假设我有一个豆子:
public class Msg {
private int code;
private Object data;
... Getter/setters...
}
我用这种测试代码将其转换为JSON或XML:
public String convert() {
Msg msg = new Msg();
msg.setCode( 42 );
msg.setData( "Are you suggesting coconuts migrate?" );
ObjectMapper mapper = new ObjectMapper();
return mapper.writeValueAsString( msg );
}
输出将以某种方式:
{"code":42,"data":"Are you suggesting coconuts migrate?"}
现在假设我想用一些动态名称替换'data'属性:
public String convert(String name) {
Msg msg = new Msg();
msg.setCode( 42 );
msg.setData( "Are you suggesting coconuts migrate?" );
ObjectMapper mapper = new ObjectMapper();
// ...DO SOMETHING WITH MAPPER ...
return mapper.writeValueAsString( msg );
}
如果我调用函数 convert(“toto”),我很想得到这个输出:
{"code":42,"toto":"Are you suggesting coconuts migrate?"}
如果我调用函数 convert(“groovy”),我很想得到这个输出:
{"code":42,"groovy":"Are you suggesting coconuts migrate?"}
当然我可以在JSON创建后进行字符串替换,但是如果你有一个程序化方法的答案,我会接受它。
由于
答案 0 :(得分:8)
您可以使用PropertyNamingStrategy
类来覆盖类属性。请参阅此类的简单实现:
class ReplaceNamingStrategy extends PropertyNamingStrategy {
private static final long serialVersionUID = 1L;
private Map<String, String> replaceMap;
public ReplaceNamingStrategy(Map<String, String> replaceMap) {
this.replaceMap = replaceMap;
}
@Override
public String nameForGetterMethod(MapperConfig<?> config, AnnotatedMethod method, String defaultName) {
if (replaceMap.containsKey(defaultName)) {
return replaceMap.get(defaultName);
}
return super.nameForGetterMethod(config, method, defaultName);
}
}
示例程序可能如下所示:
import java.io.IOException;
import java.util.Collections;
import java.util.Map;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.PropertyNamingStrategy;
import com.fasterxml.jackson.databind.cfg.MapperConfig;
import com.fasterxml.jackson.databind.introspect.AnnotatedMethod;
public class JacksonProgram {
public static void main(String[] args) throws IOException {
Msg msg = new Msg();
msg.setCode(42);
msg.setData("Are you suggesting coconuts migrate?");
System.out.println(convert(msg, "test"));
System.out.println(convert(msg, "toto"));
System.out.println(convert(msg, "groovy"));
}
public static String convert(Msg msg, String name) throws IOException {
ObjectMapper mapper = new ObjectMapper();
mapper.setPropertyNamingStrategy(new ReplaceNamingStrategy(Collections.singletonMap("data", name)));
return mapper.writeValueAsString(msg);
}
}
以上程序打印:
{"code":42,"test":"Are you suggesting coconuts migrate?"}
{"code":42,"toto":"Are you suggesting coconuts migrate?"}
{"code":42,"groovy":"Are you suggesting coconuts migrate?"}
答案 1 :(得分:4)
一种可能性是使用所谓的“任何吸气剂”:
public class Msg {
public int code;
@JsonAnyGetter
public Map<String,Object> otherFields() {
Map<String,Object> extra = new HashMap<String,Object>();
extra.put("data", findDataObject()); // or whatever mechanism you want
extra.put("name", "Some Name");
return extra;
}
}
这样您就可以返回任意一组动态属性。
您还可以使用匹配的“任何getter”(@JsonAnyGetter
)机制来接受其他属性。