java - 通过避免太多 json.getString 来组织以下代码片段的更好方法

Question

考虑以下代码片段

public static JSONObject getStudentDetails() throws Exception {
//returns the details as json
}

public static JSONObject getDepartmentDetails(String department_id) throws Exception {
//returns the details as json
}

public void doPost(HttpServletRequest req, HttpServletResponse res) throws Exception {

JSONObject json1 = getStudentDetails();
String name = json1.getString(“name”);
String role_number = json1.getString(“role_number”);
String department_id = json1.getStrring(“department_id”);
//and goes on - to get all student details


JSONObject json2 = getDepartmentDetails(department_id);
String department_hod = json2.getString(“department_hod”);
String department_name = json2.getString(“department_name”);
String department_block = json2.getString(“department_block”);
//and goes on - till i get all the department details

//Followed by main action to be performed
}

在上面的代码中，对于大多数情况，我可能会调用一个 util 方法，该方法返回一个 json 对象，我在 doGet/doPost 方法中使用该对象。因此，我的 doPost/doGet 中的大部分代码就像使用 getString 从 json 对象中获取值。有没有更好的方法来组织上面的代码片段？

Answer 1

我建议您开始使用 JSON 解析器库。有一些著名的 JSON 解析器库，例如:FastJSON、Jackson、Gson 等。

如果您不想使用任何外部库，为了使代码更简洁，您最好创建一个模型来将数据放在那里，并确保在命名变量时尽可能清晰。因为没有办法避免 json.getString 没有外部库的帮助。

class Student{
    private String name;
    private String roleNumber;
    private String departmentId;

    public String getName() {
        return name;
    }

    public Student setName(String name) {
        this.name = name;
        return this;
    }

    public String getRoleNumber() {
        return roleNumber;
    }

    public Student setRoleNumber(String roleNumber) {
        this.roleNumber = roleNumber;
        return this;
    }

    public String getDepartmentId() {
        return departmentId;
    }

    public Student setDepartmentId(String departmentId) {
        this.departmentId = departmentId;
        return this;
    }
}

public void doPost(HttpServletRequest httpServletRequest, HttpServletResponse httpServletResponse) throws Exception {
    JSONObject studentJsonObject = getStudentDetails();
    Student student = new Student()
            .setName(studentJsonObject.getString("name"))
            .setRoleNumber(studentJsonObject.getString("role_number"))
            .setDepartmentId(studentJsonObject.getString("department_id"));

    //then you can easily get each property of student
    //like student.getName(); or student.getDepartmentId();

    //the rest of your code
}