直接说结论方便一目了然:
1. 简单的直接Bean.class
2. 复杂的用 TypeReference
这样就完事了。
public class TestMain2 { public static void main(String[] args) throws JsonProcessingException { /* 首先说明 readValue 针对String 一共有3个重载,如下: public <T> T readValue(String content, Class<T> valueType);简单型,就是 直接 UserBase.class 就可。 public <T> T readValue(String content, TypeReference<T> valueTypeRef);复杂的可以 用这个 public <T> T readValue(String content, JavaType valueType);这个书写起来比较麻烦,就不说明了,不常用,前2个已经彻底满足了。 */ ObjectMapper objectMapper = new ObjectMapper(); String json1 = "{\"userName\":\"小李飞刀\",\"age\":18,\"addTime\":1591851786568}"; String json2 = "[{\"userName\":\"小李飞刀\",\"age\":18,\"addTime\":123}, {\"userName\":\"小李飞刀2\",\"age\":182,\"addTime\":1234}]"; //1.最简单的常用方法,直接将一个json转换成实体类 UserBase userBase1 = objectMapper.readValue(json1, UserBase.class); //简单类型的时候,这样最方便 System.out.println("简单: " + userBase1.getUserName()); //用 TypeReference 也可以,但是麻烦 不如第一种直接 TypeReference 主要针对繁杂类型 //UserBase userBase2 = objectMapper.readValue(json1, new TypeReference<UserBase>() {}); //2.把Json转换成map,必须使用 TypeReference , map的类型定义 可以根据实际情况来定,比如若值都是String那么就可以 Map<String, String> Map<String, Object> userBaseMap = objectMapper.readValue(json1, new TypeReference<Map<String, Object>>() {}); System.out.println("map: " + userBaseMap.get("userName")); //3.list<Bean>模式,必须用 TypeReference List<UserBase> userBaseList = objectMapper.readValue(json2, new TypeReference<List<UserBase>>() {}); System.out.println("list: " + userBaseList.get(0).getUserName()); //4.Bean[] 数组,必须用 TypeReference UserBase[] userBaseAry = objectMapper.readValue(json2, new TypeReference<UserBase[]>() {}); System.out.println("ary: " + userBaseAry[0].getUserName()); } }
==========================================================下面是详细的秒数==================================================
方法1,针对简单类型,有实体类的json,直接转换成单个实体类,上代码:
首先是一个实体类UserBase:
public class UserBase { /** * 用户名 */ private String userName; /** * 年龄 */ private Integer age; /** * 增加时间 */ private Date addTime; public String getUserName() { return userName; } public void setUserName(String userName) { this.userName = userName; } public Integer getAge() { return age; } public void setAge(Integer age) { this.age = age; } public Date getAddTime() { return addTime; } public void setAddTime(Date addTime) { this.addTime = addTime; } }
public class TestMain2 {
public static void main(String[] args) throws JsonProcessingException {
/*
1.最简单的常用方法,直接将一个json转换成实体类
*/
ObjectMapper objectMapper = new ObjectMapper();
String json = "{\"userName\":\"小李飞刀\",\"age\":18,\"addTime\":1591851786568}";
//这里需要这么写,
UserBase userBase = objectMapper.readValue(json, UserBase.class); //简单类型的时候,这样最方便
UserBase userBase1 = objectMapper.readValue(json, new TypeReference<UserBase>() {}); //这样也可以,TypeReference主要针对复杂类型
System.out.println(userBase.getUserName());
System.out.println(userBase1.getUserName());
}
}
2. 若是map呢, 应该怎么用会怎样??以下开始举例:
public class TestMain3 { public static void main(String[] args) throws JsonProcessingException { ObjectMapper objectMapper = new ObjectMapper(); //注意这里键名和键值都是String类型的 Map<String, String> map = new HashMap<>(); map.put("name", "小李飞刀"); map.put("sex", "男"); //先生成一个json方便理解 String json = objectMapper.writeValueAsString(map); System.out.println(json);//{"sex":"男","name":"小李飞刀"} /* 开始反序列化 */ Map<String,String> map1 = new HashMap<>(); //我之前是这么写的直接 Map.class 总觉得不妥,感觉他用了默认的推断,然后程序也能正常运行 map1 = objectMapper.readValue(json, Map.class); System.out.println(map1.get("name")); } }
我们调试一下代码看下,
显然这种方式,不好,1. 编译时 会有 泛型警告。2. 不完美 虽然能用,但是不要这样。那么 Map时 应该如何 反序列化呢,看如下代码:
map1 = objectMapper.readValue(json, new TypeReference<Map<String, String>>() {}); //用这个
我们来调试看下,这次是否清晰说明了 map1的类型。
好了,干完了 map,接下来还有一个常用的List<Bean> ,我们调试看下:
public class TestMain4 { public static void main(String[] args) throws JsonProcessingException { ObjectMapper objectMapper = new ObjectMapper(); String json = "[{\"userName\":\"小李飞刀\",\"age\":18,\"addTime\":123}, {\"userName\":\"小李飞刀2\",\"age\":182,\"addTime\":1234}]"; List<UserBase> userBaseList = objectMapper.readValue(json, List.class); System.out.println(userBaseList.get(0).getUserName()); } }
先不调试了,需要这么写:
List<UserBase> userBaseList = objectMapper.readValue(json, new TypeReference<List<UserBase>>() {});
上面是直接把Json, 转换成 List<bean>,关于直接转换成 Bean数组的问题即,直接把 json转换成 bean[],也是用 TypeReference 就可:看如下代码:
