我需要将某个JSON字符串转换为Java对象。我正在使用Jackson进行JSON处理。我无法控制输入JSON(我从web服务读取)。这是我的输入JSON:
{"wrapper":[{"id":"13","name":"Fred"}]}
下面是一个简化的用例:
private void tryReading() {
String jsonStr = "{\"wrapper\"\:[{\"id\":\"13\",\"name\":\"Fred\"}]}";
ObjectMapper mapper = new ObjectMapper();
Wrapper wrapper = null;
try {
wrapper = mapper.readValue(jsonStr , Wrapper.class);
} catch (Exception e) {
e.printStackTrace();
}
System.out.println("wrapper = " + wrapper);
}
我的实体类是:
public Class Student {
private String name;
private String id;
//getters & setters for name & id here
}
我的Wrapper类基本上是一个容器对象来获取我的学生列表:
public Class Wrapper {
private List<Student> students;
//getters & setters here
}
我一直得到这个错误和“包装器”返回null。我不知道少了什么。有人能帮帮我吗?
org.codehaus.jackson.map.exc.UnrecognizedPropertyException:
Unrecognized field "wrapper" (Class Wrapper), not marked as ignorable
at [Source: java.io.StringReader@1198891; line: 1, column: 13]
(through reference chain: Wrapper["wrapper"])
at org.codehaus.jackson.map.exc.UnrecognizedPropertyException
.from(UnrecognizedPropertyException.java:53)
