json文件I / O在java中使用gson的漂亮打印格式？

Question

• 我已经知道gson如何工作，也知道如何启用漂亮的打印。
• 我想使用gson而不是simplejson。
• 我遇到的问题是我无法创建一个由Employee对象列表组成的文件。
• 我已经在stackoverflow，mkyong，google的github和许多其他网站上看到了所有其他java线程，但我仍然无法完成这个简单的事情。
• 我已经知道如何读取具有此特定格式的文件，但我无法编写它。
• 问题是我无法将所有这些东西组合在一个程序中。
• 启用了漂亮打印的gson中的对象列表必须具有正确的缩进，并且每个对象必须用逗号分隔，并且这些对象必须包含在[ ]之间。
• 问题用代码解释

：

public class Employee implements Serializable {

    private String lastName;
    private String address;
    private int id;
    private String name;

}

我想创建一个具有以下内容的json文件

 [
            {
                "id":1,
                "name": "John",
                "lastName": "Doe",
                "address": "NY City"
            },
            {
                "id":2,
                "name": "John",
                "lastName": "Doe",
                "address": "Canada"
            },
            {
                "id":3,
                "name": "John",
                "lastName": "Doe",
                "address": "Las Vegas"
            },
    ]

• 我设法创建并编写了一个Person对象的json文件（作为gson json Person对象），然后读取它，但又只作为Person对象，其中每一行都是一个独立的对象，而不是List或Person对象的一部分， 像这样

   {"id":1,"name": "John","last": "Doe","address": "NY City"} {"id":2,"name": "John","last": "Doe","address": "Canada"} {"id":3,"name": "John","last": "Doe","address": "Las Vegas"} 

但这不是我想要的最终计划。

• 我还能够使用以下信息和格式对文件进行硬编码，并成功获取Person对象

  [ { "id":1, "name": "John", "lastName": "Doe", "address": "NY City" }, { "id":2, "name": "John", "lastName": "Doe", "address": "Canada" }, { "id":3, "name": "John", "lastName": "Doe", "address": "Las Vegas" }, ] 

但我不知道如何使用java程序创建和编写这个json文件作为Person对象的数组，其中每个Person对象都是这个列表或数组的一部分，具有漂亮的打印格式，就像我硬编码的那个能够阅读。 我怎么能以优雅的方式做到这一点？

编辑---非常感谢@Shyam！

这是我的最终代码，希望它对某人有帮助。

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.reflect.TypeToken;

public class TestFileOfGsonWriter {

    Gson gson ;
    String filePath ;
    BufferedReader bufferToReader ;


    public TestFileOfGsonWriter()
    {
        this.filePath = 
                "C:\\FileOfGsonSingleListOfEmployees\\employees.json" ;
    }

    public List<Employee> createEmployees()
    {
        Employee arya = new Employee("Stark", "#81, 2nd main, Winterfell", 2, "Arya");
        Employee jon = new Employee("Snow", "#81, 2nd main, Winterfell", 1, "Jon");
        Employee sansa = new Employee("Stark", "#81, 2nd main, Winterfell", 3, "Sansa");

        List<Employee> employees = new ArrayList<>();
        employees.add(jon);
        employees.add(arya);
        employees.add(sansa);
        return employees ;
    }

    public void jsonWriter(List<Employee> employees, String filePath)
    {
        this.gson = new GsonBuilder().setPrettyPrinting().create();
        try(FileWriter writer = new FileWriter(filePath))
        {
            gson.toJson(employees,writer);
            writer.close();
        }
        catch(IOException e)
        {
            e.printStackTrace();
        }
    }

    public void showEmployeeObjects()
    {
        try {
            List<Employee> employees = this.getAllEmployees();
            employees.forEach(e -> Employee.showEmployeeDetails(e));
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    public ArrayList<Employee> getAllEmployees() throws IOException
    {
        FileReader reader = new FileReader(this.filePath);
        this.bufferToReader = new BufferedReader(reader) ;
        ArrayList <Employee> employees = this.gson.fromJson(getJson(), 
                new TypeToken<ArrayList<Employee>>(){}.getType());
        return employees ;
    }

    private String getJson() throws IOException
    {
        StringBuilder serializedEmployee = new StringBuilder();
        String line ;
        while ( (line = this.bufferToReader.readLine()) != null )
        {
            serializedEmployee.append(line);
        }
        this.bufferToReader.close();
        return serializedEmployee.toString();
    }

    public static void main(String[] args) {
        TestFileOfGsonWriter testWriting = new TestFileOfGsonWriter() ;
        List<Employee> employees = testWriting.createEmployees();
        testWriting.jsonWriter(employees, testWriting.filePath);
        testWriting.showEmployeeObjects();
    }
}

