以下是java代码。我试图从problem1.txt获取数组并将其保存到arraylist中。
import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;
class CSP {
public int nMeetings;
public static int nEmployees;
public int nTimeSlots;
public static int probNumber;
public ArrayList<ArrayList<Integer>> nMeetingsPerEmployee;
//public int[][] travelTime;
public CSP(int nMeetings, int nEmployees, int nTimeSlots, int probNumber) throws FileNotFoundException {
this.nMeetings = nMeetings;
this.nEmployees = nEmployees;
this.nTimeSlots = nTimeSlots;
this.probNumber = probNumber;
this.nMeetingsPerEmployee = getnMeetingsPerEmployeeArrayList();
}
//gets nMeetingsPerEmployee Arrylist
public static ArrayList<ArrayList<Integer>> getnMeetingsPerEmployeeArrayList() throws FileNotFoundException
{
ArrayList<ArrayList<Integer>> arraylist = new ArrayList<ArrayList<Integer>>();
File openFile = null;
if(probNumber == 1){
openFile = new File("problem1.txt");
}else if(probNumber == 2){
openFile = new File("problem2.txt");
}else if(probNumber ==3){
openFile = new File("problem3.txt");
}else
System.out.println("File Not Found");
Scanner in = new Scanner(openFile);
for(int i=0; i<5; i++)
in.nextLine(); //skip 5 lines until matrix
while(true){
for(int i=0; i<nEmployees; i++) //nEmployees = 33
{
for(int j=0; j<6 ;j++)
{
for(int k=0; k<3; k++)
in.next(); //skip 3 characters until individual number
arraylist.get(nEmployees).set(nEmployees, Integer.parseInt(in.next()));
if(in.hasNext() == false);
break;
}
}
return arraylist;
}
}
}
public class CSP_Main
{
public static void main(String[] args) throws FileNotFoundException
{
CSP problem1 = new CSP(20, 33, 12, 1);
//CSP problem2 = new CSP(20, 33, 12, 2);
//CSP problem3 = new CSP(40, 62, 12, 3);
}
}
以下是我不断得到的错误。
Exception in thread "main" java.io.FileNotFoundException: problem1.txt
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(FileInputStream.java:138)
at java.util.Scanner.<init>(Scanner.java:611)
at CSP.getnMeetingsPerEmployeeArrayList(CSP_Main.java:43)
at CSP.<init>(CSP_Main.java:22)
at CSP_Main.main(CSP_Main.java:70)
以下是我试图阅读并保存在arraylist中的problem1.txt文件。
Number of meetings: 20
Number of employees: 33
Number of time slots: 12
Meetings each employee must attend:
1: 2 6 7 9 19
2: 2 5 6 12 16
3: 1 3 8 9 16
4: 1 6 15 16 18
5: 1 3 8 13 18
6: 8 10 11 17 20
7: 3 8 10 13 20
8: 1 3 14 16 20
9: 6 7 9 16 19
10: 2 6 7 12 17 19
11: 2 6 7 9 13
12: 2 4 7 12 16
13: 2 6 7 9 16 18
14: 2 5 11 17 18
15: 1 5 11 17 18 20
16: 6 8 12 16 18
17: 6 7 15 17 19
18: 1 7 11 18 20
19: 4 5 9 10 13
20: 4 7 9 17 18
21: 7 10 11 12 17
22: 5 6 9 13 18
23: 1 9 11 17 18
24: 2 11 14 15 17
25: 1 3 14 15 16
26: 5 6 10 13 15
27: 8 11 15 17 18
28: 1 4 7 9 16
29: 1 11 13 18 20
30: 2 5 12 13 18
31: 2 6 8 12 16
32: 2 7 15 17 19
33: 6 7 15 17 18
Travel time between meetings:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1: 0 1 1 1 2 1 1 1 1 1 1 2 1 2 1 1 1 1 2 1
2: 1 0 1 1 2 2 1 1 1 1 1 1 1 1 1 2 2 1 1 2
3: 1 1 0 1 1 2 2 1 1 2 2 2 2 2 2 1 2 2 1 1
4: 1 1 1 0 1 2 2 2 2 2 1 1 1 2 2 1 1 2 2 2
5: 2 2 1 1 0 2 2 2 1 1 1 2 1 2 1 2 2 1 2 2
6: 1 2 2 2 2 0 1 1 1 1 2 1 2 1 2 1 1 2 2 1
7: 1 1 2 2 2 1 0 1 2 1 1 1 1 1 2 2 2 1 1 2
8: 1 1 1 2 2 1 1 0 2 1 1 2 1 1 2 1 2 1 2 1
9: 1 1 1 2 1 1 2 2 0 2 2 1 1 1 2 2 2 2 1 2
10: 1 1 2 2 1 1 1 1 2 0 1 2 2 2 2 2 1 1 2 2
11: 1 1 2 1 1 2 1 1 2 1 0 1 2 1 1 2 1 2 1 1
12: 2 1 2 1 2 1 1 2 1 2 1 0 1 2 2 2 2 1 2 2
13: 1 1 2 1 1 2 1 1 1 2 2 1 0 1 1 2 2 2 1 1
14: 2 1 2 2 2 1 1 1 1 2 1 2 1 0 2 1 1 2 1 1
15: 1 1 2 2 1 2 2 2 2 2 1 2 1 2 0 2 1 2 2 1
16: 1 2 1 1 2 1 2 1 2 2 2 2 2 1 2 0 1 1 2 2
17: 1 2 2 1 2 1 2 2 2 1 1 2 2 1 1 1 0 2 2 2
18: 1 1 2 2 1 2 1 1 2 1 2 1 2 2 2 1 2 0 1 1
19: 2 1 1 2 2 2 1 2 1 2 1 2 1 1 2 2 2 1 0 2
20: 1 2 1 2 2 1 2 1 2 2 1 2 1 1 1 2 2 1 2 0
我试图打开的problem1.txt文件肯定在我的java代码所在的目录中。所以我不知道为什么我一直收到这个错误。
感谢您的时间。
答案 0 :(得分：0)
ProjectRoot是Eclipse的工作目录。
将文件移到src目录上方，即项目根目录（从eclipse运行）
openFile = new File("problem1.txt");
提供从src或bin到文件位置的路径（从eclipse运行）：
openFile = new File("src/superbase/problem1.txt");
答案 1 :(得分：0)
您可以看到以下示例。您将从同一个包中获取文件。
public class Hello {
public static void main(String[] args) {
Hello obj = new Hello();
System.out.println(obj.getFile("file/p1.txt"));
}
private String getFile(String fileName) {
StringBuilder result = new StringBuilder("");
//Get file from resources folder
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource(fileName).getFile());
try (Scanner scanner = new Scanner(file)) {
while (scanner.hasNextLine()) {
String line = scanner.nextLine();
result.append(line).append("\n");
}
scanner.close();
} catch (IOException e) {
e.printStackTrace();
}
return result.toString();
}
}
答案 2 :(得分：0)
当具有指定路径名的文件不存在时，FileInputStream，FileOutputStream和RandomAccessFile构造函数将抛出此异常。如果文件确实存在但由于某种原因无法访问，例如当尝试打开只读文件进行写入时，这些构造函数也会抛出它。
更多信息：https://docs.oracle.com/javase/7/docs/api/java/io/FileNotFoundException.html
答案 3 :(得分：0)
这是你可以给出正确的路径，你没有添加反斜杠，反斜杠应该是两个，因为一个反斜杠被认为是转义序列。
File file = File("C:\\abcfolder\\textfile.txt");