如何在java中解析多個字符串到int中

Question

好的，所以我在編程任務時遇到了很多麻煩。我們將從文本文件中讀取信息並使用我們創建的某些方法對其進行格式化。我能夠從文本文件的第一行中讀取信息，但後綴出現錯誤。我的代碼如下：

String name1 = scan.nextLine(); 
    String name2 = scan.nextLine(); 
    scan.close(); 
    int length = name1.length(); 
    int count = 0; 
    int a = 0; 
    int b = 1; 
    while(end != true) 
    { 

     String check = name1.substring(a,b); 
     a++; 
     b++; 
     count++; 
     char z = check.charAt(0); 
     if(z == '0' || z == '1' || z == '2' || z == '3' || z == '4' || z == '5' || z == '6' || z == '7' || z == '8' || z == '9') 
     { 
      end = true; 

     } 
     if(count == length) 
     { 
      end = true; 

     } 
    } 
    String number1 = name1.substring(count,length); 
    int number01 = Integer.parseInt(number1); 
    name1=name1.substring(0, count-1); 
    int d = name1.indexOf(" "); 
    int length1 = name1.length(); 
    String name1first = name1.substring(0,d); 
    name1first = name1first.trim(); 
    String name1last = name1.substring(d,length1); 
    name1last = name1last.trim(); 

    System.out.println(name1first); 
    System.out.println(name1last); 
    System.out.println(number01); 


    length = name2.length(); 
    int countt = 0; 
int aa = 0; 
int bb = 1; 
    while(end != true) 
    { 

     String check = name2.substring(aa,bb); 
     aa++; 
     bb++; 
     countt++; 
     char z = check.charAt(0); 
     if(z == '0' || z == '1' || z == '2' || z == '3' || z == '4' || z == '5' || z == '6' || z == '7' || z == '8' || z == '9') 
     { 
      end = true; 

     } 
     if(countt == length) 
     { 
      end = true; 

     } 
    } 
    String number2 = name2.substring(countt,length); 
    int number02 = Integer.parseInt(number2); 
    name2=name2.substring(0, countt-1); 
    d = name2.indexOf(" "); 
    int length2 = name2.length(); 
    String name2first = name2.substring(0,d); 
    name2first = name2first.trim(); 
    String name2last = name2.substring(d,length2); 
    name2last = name2last.trim(); 

    System.out.println(name2first); 
    System.out.println(name2last); 
    System.out.println(number02); 

，我得到這個錯誤：

Exception in thread "main" java.lang.NumberFormatException: For input string: "Jennifer Sutter 52114" 
at java.lang.NumberFormatException.forInputString(Unknown Source) 
at java.lang.Integer.parseInt(Unknown Source) 
at java.lang.Integer.parseInt(Unknown Source) 
at Salary.main(Salary.java:101) 

Answer 1

這可是顯而易見的，爲什麼你得到例外：

你想轉換「珍妮弗·薩特52114「爲一個整數。

解決方法是首先使用try{}catch()塊，同時嘗試呼叫Integer.parseInt()。

我認爲什麼是你想要得到的字符串的最後一點，並將其轉換爲整數：既然你使用Scanner你可以在上面nextInt()像這樣：

new Scanner(theStringWhichMightHaveANumber).nextInt();