ava 序列化对象如何排除指定属性呢?
java 中序列化对象有多种方式:struts2 ,jackson,json-lib
(1)使用struts2 json插件
依赖的jar包:struts2-json-plugin-2.3.15.3.jar,xwork-core-2.3.15.3.jar,当然还有servlet-api.jar
范例:
private String getMessageJson(PushMessage message) { List<Pattern> excludeProperties = new ArrayList<Pattern>(); Pattern pattern1 = Pattern.compile("description"); Pattern pattern2 = Pattern.compile("creator");// 创建者ID Pattern pattern3 = Pattern.compile("modifier");// 修改者ID Pattern pattern4 = Pattern.compile("deliverTime");// Pattern pattern5 = Pattern.compile("description");// Pattern pattern6 = Pattern.compile("createTime");// Pattern pattern7 = Pattern.compile("modifyTime");// excludeProperties.add(pattern1); excludeProperties.add(pattern2); excludeProperties.add(pattern3); excludeProperties.add(pattern4); excludeProperties.add(pattern5); excludeProperties.add(pattern6); excludeProperties.add(pattern7); String pushJsonStr = null; try { PushMessage pushMessage = null; try { pushMessage = message.clone(); } catch (CloneNotSupportedException e) { logger.error("pushmessage clone failed.", e); } pushJsonStr = JSONUtil.serialize(pushMessage, excludeProperties, null, false, false); logger.info("after struts serialize:" + pushJsonStr); } catch (JSONException e) { logger.error("struts serialize failed.", e); }// TOOD 判断json字符串的长度是否超过了256 return pushJsonStr; }
注意:Pattern.compile 的参数就是要排除的成员变量名称(即description,creator,modifier都是成员变量名称)
(2)使用Jackson
官网：http://jackson.codehaus.org/
参考：http://blog.csdn.net/sciurid/article/details/8624107
http://www.cnblogs.com/hoojo/archive/2011/04/22/2024628.html
依赖的jar:jackson-mapper-lgpl-1.9.9.jar,jackson-core-lgpl-1.9.9.jar
如果使用maven,则在pom.xml中添加依赖
- <!-- Json转化模块 -->
- <dependency>
- <groupId>org.codehaus.jackson</groupId>
- <artifactId>jackson-mapper-lgpl</artifactId>
- <version>1.9.9</version>
- </dependency>
如何排除指定属性呢?
方式一:
先把要准备排除的属性的值设置为null
然后设置mapper的包含策略,看下面的实例:
public void test_jackson(){ // Map map=new HashMap(); // map.put("name", "黄威"); List<Student2> stus=null; stus=new ArrayList<Student2>(); Student2 stu=new Student2(); stus.add(stu); stu.setAddress(null); ObjectMapper mapper = new ObjectMapper(); mapper.setSerializationInclusion(Inclusion.NON_NULL); String content = null; try { content = mapper.writeValueAsString(stus); System.out.println(content); } catch (JsonGenerationException e) { e.printStackTrace(); } catch (JsonMappingException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } }
我把Student2对象的属性address设置为null,那么序列化时就会排除address属性.
注意:mapper.setSerializationInclusion(Inclusion.NON_NULL); 表示排除值为null的属性(成员变量)
方式二:使用FilterProvider
@Test public void test_jackson2(){ List<Student2> stus=null; stus=new ArrayList<Student2>(); Student2 stu=new Student2(); stus.add(stu); stu.setClassroom("36班"); ObjectMapper mapper = new ObjectMapper(); String content = null; try { // content = mapper.writeValueAsString(stus); SimpleBeanPropertyFilter theFilter = SimpleBeanPropertyFilter.serializeAllExcept("schoolNumber"); FilterProvider filters = new SimpleFilterProvider().addFilter("myFilter", theFilter); content = mapper.writer(filters).writeValueAsString(stu); System.out.println(content); } catch (JsonGenerationException e) { e.printStackTrace(); } catch (JsonMappingException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } }
注意:在排除属性的对象上面增加注解:@JsonFilter("myFilter")
参考:http://www.baeldung.com/jackson-ignore-properties-on-serialization
http://stackoverflow.com/questions/11757487/how-to-tell-jackson-to-ignore-a-field-during-serialization-if-its-value-is-null
http://www.cnblogs.com/yangy608/p/3936848.html
附件是json学习笔记