[英]Encode/decode JSON keys?
I want to send a minified version of my JSON by minifying the keys . 我想通过缩小键来发送JSON的缩小版本。
The Input JSON string obtained after marshalling my POJO to JSON: 将我的POJO编组为JSON后获得的Input JSON字符串 :
{
"stateTag" : 1,
"contentSize" : 10,
"content" : {
"type" : "string",
"value" : "Sid"
}
}
Desired JSON STRING which I want to send over the network to minimize payload: 我希望通过网络发送以减少有效负载的所需JSON STRING :
{
"st" : 1,
"cs" : 10,
"ct" : {
"ty" : "string",
"val" : "Sid"
}
}
Is there any standard way in java to achieve this ?? Java中有什么标准方法可以实现这一目标?
PS: My json string can be nested with other objects which too I will have to minify. PS:我的json字符串可以与其他对象嵌套在一起,我也必须缩小这些对象。
EDIT : 编辑 :
I cannot change my POJOs to provide annotations. 我无法更改POJO来提供注释。 I have XSD files from which I generate my java classes.
我有从中生成Java类的XSD文件。 So changing anything there is not an option.
因此,更改任何内容都是没有选择的。
在gson中使用批注...:在Class Member上添加@SerializedName("st")
会将变量stateTag序列化为"st" : 1
,将对象嵌套在json中的深度没有关系。
You can achieve this in Jackson by using @JsonProperty
annotation. 您可以使用
@JsonProperty
批注在Jackson中实现。
public class Pojo {
@JsonProperty(value = "st")
private long stateTag;
@JsonProperty(value = "cs")
private long contentSize;
@JsonProperty(value = "ct")
private Content content;
//getters setters
}
public class Content {
@JsonProperty(value = "ty")
private String type;
@JsonProperty(value = "val")
private String value;
}
public class App {
public static void main(String... args) throws JsonProcessingException, IOException {
ObjectMapper om = new ObjectMapper();
Pojo myPojo = new Pojo(1, 10, new Content("string", "sid"));
System.out.print(om.writerWithDefaultPrettyPrinter().writeValueAsString(myPojo));
}
Outputs: 输出:
{
"st" : 1,
"cs" : 10,
"ct" : {
"ty" : "string",
"val" : "sid"
}
}
SOLUTION 2 (Using Custom Serializer): 解决方案2(使用自定义序列化程序):
This solution is specific to your pojo, it means for every pojo you will need a new serializer. 此解决方案特定于您的pojo,这意味着对于每个pojo,您都需要一个新的序列化器。
public class PojoSerializer extends JsonSerializer<Pojo> {
@Override
public void serialize(Pojo pojo, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException {
/* your pojo */
jgen.writeStartObject();
jgen.writeNumberField("st", pojo.getStateTag());
jgen.writeNumberField("cs", pojo.getContentSize());
/* inner object */
jgen.writeStartObject();
jgen.writeStringField("ty", pojo.getContent().getType());
jgen.writeStringField("val", pojo.getContent().getValue());
jgen.writeEndObject();
jgen.writeEndObject();
}
@Override
public Class<Pojo> handledType() {
return Pojo.class;
}
}
ObjectMapper om = new ObjectMapper();
Pojo myPojo = new Pojo(1, 10, new Content("string", "sid"));
SimpleModule sm = new SimpleModule();
sm.addSerializer(new PojoSerializer());
System.out.print(om.registerModule(sm).writerWithDefaultPrettyPrinter().writeValueAsString(myPojo));
SOLUTION 3 (Using a naming strategy): This solution is a general solution. 解决方案3(使用命名策略):此解决方案是常规解决方案。
public class CustomNamingStrategy extends PropertyNamingStrategyBase {
@Override
public String translate(String propertyName) {
// find a naming strategy here
return propertyName;
}
}
ObjectMapper om = new ObjectMapper();
Pojo myPojo = new Pojo(1, 10, new Content("string", "sid"));
om.setPropertyNamingStrategy(new CustomNamingStrategy());
System.out.print(om.writerWithDefaultPrettyPrinter().writeValueAsString(myPojo));
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