Question

我正在研究的这个项目，这里是将数据写入文件的函数：

    public void writetofile(){
    String bucky[]={custname,custlname,agee,address,id};
    for(int i = 0; i < bucky.length; i++) {
    try (BufferedWriter bw = new BufferedWriter(new        FileWriter("records.txt", true))) {
             String s;
             s = bucky[i];
             bw.write(s);
             bw.newLine();
             bw.flush();
         }

         catch(IOException ex) {
            JOptionPane.showMessageDialog(null, "Error In File");
      }
     }
   };

并且定义了另一个从文件中读取数据的函数

    public void filereader (){
    int i=0;
    Object[] options = {"OK"};
    try
    {
    FileInputStream in = new FileInputStream("records.txt");
    BufferedReader br = new BufferedReader(new InputStreamReader(in));
    String strLine;
    StringBuffer sb = new StringBuffer();

    while((strLine = br.readLine())!= null)
   {
   sb.append(strLine +"\n");
   }
   JOptionPane.showMessageDialog( null, sb.toString());
   }catch(Exception e){
   JOptionPane.showOptionDialog(null, "Error", "Customers",    JOptionPane.PLAIN_MESSAGE,JOptionPane.QUESTION_MESSAGE, null, options, options[0]);
   }

   };

问题是，假设我想在我的项目中为客户创建一个帐户，所以很少问到诸如姓名，姓氏，性别等问题，所有这些信息都已保存，但是当我尝试读取文件所有数据都将以无格式显示，我希望它们以正确的方式显示：

customer name : Michelle , 
gender : Female ;

但是现在它看起来像这样：

michelle , 
female ,

Answer 1

您可以先构建所需的输出，然后只调用bw.write(s);

或遵循R.J的建议。将所有内容放入对象并覆盖该类的toString methid。这将是解决问题的最佳方式

        import java.io.BufferedWriter;
    import java.io.FileWriter;
    import java.io.IOException;

    import javax.swing.JOptionPane;

    public class Customer
    {

        private final String firstName;

        private final String lastName;

        private final String address;

        private final int    age;

        private final int    id;

        public Customer(String firstName, String lastName, String address, int age,
                int id)
        {
            super();
            this.firstName = firstName;
            this.lastName = lastName;
            this.address = address;
            this.age = age;
            this.id = id;
        }

        @Override
        public String toString ()
        {
            return "Customer [firstName=" + firstName + ", lastName=" + lastName
                    + ", address=" + address + ", age=" + age + "]";
        }

        public static void main (String[] args)
        {

        }

        public void writetofile (String custname, String custlname, int agee,
                String address, int age)
        {

            Customer customer = new Customer(custname, custlname, address, age, id);
            try (BufferedWriter bw = new BufferedWriter(new FileWriter(
                    "records.txt", true)))
            {
                String s = customer.toString();
                bw.write(s);
                bw.newLine();
                bw.flush();
            }

            catch (IOException ex)
            {
                JOptionPane.showMessageDialog(null, "Error In File");
            }
        }

    }

文件阅读器格式

1 个答案: