java - 正确输入解析json

Question

解析 json 响应时，其中一个键可以是字符串“value”或 null。我将响应解析为:

  @JsonInclude(JsonInclude.Include.NON_NULL)
    @JsonPropertyOrder({
           "status",
           "code",
           "exception"
    })

    public class Response {

    private String status;        
    private long code; 
    private String exception;


    public String getStatus() {
        return status;
    }

    public void setStatus(String status) {
        this.status = status;
    }


    public long getCode() {
        return code;
    }

    public void setCode(long code) {
        this.code = code;
    }


    public String getException() {
        return exception;
    }

    public void setException(String exception) {
        this.exception = exception;
    }

当我得到 null 时，我有错误。我怎样才能正确地做到这一点？

我尝试@JsonInclude(JsonInclude.Include.NON_NULL)，但它不起作用

public class JsonParserResponse  throws UnsupportedEncodingException, JsonParseException, IOException {
    Response resp= new Response();
try{
        JSONObject outerObject = new JSONObject({"status":null,"code":0,"exception":""} );
       resp.setStatus(outerObject.getString("status")); System.out.printf("name=%s , value=%s\n", "status", outerObject.getString("status"));
             resp.setCode(outerObject.getInt("code")); System.out.printf("name=%s , value=%s\n", "code",outerObject.getInt("code"));
             resp.setException(outerObject.getString("exception")); System.out.printf("name=%s , value=%s\n","exception" ,outerObject.getString("exception"));
}catch(Exception e) {
            e.printStackTrace();
        }        

我有错误^

JSONObject["status"] not a string.
    at org.json.JSONObject.getString(JSONObject.java:639)
    at ru.JsonParserResponse.parseResult(JsonParserResponse.java:161)
    at ru.JsonParserResponse.main(JsonParserResponse.java:74)

Answer 1

这会有所帮助

在 Response 类上添加 @JsonIgnoreProperties(ignoreUnknown =true)，这将忽略空值 (com.fasterxml.jackson.annotation。)

当变量名称与 JsonProperty 相同时，无需使用 JsonProperty