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");
- Pattern pattern3 = Pattern.compile("modifier");
- 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);
- }
- 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
-
- <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(){
-
-
- 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 {
-
- 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