java while while try catch，can not print to a new line

有一個問題在這裏，我需要這個循環來打印新的代碼行到一個文件，直到它做什麼是打印1行然後第二次失敗，java while while try catch，can not print to a new line

永遠無法得到它打印到另一條線，下面是代碼

public class study { 
    public static void main(String[] args) throws IOException{ 
     BufferedWriter post = null; 
     File file = new File("text.txt"); 

      if(!file.exists()){ 
       file.createNewFile(); 
      } 
      boolean promptUser = true; 
      FileWriter fileWriter = new FileWriter(file); 
      post = new BufferedWriter(fileWriter); 
      try { 
       while(promptUser){ 
        System.out.println("enter age "); //get age 
        Scanner getage = new Scanner(System.in); 
        int age= getage.nextInt(); 

        if(age <20 || age>50){ //age range 
         System.out.println("age must be between 20 and 50"); 
         System.exit(0); 
        } 
        System.out.println("enter name ");  //get name 
        Scanner getname = new Scanner(System.in);  
        String name= getname.nextLine(); 

        System.out.println("enter email "); //get email 
        Scanner getarea = new Scanner(System.in);  
        String email= getarea.nextLine(); 

        post.write(age + "\t"); <===== fails here on second run 
        post.write(name + "\t");      
        post.write(email + "\t"); 
        post.newLine(); 
        post.close(); 

        System.out.println("enter quit to quit or any key to continue"); 
        Scanner options = new Scanner(System.in); 
        String option = options.nextLine();  

        if(option.equalsIgnoreCase("quit")){    
          System.out.println("goodbye!"); 
          System.exit(0); 
        } 
       } 
      } catch (IOException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 
     } 
    }

來源

2014-10-29 user3766275

爲什麼不使用bufferedWriter.newLine（）的;每個條目之後，而在你的情況寫入文件，即post.newLine（）; – 2014-10-29 05:12:55

我在關閉之前做了 – user3766275 2014-10-29 05:15:51

爲什麼不在每個post.write（）後都在嘗試 – 2014-10-29 05:17:19

  post.write(age + "\t"); 
      post.newLine(); 
      post.write(name + "\t");      
      post.newLine(); 
      post.write(email + "\t"); 
      post.newLine();

//刪除post.close（）;從這裏現在它可能會解決您的問題

來源

2014-10-29 05:29:09

我的問題是，我需要程序循環，並採取數據，直到我說退出，文件只顯示1行數據，這是最後一個輸入的數據，我需要它繼續每一組數據的下一行年齡，姓名，電子郵件（THEN NEXT LINE）年齡，姓名，電子郵件 – user3766275 2014-10-29 05:31:42

好吧，那麼你必須將輸入值存儲到一組數組中，然後寫入或調用任何在用戶輸入後準確寫入這些數據的函數。QUIT – 2014-10-29 05:34:33

對陣列不好，你能舉個例子嗎？ – user3766275 2014-10-29 05:37:13

替換post.close();與post.flush();，你應該沒問題。

輸入退出條件時關閉流。

來源

2014-10-29 05:38:52 Tirath

同樣的東西，文件中有1行數據 – user3766275 2014-10-29 05:45:40

@ user3766275我照原樣複製，只改變了上面提到的那一行，它就起作用了。 – Tirath 2014-10-29 06:08:59

固定它GUYS

我需要的FileWriter的線遷出TRY

import java.io.*; 
import java.util.*; 

public class study { 
    public static void main(String[] args) throws IOException { 
     BufferedWriter post = null; 

     File file = new File("text.txt"); //create file 

     if (!file.exists()) 
     { 
       file.createNewFile(); 
     } 

     boolean promptUser = true; 
     FileWriter fileWriter = new FileWriter(file); 

     try { 
      while (promptUser) { 
       post = new BufferedWriter(fileWriter); 
       System.out.println("enter age "); // get age  
       Scanner getage = new Scanner(System.in);   
       int age = getage.nextInt(); 

       if (age < 20 || age > 50){ //age range 
        System.out.println("age must be between 20 and 50"); 
        System.exit(0); 
       } 

       System.out.println("enter name "); //get name 
       Scanner getname = new Scanner(System.in);  
       String name= getname.nextLine(); 

       System.out.println("enter email "); // get email  
       Scanner getarea = new Scanner(System.in);  
       String email= getarea.nextLine(); 

       //send data to file 
       post.write(age + ";"); 
       post.write(name + ";");     
       post.write(email + ";"); 
       post.newLine(); 
       post.flush(); 

       System.out.println("enter quit to quit or any key to continue"); 
       Scanner options = new Scanner(System.in); 
       String option = options.nextLine();  

       if (option.equalsIgnoreCase("quit")) { 
        System.out.println("goodbye!"); 
        post.close(); // close file upon quitting 
        System.exit(0); 
       } 

      } 
     } 
     catch (IOException e) { 
      // TODO Auto-generated catch block 
      e.printStackTrace(); 
     }   
    } 
}

來源

2014-10-29 05:56:49 user3766275

java while while try catch，can not print to a new line

回答

相關問題