public class TestMain4 { public static void main(String[] args) throws JsonProcessingException { ObjectMapper objectMapper = new ObjectMapper(); String json = "[{\"userName\":\"小李飞刀\",\"age\":18,\"addTime\":123}, {\"userName\":\"小李飞刀2\",\"age\":182,\"addTime\":1234}]"; UserBase[] userBaseAry = objectMapper.readValue(json, new TypeReference<UserBase[]>() {}); System.out.println(userBaseAry[0].getUserName()); } }
===================================================================================================
由于 objectMapper 一共有3个重载,我们已经讲了 2个,还有一个 我们看下他的用法,这个不常用,以后尽量少用 会不用,写代码 没有上面2种 来的直接和方便。
不浪费时间了,直接粘贴网上的文章:
来源:https://www.cnblogs.com/gaomanito/p/9591730.html
复制代码 ObjectMapper mapper = new ObjectMapper(); // 排除json字符串中实体类没有的字段 objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES,false); String json = "[{\"name\":\"a\",\"password\":\"345\"},{\"name\":\"b\",\"password\":\"123\"}]"; //第一种方法 List<User> list = mapper.readValue(json, new TypeReference<List<User>>(){/**/}); //第二种方法 JavaType javaType = mapper.getTypeFactory().constructCollectionType(List.class, User.class); List<User> list2 = mapper.readValue(json, javaType); 复制代码 Jackson,我感觉是在Java与Json之间相互转换的最快速的框架,当然Google的Gson也很不错,但是参照网上有人的性能测试,看起来还是Jackson比较快一点 Jackson处理一般的JavaBean和Json之间的转换只要使用ObjectMapper 对象的readValue和writeValueAsString两个方法就能实现。但是如果要转换复杂类型Collection如 List<YourBean>,那么就需要先反序列化复杂类型 为泛型的Collection Type。 如果是ArrayList<YourBean>那么使用ObjectMapper 的getTypeFactory().constructParametricType(collectionClass, elementClasses); 如果是HashMap<String,YourBean>那么 ObjectMapper 的getTypeFactory().constructParametricType(HashMap.class,String.class, YourBean.class); 复制代码 public final ObjectMapper mapper = new ObjectMapper(); public static void main(String[] args) throws Exception{ JavaType javaType = getCollectionType(ArrayList.class, YourBean.class); List<YourBean> lst = (List<YourBean>)mapper.readValue(jsonString, javaType); } /** * 获取泛型的Collection Type * @param collectionClass 泛型的Collection * @param elementClasses 元素类 * @return JavaType Java类型 * @since 1.0 */ public static JavaType getCollectionType(Class<?> collectionClass, Class<?>... elementClasses) { return mapper.getTypeFactory().constructParametricType(collectionClass, elementClasses); } 复制代码 复杂类型转换 复制代码 ObjectMapper objectMapper = new ObjectMapper(); // 排除json字符串中实体类没有的字段 objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false); PledgeCertificate pledgeCertificate; Pledge pledge = new Pledge(); try { pledgeCertificate = objectMapper.readValue(requestBody, PledgeCertificate.class); pledge = objectMapper.readValue(requestBody, Pledge.class); Map<String, Object> map = objectMapper.readValue(requestBody, Map.class); String writeValueAsString = objectMapper.writeValueAsString(map.get("obligee")); JavaType javaType = objectMapper.getTypeFactory().constructParametricType(List.class, Obligee.class); List<Obligee> obligee = objectMapper.readValue(writeValueAsString,javaType); }catch (IOException e) { return “转换错误”; } 数据格式:(实体类没有obligee字段,先排除) { “certificate”:"豫(2016)郑州市不动产权第0026369号", “debtEnd”: "yyyy-mm-dd", “debtStart”: "yyyy-mm-dd", “pledgeType”: "2", “maxDebtAmount”:88, “registType”:"0201", “obligee” :[{ "obligeeType":"1","name":张三","certType":"1","certNo":"4114211..."},{ "obligeeType":"1","name":"李四","certType":"1","certNo":"4114211..."}] } 复制代码 { “certificate”:"豫(2016)郑州市不动产权第0026369号", “debtEnd”: "yyyy-mm-dd", “debtStart”: "yyyy-mm-dd", “pledgeType”: "2", “maxDebtAmount”:88, “registType”:"0201", “obligee” :[{ "obligeeType":"1","name":张三","certType":"1","certNo":"4114211..."},{ "obligeeType":"1","name":"李四","certType":"1","certNo":"4114211..."}] } 既然我已经踏上这条道路,那么,任何东西都不应妨碍我沿着这条路走下去!!!!!!!!!! !!! ! !! ! 个人公众号《后端技术开发之路》,欢迎您关注!
来源:https://www.cnblogs.com/surge/p/9046223.html
Jackson 处理复杂类型(List,map)两种方法 方法一: String jsonString="[{‘id‘:‘1‘},{‘id‘:‘2‘}]"; ObjectMapper mapper = new ObjectMapper(); JavaType javaType = mapper.getTypeFactory().constructParametricType(List.class, Bean.class); //如果是Map类型 mapper.getTypeFactory().constructParametricType(HashMap.class,String.class, Bean.class); List<Bean> lst = (List<Bean>)mapper.readValue(jsonString, javaType); 方法二: String jsonString="[{‘id‘:‘1‘},{‘id‘:‘2‘}]"; ObjectMapper mapper = new ObjectMapper(); List<Bean> beanList = mapper.readValue(jsonString, new TypeReference<List<Bean>>() {});