version>1.17</version>
</dependency>
下面是程序的主类,其答案由flyx
public class Demo {
public static void main(String[] args) {
Person person = new Person();
person.setFirstName("Kasun");
person.setLastName("Siyambalapitiya");
Constructor constructor = new Constructor(Person.class);
Representer representer = new Representer();
TypeDescription personDesc = new TypeDescription(Person.class);
personDesc.substituteProperty("first-name", Person.class, "getFirstName", "setFirstName");
personDesc.substituteProperty("last-name", Person.class, "getLastName", "setLastName");
constructor.addTypeDescription(personDesc);
representer.addTypeDescription(personDesc);
Yaml yaml = new Yaml(constructor, representer);
String yamlString = yaml.dumpAs(person, Tag.MAP, DumperOptions.FlowStyle.BLOCK);
Path updateDescriptorFilePath =
Paths.get(File.separator + "tmp" + File.separator + "sample.json");
try {
FileUtils.writeStringToFile(updateDescriptorFilePath.toFile(), yamlString);
} catch (IOException e) {
System.out.println(e.getCause());
}
}
}
上述结果产生以下输出
/tmp cat sample.json
first-name: Kasun
last-name: Siyambalapitiya
firstName: Kasun
lastName: Siyambalapitiya
/tmp