使用jackson将java对象转换为json字符串时复制json属性

我有Pojo对象，并带有getAsJson函数来为此对象返回Json字符串。我使用JsonProperty在此对象中定义json属性。使用ObjectMapper的writeValueAsString为此对象编写json字符串。使用jackson将java对象转换为json字符串时复制json属性

import org.codehaus.jackson.JsonGenerationException; 
import org.codehaus.jackson.annotate.JsonIgnore; 
import org.codehaus.jackson.annotate.JsonIgnoreProperties; 
import org.codehaus.jackson.annotate.JsonProperty; 
import org.codehaus.jackson.map.JsonMappingException; 
import org.codehaus.jackson.map.ObjectMapper; 

@JsonIgnoreProperties(ignoreUnknown=true) 
public class LogLikeArticleDetail extends BaseObject { 
    private static final long serialVersionUID = -2018373118257019033L; 
    @JsonProperty("LikeArticleGUId") 
    private String likeArticleGUId; 
    @JsonProperty("UserId") 
    private String userID; 
    @JsonProperty("UserName") 
    private String userName; 
    @JsonProperty("IP") 
    private String ip; 
    @JsonProperty("OS") 
    private String os; 
    @JsonProperty("UserAgent") 
    private String userAgent; 
    @JsonProperty("WebsiteCode") 
    private String websiteCode; 
    @JsonProperty("ArticleId") 
    private String articleID; 
    @JsonProperty("ATitle") 
    private String aTitle; 
    @JsonProperty("CateAlias") 
    private String cateAlias; 
    @JsonProperty("LikeStatus") 
    private String likeStatus; 
    @JsonProperty("TimeStamp") 
    private Date timeStamp; 
     //get, set.... 
     //.... 
     @JsonIgnore 
    public String getAsJSON() throws JsonGenerationException, JsonMappingException, IOException{ 
     ObjectMapper mapper = new ObjectMapper(); 
     return mapper.writeValueAsString(this) ; 
    } 
}

现在，我得到的结果

public static void main(String[] args) throws JsonGenerationException, JsonMappingException, IOException { 
     Calendar calendar = Calendar.getInstance(); 
     LogLikeArticleDetail logLikeArticle = new LogLikeArticleDetail("1","2","3","4","5","6","7","8","what thing \"nothing\" show","10","11",calendar.getTime()); 
     System.out.println(logLikeArticle.getAsJSON()); 
    }

但结果的重复属性：

{"LikeArticleGUId":"1","UserId":"2","UserName":"3","IP":"4","OS":"5","UserAgent":"6","WebsiteCode":"7","ArticleId":"8","ATitle":"what thing \"nothing\" show","CateAlias":"10","LikeStatus":"11","TimeStamp":1352256727062,"_likeArticleGUId":"1","websiteCode":"7","likeStatus":"11","userID":"2","userName":"3","ip":"4","os":"5","userAgent":"6","articleID":"8","aTitle":"what thing \"nothing\" show","cateAlias":"10","timeStamp":1352256727062}

什么的发生在这个问题让我看看？

来源

2012-11-07 Sonrobby

我想这可能是答案：http://stackoverflow.com/questions/7105745/how-to-specify-jackson-to-only-use-fields-preferably-globally – pawelzieba

哦，是的，谢谢@pawelzieba这是答案。 – Sonrobby

，所以我遵循： how to specify jackson to only use fields - preferably globally

我之前

public class LogLikeArticleDetail extends BaseObject

和我想要的结果添加

@JsonAutoDetect(fieldVisibility = Visibility.ANY, getterVisibility = Visibility.NONE, setterVisibility = Visibility.NONE)

。

所以可以在另一个解决在getAsJson（）函数，如：

ObjectMapper mapper = new ObjectMapper(); 
mapper.setVisibilityChecker(mapper.getSerializationConfig().getDefaultVisibilityChecker() 
       .withFieldVisibility(JsonAutoDetect.Visibility.ANY) 
       .withGetterVisibility(JsonAutoDetect.Visibility.NONE) 
       .withSetterVisibility(JsonAutoDetect.Visibility.NONE) 
       .withCreatorVisibility(JsonAutoDetect.Visibility.NONE)); 
return mapper.writeValueAsString(this) ;

感谢@Sean匠的问题，@链路kmb385回答以上。

来源

2012-11-09 07:54:31 Sonrobby

您也可以使用注释对每个POJO执行此操作。这个字符串添加到您的类的顶部，你想上没有自动检测：

@JsonAutoDetect（fieldVisibility = JsonAutoDetect.Visibility.ANY， getterVisibility = JsonAutoDetect.Visibility.NONE， setterVisibility = JsonAutoDetect.Visibility .NONE， creatorVisibility = JsonAutoDetect.Visibility.NONE）

例如：

@JsonAutoDetect(fieldVisibility=JsonAutoDetect.Visibility.ANY, getterVisibility=JsonAutoDetect.Visibility.NONE, 
     setterVisibility=JsonAutoDetect.Visibility.NONE, creatorVisibility=JsonAutoDetect.Visibility.NONE) 
class Play { 
    @JsonProperty("Name") 
    private String name; 

    @JsonProperty("NickName") 
    private String nickName; 

    public Play(){ 

    } 

    public String getName() { 
     return name; 
    } 

    public void setName(String name) { 
     this.name = name; 
    } 

    public String getNickName() { 
     return nickName; 
    } 

    public void setNickName(String nickName) { 
     this.nickName = nickName; 
    } 
}

这将返回I”的属性已经定义并且不会自动检测字段名称并将它们添加到我返回的JSON结果中。

来源

2016-11-28 11:11:39 mkwyche

使用jackson将java对象转换为json字符串时复制json属性

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