将json映射到Java对象的最佳方法 [英] Best way to map json to a Java object
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问题描述
我正在使用restTemplate向servlet发出请求,该servlet返回json中对象的非常简单的表示形式.
I'm using restTemplate to make a rquest to a servlet that returns a very simple representation of an object in json.
{
"id":"SomeID"
"name":"SomeName"
}
我有一个DTO,其中包含这2个字段以及相应的setter和getter. 我想知道的是如何使用该json响应创建对象 无需解析"响应.
And I have a DTO with those 2 fields and the corresponding setters and getters. What I would like to know is how to create the object using that json response without having to "parse" the response.
推荐答案
我个人会推荐Jackson.它相当轻巧,速度非常快,并且只需要很少的配置.这是反序列化的示例:
Personally I would recommend Jackson. Its fairly lightweight, very fast and requires very little configuration. Here's an example of deserializing:
@XmlRootElement
public class MyBean {
private String id;
private String name;
public MyBean() {
super();
}
// Getters/Setters
}
String json = "...";
MyBean bean = new ObjectMapper().readValue(json, MyBean.class);
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