我正在做一个处理课程和对象的家庭作业。在其中，我有一个Address类，Letter类和PostOffice类，在调用PostOffice时，获取输入文件并对其进行扫描，基本上打印输入文件中的所有内容，就像它应该出现在字母上一样（to：bla，from ：bla，邮资金额，往返地址等）
我收到的错误是：
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)
我真的不明白为什么......
这是我的邮局课程：
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)
}
}
我的信件类：
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;
}
}
和我的地址类：
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;
}
}
这是文本文件的内容
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)
一切都在编译，我只需要它输出如下：
----------------------------------------------------------------------------
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)
----------------------------------------------------------------------------
答案 0 :(得分：4)
您在s.readline()循环的1次迭代中多次执行while次。这就是你收到错误的原因。
您只需在while循环中调用readLine()
代码中的示例：
while(s.hasNext()){
toName = s.nextLine();
toStreet = s.nextLine();
temp = s.nextLine();
// ...
}
这是对nextLine的3次调用，您如何确定仍有行？
<强> SOLUTION：
在除第一个之外的每个s.nextLine（）之前放置一个if语句。 例如：
while(s.hasNext()){
toName = s.nextLine();
if (s.hasNext()) {
toStreet = s.nextLine();
}
if (s.hasNext()) {
temp = s.nextLine();
}
// ...
}
答案 1 :(得分：2)
答案 2 :(得分：0)
问题在于错误匹配（hasNext()和nextLine()）。
使用nextLine()，您应该使用hasNextLine()。这可能是您需要的组合，因为您似乎逐行阅读。
e.g。查看你的文本文件应该如下所示，但是如果文本文件并不总是很好的话，必须进行修改：
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
虽然hasNext()应与next()结合使用。
请完成Scanner doc。
答案 3 :(得分：0)
您的怀疑是对的：您没有打开文件，只是将字符串“letter.in”传递给扫描仪。正确的代码是：
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) {
....
}
您的程序仍然不会打印任何内容，因为printLetters（）方法没有实现，但它不会再崩溃（至少如果文件总是有7行的倍数则不会崩溃）。 输入格式选择不是很好，但由于它是作业，我想这不是你真正的选择。如果确实存在下一行（这不是一个好的解决方案，而是一个没有太多工作的方法），那么在每个nextLine之前，你可以提出至少一点点容易出错的问题。