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Java 无法读取文件

转载 作者:行者123 更新时间:2023-12-01 07:32:59 25 4
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我正在编写一个Java程序,它可以获取用户条目并将它们保存到ArrayList中,然后使用ArrayList打开一系列网页。该程序还应该能够从文件中读取网址。这就是我遇到问题的地方。

我目前得到:找不到文件 bills.txt。//该文件位于我的 src 文件夹中

Exception in thread "main" java.lang.NullPointerException
at PayBills.main(PayBills.java:92) //this is when the BufferdReader is closed

这不是作业,但该程序与我即将要做的作业共享概念,因此我不想改变我在文本中阅读方式的任何基本内容。如有任何建议,我们将不胜感激!

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

public class PayBills implements Serializable
{
/*The purpose of this program is to assist the user in opening a series of webpages to
* pay bills online. The user will be able to save and edit a list of webpages,
* open the newly created list, or read a list from file.
*/
public static void main(String[] args) throws IOException
{
char input1;
String line = new String();
ArrayList<String> list1 = new ArrayList<String>();
Runtime rt = Runtime.getRuntime();
String filename = new String();

try
{
// print out the menu
rt.exec( "rundll32 url.dll,FileProtocolHandler " + "http://www.google.com");
printMenu();

// create a BufferedReader object to read input from a keyboard
InputStreamReader isr = new InputStreamReader (System.in);
BufferedReader stdin = new BufferedReader (isr);

do
{
System.out.println("\nWhat action would you like to perform?");
line = stdin.readLine().trim(); //read a line
input1 = line.charAt(0);
input1 = Character.toUpperCase(input1);
if (line.length() == 1) //check if a user entered only one character
{
switch (input1)
{
case 'A': //Add address process to array
System.out.println("\nPlease enter a web address to add to list:\n");
String str1 = stdin.readLine().trim();
if(str1.startsWith("http://www.") || str1.startsWith("https://www."))
list1.add(str1);
else
{
System.out.println("Please enter a valid web address (starting with http:// or https://).");
}
break;
case 'D': //Show current list
System.out.println(list1.toString());
break;
case 'E': //Execute current list
//rt.exec( "rundll32 url.dll,FileProtocolHandler " + "http://www.speedtest.net");
for(int i = 0; i < list1.size(); i++)
{
Process p1 = Runtime.getRuntime().exec("cmd /c start " + list1.get(i));
}
break;
case 'R': //Read list from file
System.out.println("\nPlease enter the filename to read: ");
{
filename = stdin.readLine().trim();
}
FileReader fr = null;
BufferedReader inFile = null;
try
{
fr = new FileReader(filename);
inFile = new BufferedReader(fr);
line = inFile.readLine();
System.out.println("Test2");
while(line != null)
{
System.out.println("Test3");
if(line.startsWith("http://www.") == false || line.startsWith("https://www.") == false)
System.out.println("Error: File not in proper format.");
else
list1.add(line);
}
System.out.println(filename + " was read.");
}
catch(FileNotFoundException exception)
{
System.out.println("The file " + filename + " was not found.");
break;
}
catch(IOException exception)
{
System.out.println("Error. " + exception);
}
finally
{
inFile.close();
}

break;
case '?': //Display Menu
printMenu();
break;
case 'Q': //Quit
System.out.println("Goodbye!");
System.exit(0);
}//end switch
}//end if
else
System.out.print("Unknown action\n");
}//end do
while (input1 != 'Q' || line.length() != 1);
}

catch(IOException e1)
{
System.out.println("Error: " + e1);
}
}//end main
public static void printMenu()
{
System.out.print("Choice\t\tAction\n" +
"------\t\t------\n" +
"A\t\tAdd Web Address to List\n" +
"D\t\tDisplay Current List\n" +
"E\t\tExecute Current List\n" +
"R\t\tRead List from File\n" +
"?\t\tDisplay Menu\n" +
"Q\t\tQuit\n");
}//end of printMenu()

}//end PayBills

编辑:好的,程序不再崩溃,因为我修复了 NPE,但我仍然收到“找不到文件 bills.txt。”,这是捕获的异常。如上所述,该文件位于我的 src 文件夹中,因此路径应该是正确的。

最佳答案

如果您只是传递文件名,则必须在读取文件之前附加文件的位置

File file = new File(location + filename);

这就是它的工作原理 如果您仅将文件名作为参数传递给 File 构造函数,它将尝试在/Project/目录中查找文件(例如 c:/workspace/project/test ,如果您的项目名称是 test 并且位于 c:/工作区/项目)

因此,为了传递正确的位置,您必须指定您尝试读取的文件的完整路径

因此创建字符串位置,该位置将保存文件所在文件夹的位置,然后附加文件名

String location = "C:/opt/files/";
String filename = "abc.txt";

File file = new File(location + filename);

此 abc.txt 应位于“C:/opt/files/”

记住我已经添加了驱动器名称,因为您正在尝试运行 main

但是如果相同的代码在服务器上运行,则必须指定相对于服务器运行的根目录的路径

String location = "/opt/files/";

如果您只传递文件名,代码将在项目文件夹中查找文件,而不是在 java 类所在的文件夹中查找文件。

希望这有帮助。

关于Java 无法读取文件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15840554/

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