Question

我是Java编程新手，我正在开发一个带有REST服务的Spring Boot应用程序，该服务将调用另一个服务并返回JSON响应。我正在使用OkHttpClient来处理这个调用。

但是，从JSON响应中，我只需要很少的属性作为List格式的最终输出。

如何从okHttpCliwnt响应中仅提取所需的属性？

我对第三方服务的回复如下：

{
    "employeeDetail": [{
            "employee": {
                "name": "abc",
                "age": "30",
                "details": {
                    "role": "developer",
                    "phone": "123"
                }
            }
        },
        {
            "employee": {
                "name": "abc",
                "age": "30",
                "details": {
                    "role": "sr.developer",
                    "phone": "1234"
                }
            }
        }
    ]
}

从这个回复中，我的最终回复只需要如下：

{
    "employeeDetail": [{
            "name": "abc",
            "age": "30",
            "role": "developer"
        },
        {
            "name": "abc",
            "age": "30",
            "role": "sr.developer"
        }
    ]
}

请帮助我。

Answer 1

我不知道如何将JSON转换为List＆lt;＆gt;但您可以使用Gson将JSON转换为Java对象。

之后，您可以将对象或对象本身的内容添加到列表中。

这是https://www.mkyong.com/java/how-do-convert-java-object-to-from-json-format-gson-api/

的摘录

Gson gson = new Gson();

// 1. JSON to Java object, read it from a file.
Staff staff = gson.fromJson(new FileReader("D:\\file.json"), Staff.class);

// 2. JSON to Java object, read it from a Json String.
String jsonInString = "{'name' : 'mkyong'}";
Staff staff = gson.fromJson(jsonInString, Staff.class);

// JSON to JsonElement, convert to String later.
JsonElement json = gson.fromJson(new FileReader("D:\\file.json"), 
JsonElement.class);
String result = gson.toJson(json);

Answer 2

我搜索但是为了这样的筑巢，我找不到任何具体的东西。然而，我尝试了JsonNode，我得到了这个。

    ObjectMapper mapper = new ObjectMapper();
    JsonNode rootNode = mapper.readTree(str);
    JsonNode empDetNode = rootNode.path("employeeDetail");
    Iterator<JsonNode> elements = empDetNode.elements();
    List<Employee> empList = new ArrayList<Employee>();
    Gson gson = new Gson();
    while (elements.hasNext()) {
        Employee emp1 = new Employee(); 
        JsonNode emp= elements.next();
        JsonNode empl= emp.path("employee");
        JsonNode name= empl.path("name");
        JsonNode age= empl.path("age");
        JsonNode details= empl.path("details");
        JsonNode role= details.path("details");
        emp1.setAge(age.toString());
        emp1.setName(name.toString());
        emp1.setRole(role.toString());
        empList.add(emp1);
    }
    EmpDetl empdetl = new EmpDetl();
    empdetl.setEmployeeDetl(empList);

员工类

public class Employee {
    private String name;
    private String age;
    private String role;
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getAge() {
        return age;
    }
    public void setAge(String age) {
        this.age = age;
    }
    public String getRole() {
        return role;
    }
    public void setRole(String role) {
        this.role = role;
    }
}

EmployeeDetl

import java.util.List;

public class EmpDetl {

    private List<Employee> employeeDetl;
    public List<Employee> getEmployeeDetl() {
        return employeeDetl;
    }
    public void setEmployeeDetl(List<Employee> empLists) {
        this.employeeDetl = empLists;
    }
    @Override
    public String toString() {
        return "EmpDetl [empLists=" + employeeDetl + "]";
    }

}

Answer 3

杰克逊可能是您正在寻找的工具。你只需要创建一个类，比如说Employee.java：

public class Employee {
    @JsonProperty("name")
    private String name;

    @JsonProperty("age")
    private String age;

    @JsonProperty("role")
    private String role;

    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getAge() {
        return age;
    }
    public void setAge(String age) {
        this.age = age;
    }
    public String getRole() {
        return role;
    }
    public void setRole(String role) {
        this.role = role;
    }
}

和EmployeeDetail.java

import java.util.List;

@JsonRootName(value = "employeeDetail")
public class EmployeeDetail {

    private List<Employee> employees;
    public List<Employee> getEmployees() {
        return employees;
    }
    public void setEmployees(List<Employee> employees) {
        this.employees = employees;
    }
}

更多注释，请refer

Answer 4

您必须使用org.json或其他一些json框架手动进行解析。

将收到的JSON响应转换为Java中的List <t>

4 个答案:

员工类

EmployeeDetl