我修改了我的Employee类，所以它会与他的虚拟对象相匹配，我觉得这更好，这就是它现在的样子。

import java.io.Serializable;

public class Employee implements Serializable {

    private static final long serialVersionUID = 1L;
    String name ;
    String address;
    String lastName ;
    int id ;

    public static void showEmployeeDetails(Employee e)
    {
        System.out.println();
        System.out.println("Employee's name: "+e.name);
        System.out.println("Employee's last name"+e.lastName);
        System.out.println("Employee's address: "+e.address);
        System.out.println("Employee's ID: "+e.id);
    }

    public Employee(String myName, String myAddress, int myId, String myLastName)
    {
        this.name = myName ;
        this.address = myAddress;
        this.lastName = myLastName;
        this.id = myId ;
    }
}

因此，程序创建的json文件看起来正是我想要的：

[
  {
    "name": "Snow",
    "address": "#81, 2nd main, Winterfell",
    "lastName": "Jon",
    "id": 1
  },
  {
    "name": "Stark",
    "address": "#81, 2nd main, Winterfell",
    "lastName": "Arya",
    "id": 2
  },
  {
    "name": "Stark",
    "address": "#81, 2nd main, Winterfell",
    "lastName": "Sansa",
    "id": 3
  }
]

最后，这是输出：

Employee's name: Snow
Employee's last nameJon
Employee's address: #81, 2nd main, Winterfell
Employee's ID: 1

Employee's name: Stark
Employee's last nameArya
Employee's address: #81, 2nd main, Winterfell
Employee's ID: 2

Employee's name: Stark
Employee's last nameSansa
Employee's address: #81, 2nd main, Winterfell
Employee's ID: 3

Answer 1

在您是新手的时候，我将快速引导您完成将Employee对象List编写为具有漂亮打印的JSON文件的过程：

步骤1：创建一个接受List和String filePath ：

public void jsonWriter(List<Employee> employees, String filePath)

第2步：使用漂亮的打印功能构建Gson对象：

Gson gson = new GsonBuilder().setPrettyPrinting().create();

步骤3：使用FileWriter将List<Employee>写入给定filePath的JSON文件：

       try(FileWriter writer = new FileWriter(filePath)) {
            gson.toJson(employees, writer);
            writer.close();
        } catch (IOException e) {
            e.printStackTrace();
        }

最后整个方法看起来像这样：

public void jsonWriter(List<Employee> employees, String filePath) {
        Gson gson = new GsonBuilder().setPrettyPrinting().create();
        try(FileWriter writer = new FileWriter(filePath)) {
            gson.toJson(employees, writer);
            writer.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

第4步：现在，构建您的Employee对象，将它们添加到List并使用适当的filePath调用此方法

        Employee arya = new Employee("Stark", "#81, 2nd main, Winterfell", 2, "Arya");
        Employee jon = new Employee("Snow", "#81, 2nd main, Winterfell", 1, "Jon");
        Employee sansa = new Employee("Stark", "#81, 2nd main, Winterfell", 3, "Sansa");

        List<Employee> employees = new ArrayList<>();
        employees.add(jon);
        employees.add(arya);
        employees.add(sansa);

        jsonWriter(employees, "C:/downloads/employees.json");

运行此代码后，JSON文件的内容将如下所示：

[
  {
    "lastName": "Snow",
    "address": "#81, 2nd main, Winterfell",
    "id": 1,
    "name": "Jon"
  },
  {
    "lastName": "Stark",
    "address": "#81, 2nd main, Winterfell",
    "id": 2,
    "name": "Arya"
  },
  {
    "lastName": "Stark",
    "address": "#81, 2nd main, Winterfell",
    "id": 3,
    "name": "Sansa"
  }
]

我希望这会对你的学习过程有所帮助。

注意：我使用了一些随机的Employee姓名和详细信息。 您可以将其替换为所需的详细信息。

Answer 2

首先将这些对象存储到对象列表中。 然后添加这行代码以进行漂亮的打印。

Gson gson = new GsonBuilder().setPrettyPrinting().create();
System.out.println(gson.toJson (list));