Java。閱讀輸入時遇到的問題

Question

我正在做一個處理類和對象的作業。在它中，我有一個Address類，Letter類和PostOffice類，它們在調用PostOffice時會接收一個輸入文件並掃描它，基本上以輸入文件中所有內容的方式將其顯示在一個字母上（如：bla，from ：BLA，郵資金額，並從地址等）

我得到一條錯誤：

Exception in thread "main" java.util.NoSuchElementException: No line found 
at java.util.Scanner.nextLine(Scanner.java:1516) 
at PostOffice.readLetters(PostOffice.java:33) 
at PostOffice.main(PostOffice.java:14) 

和我真的不明白爲什麼....

這裏是我的崗位辦公室班級：

import java.util.Scanner; 
import java.io.File; 
import java.io.FileNotFoundException; 
import java.io.IOException; 

public class PostOffice { 

private final int MAX = 1000; 
private Letter [] letterArray = new Letter[MAX]; 
private int count; 

public static void main(String [] args) { 
    PostOffice postOffice = new PostOffice(); 
    postOffice.readLetters("letters.in"); 
    postOffice.sortLetters(); 
    postOffice.printLetters(); 
} 

public PostOffice() { 
    Letter [] myLetters = letterArray; 
    this.count = 0; 
} 

public void readLetters(String filename) { 
    String toName, toStreet, toCity, toState, toZip; 
    String fromName, fromStreet, fromCity, fromState, fromZip, temp; //, weight; 
    double weight; 
    int index; 
    Scanner s = new Scanner(filename); 
    if (s != null) { 
     while(s.hasNext()){ 
      toName = s.nextLine(); 
      toStreet = s.nextLine(); 
      temp = s.nextLine(); 
      index = temp.indexOf(","); 
      toCity = temp.substring (0, index); 
      index = index + 2; 
      toState = temp.substring (index, index + 2); 
      toZip = temp.substring (index); 
      fromName = s.nextLine(); 
      fromStreet = s.nextLine(); 
      temp = s.nextLine(); 
      index = temp.indexOf(","); 
      fromCity = temp.substring (0, index); 
      index = index + 2; 
      fromState = temp.substring (index, index + 2); 
      fromZip = temp.substring (index); 
      String var = s.nextLine(); 
      weight = Double.parseDouble(var); 
      //weight = s.nextLine(); 
      Letter l = new Letter(toName, toStreet, toCity, toState, toZip, fromName, fromStreet, fromCity, fromState, fromZip, weight); 
      this.count += 1; 
      this.letterArray[count - 1] = l; 
     } 
    } 
    s.close(); 
} 

public static void sortLetters() { 
//call sortSearchUtil This method should call the compareTo method provided by the Letter class to sort. 
//You may use any sorting routine you wish (see SortSearchUtil.java) 
} 

public static void printLetters() { 
//call tostring of letter class. print the count of Letters followed by the total postage followed 
//by each Letter (make sure you use the toString method provided by the Address and Letter classes for this) 
} 

}

我的信類：

public class Letter extends PostOffice implements Comparable<Letter> { 
private static final double POSTAGE_RATE = 0.46; 
private String fromName; 
private Address fromAddr; 
private String toName; 
private Address toAddr; 
private double weight; 


    public Letter (String fromName, String fromStreet, String fromCity, String fromState,  String fromZip, String toName, 
String toStreet, String toCity, String toState, String toZip, double weight) { 
this.fromName = fromName; 
this.fromAddr = new Address(fromStreet, fromCity, fromState, fromZip); 
this.toName = toName; 
this.toAddr = new Address(toStreet, toCity, toState, toZip); 
this.weight = weight; 
} 

public String toString() { 
String result; 
result = String.format("from: %s\t\t\t%5.2f\n%s", fromName, getPostage(weight), fromAddr); 
result = result + String.format("\t\t To: %s\n\t\t%s", toName, toAddr); 
return result; 
} 

    public int compareTo(Letter that) { 
int value; 
value = this.toAddr.getZip().compareTo(that.toAddr.getZip()); 
return value; 
} 


public static double getPostage(double weight) { 
double workWeight; 
workWeight = weight + 0.999; 
workWeight = (int)workWeight; 
return workWeight * POSTAGE_RATE; 
    } 
} 

