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java 序列化时排除指定属性

时间:2015-03-10 02:08:23      阅读:1098      评论:0      收藏:0      [点我收藏+]

java 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)

?

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(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

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<!-- 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

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@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")
bubuko.com,布布扣
参考:
http://www.baeldung.com/jackson-ignore-properties-on-serialization?

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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

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java 序列化时排除指定属性

原文:http://hw1287789687.iteye.com/blog/2190768

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