GSON解析JSON中的问题

我在GSON的JSON解析中遇到了问题

我的回答是

    {"services":[{"service":{"name":"asd","id":"1"}},
    {"service":{"name":"asdf","id":"2"}},
    {"service":{"name":"asdfg","id":"3"}}]}

如何解析这个响应

意味着我在创建上述响应类时遇到问题 我创建了服务类，但我对如何创建服务类感到困惑

public class services {

@SerializedName("service")
ArrayList<service> list;

public services(){
    System.out.println("services constructor stuff");
    list= new ArrayList<service>();
}

/**
 * @return the list
 */
public ArrayList<service> getList() {
    return list;
}

/**
 * @param list the list to set
 */
public void setList(ArrayList<service> list) {
    this.list = list;
}

注意：我无法更改响应，因此不建议更改

谢谢您

Services.java

public class Services {

    private List<Service> services;

    public List<Service> getServiceList() {
        return services;
    }

    public void setServiceList(List<Service> services) {
        this.services = services;
    }
}

您的json解析逻辑如下：

Gson gson = new Gson();
String s = "{\"services\":[{\"service\":{\"name\":\"asd\",\"id\":\"1\"}},{\"service\":{\"name\":\"asdf\",\"id\":\"2\"}},{\"service\":{\"name\":\"asdfg\",\"id\":\"3\"}}]}";
Services services = gson.fromJson(s, Services.class);

---

基本上，json字符串不是简单的json数组格式，它实际上是json对象中的json数组。因此，这意味着您需要两个类，一个表示json数组中的每个项-Service.java，另一个充当包含项列表的包装器。

好的，我们需要创建一个中间层类来实现它。 Services.java

public class Services {
    private ArrayList<ServiceWrapper> services = new ArrayList<ServiceWrapper>();
    public ArrayList<ServiceWrapper> getServices() {
        return services;
    }
    public void setServices(ArrayList<ServiceWrapper> services) {
        this.services = services;
    }
}

public class ServiceWrapper {
    private Service service;
    public Service getService() {
        return service;
    }
    public void setService(Service service) {
        this.service = service;
    }
}

public class Service {
    private int id;
    private String name;
    public int getId() {
        return id;
    }
    public void setId(int id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
}

Service.java

public class Services {
    private ArrayList<ServiceWrapper> services = new ArrayList<ServiceWrapper>();
    public ArrayList<ServiceWrapper> getServices() {
        return services;
    }
    public void setServices(ArrayList<ServiceWrapper> services) {
        this.services = services;
    }
}

public class ServiceWrapper {
    private Service service;
    public Service getService() {
        return service;
    }
    public void setService(Service service) {
        this.service = service;
    }
}

public class Service {
    private int id;
    private String name;
    public int getId() {
        return id;
    }
    public void setId(int id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
}

下面是测试代码

Gson gson = new Gson();
String s = "{\"services\":[{\"service\":{\"name\":\"asd\",\"id\":\"1\"}},{\"service\":{\"name\":\"asdf\",\"id\":\"2\"}},{\"service\":{\"name\":\"asdfg\",\"id\":\"3\"}}]}";
Services services = gson.fromJson(s, Services.class);
for(ServiceWrapper serviceWrapper : services.getServices()){
    System.out.println(serviceWrapper.getService().getId());
    System.out.println(serviceWrapper.getService().getName());
}

---

@T.J.Crowder我已经编辑了问题thank you@test:引用您迄今为止尝试过的代码，这将帮助人们帮助您。@spiritwlker您检查过getServiceList（）的大小了吗？它为我打印了3。您使用的是哪个版本的GSON？您是对的，谢谢，还有一个问题是我们可以使用ArrayList代替Listplz吗？您可以尝试获取值，因为它在all@test，是的，您是对的，它无法检索服务中的值。我已经在另一个答案框中发布了新的解决方案。它经过全面测试，并按预期打印值。