和我的地址類：

import java.awt.*; 
import java.util.*; 

public class Address { 
private String street; 
private String city; 
private String state; 
private String zip; 

    public Address (String street, String city, String state, String zip) { 
    this.street = street; 
    this.city = city; 
    this.state = state; 
    this.zip = zip; 
    } 

    public String getStreet() { 
    return street; 
    } 

    public void setStreet(String street) { 
    this.street = street; 
    } 

    public String getCity() { 
    return city; 
    } 

    public void setCity(String city) { 
    this.city = city; 
    } 

    public String getState() { 
    return state; 
    } 

    public void setState(String state) { 
    this.state = state; 
    } 

    public String getZip() { 
    return zip; 
    } 

    public void setZip(String zip) { 
    this.zip = zip; 
    } 

    public String toString() { 
    String result; 
    result = String.format("%s\n%s, %s %s", street, city, state, zip); 
    return result; 
    } 
} 

，這是文本文件

 
Stu Steiner (NEW LINE) 
123 Slacker Lane (NEW LINE) 
Slackerville, IL 09035 (NEW LINE) 
Tom Capaul(NEW LINE) 
999 Computer Nerd Court (NEW LINE) 
Dweebsville, NC 28804-1359 (NEW LINE) 
0.50 (NEW LINE) 
Tom Capaul (NEW LINE) 
999 Computer Nerd Court (NEW LINE) 
Dweebsville, NC 28804-1359 (NEW LINE) 
Chris Peters (NEW LINE) 
123 Some St. (NEW LINE) 
Anytown, CA 92111-0389 (NEW LINE) 
1.55 (NEW LINE) 

一切編譯的內容，我只需要它像這樣輸出：

 
---------------------------------------------------------------------------- 

From: From value            Postage value 
From Address value (be sure and use Address toString) 

        To: To value 
        To Address value (be sure and use Address toString) 

---------------------------------------------------------------------------- 

Answer 1

你做s.readline()多次內1次迭代while循環。這就是你遇到錯誤的原因。

你需要調用readLine()只有一次while循環

例中在你的代碼：

while(s.hasNext()){ 
     toName = s.nextLine(); 
     toStreet = s.nextLine(); 
     temp = s.nextLine(); 
     // ... 
    } 

這些都是3調用nextLine，怎麼你肯定還是有行？

SOLUTION：

放的，如果每個s.nextLine前聲明（）除了第一個。 實施例：

while(s.hasNext()){ 
    toName = s.nextLine(); 
    if (s.hasNext()) { 
     toStreet = s.nextLine(); 
    } 
    if (s.hasNext()) { 
     temp = s.nextLine(); 
    } 
    // ... 
} 

Answer 2

也許你錯過了測試，如果你有下一行：

http://docs.oracle.com/javase/1.5.0/docs/api/java/util/Scanner.html#hasNext%28%29

Answer 3

的問題是mistmatch（hasNext()和nextLine()）。您可以使用hasNextLine()。這可能是您需要的組合，因爲您似乎逐行讀取它。

例如看你的文本文件的東西如下而是應該修改時，必須注意進行，如果文本文件並不總是很好地形成：

while(s.hasNextLine()) { // check if there is next line 
    String toName = s.nextLine(); 
    String toAddrLine1 = s.nextLine(); 
    String toAddrLine2 = s.nextLine(); 
    String fromName = s.nextLine(); 
    String fromAddrLine1 = s.nextLine(); 
    String fromAddrLine2 = s.nextLine(); 
    String postageValueStr = s.nextLine(); 

    //do further processing down here 

雖然hasNext()應結合使用next()。

請通過Scannerdoc。

Answer 4

您的懷疑是對的：您沒有打開文件，只是將字符串「letter.in」傳遞給掃描儀。正確的代碼是：

Scanner s = null; 
    try { 
     s = new Scanner(new File(filename)); 
    } catch (FileNotFoundException ex) { 
     Logger.getLogger(PostOffice.class.getName()).log(Level.SEVERE, null, ex); 
    } 
    if (s != null) { 
     .... 
    } 

你的程序仍然不會打印出任何東西作爲printLetters（）方法未實現，但它不會了（至少不是如果文件總是有7行的倍數崩潰）。 輸入格式不是很好選擇，但因爲它是作業，我想這不是你的選擇。如果實際上有下一行（這不是一個好的解決方案，而是一個沒有太多工作的方法），那麼在每個nextline之前，可以讓代碼至少減少一點錯誤。