Java 未处理的异常：FileNotFoundException
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Unhandled exception: FileNotFoundException
提问by user1841492
I have some problems reading file in java: my file is for example:
我在 java 中读取文件时遇到一些问题：例如，我的文件是：
3,4
2
6
4
1
7
3
8
9
where first line 3 and 4 are the lenght of array A and B and then the element of each array. I made this
其中第一行 3 和 4 是数组 A 和 B 的长度，然后是每个数组的元素。我做的
import java.io.*;
import java.util.Arrays;
public class Progetto {
public static void main(String args[])
{
// Open the file that is the first
// command line parameter
FileInputStream fstream = new FileInputStream("prova.txt");
BufferedReader br = new BufferedReader(new InputStreamReader(fstream));
String strLine = br.readLine(); // step 1
if (strLine != null) {
String[] delims = strLine.split(","); // step 2
// step 3
int[] a = new int[Integer.parseInt(delims[0])];
int[] b = new int[Integer.parseInt(delims[1])];
// step 4
for (int i=0; i < a.length; i++)
a[i] = Integer.parseInt(br.readLine());
// step 5
for (int i=0; i < b.length; i++)
b[i] = Integer.parseInt(br.readLine());
br.close(); // step 6
// step 7
System.out.println(Arrays.toString(a));
System.out.println(Arrays.toString(b));
}
}
}
But it gives me error: -Unhandled exception type FileNotFoundException (line 11) -Unhandled exception type IOException (lines 15 26 30 32) but i don't know why. Someone can help me. Thanks in advance
但它给了我错误：-未处理的异常类型 FileNotFoundException（第 11 行）-未处理的异常类型 IOException（第 15 26 30 32 行）但我不知道为什么。有人可以帮助我。提前致谢
回答by BobTheBuilder
All you have to do is add try-catch blocks for the unhandled exceptions.
This happens because FileInputStreamthrows FileNotFoundExceptionwhich is a checked exception You can readhere more)
您所要做的就是为未处理的异常添加 try-catch 块。发生这种情况是因为FileInputStreamthrowsFileNotFoundException这是一个已检查的异常你可以在这里阅读更多）
The same issue happens here with
同样的问题在这里发生
String strLine = br.readLine()
回答by Ruchira Gayan Ranaweera
Change the way your main method throws IOException. Since these operations may cause either FileNotFoundExceptionor IOException.
更改您的主要方法抛出的方式IOException。由于这些操作可能导致FileNotFoundException或IOException。
public static void main(String[] args) throws FileNotFoundException {
}
Or add a try-catchblock
或者添加一个try-catch块
try {
FileInputStream fstream = new FileInputStream("prova.txt");
String strLine = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
After all these thing make sure that file is exist.
毕竟这些事情确保文件存在。
回答by Sunny Gupta
Please add throws IOExceptionafter main method's signature
请throws IOException在main方法的签名后添加
public static void main(String[] args) throws IOException
For your Next Question :
对于您的下一个问题：
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
public class Progetto {
public static void main(String args[]) throws IOException {
int a[] = null;
int b[] = null;
try {
// Open the file that is the first
// command line parameter
FileInputStream fstream = new FileInputStream("prova.txt");
BufferedReader br = new BufferedReader(new InputStreamReader(fstream));
String strLine = br.readLine(); // step 1
if (strLine != null) {
String[] delims = strLine.split(","); // step 2
// step 3
a = new int[Integer.parseInt(delims[0])];
b = new int[Integer.parseInt(delims[1])];
// step 4
for (int i = 0; i < a.length; i++)
a[i] = Integer.parseInt(br.readLine());
// step 5
for (int i = 0; i < b.length; i++)
b[i] = Integer.parseInt(br.readLine());
br.close();
}// step 6
} catch (Exception e) {// Catch exception if any
System.err.println("Error: " + e.getMessage());
}
// step 7
System.out.println(Arrays.toString(a));
System.out.println(Arrays.toString(b));
}
}
回答by user1841492
Thank you Ruchira, i made this
谢谢鲁奇拉，我做了这个
import java.io.*;
import java.util.Arrays;
public class Progetto {
public static void main(String args[]) throws IOException {
}
{
try {
// Open the file that is the first
// command line parameter
FileInputStream fstream = new FileInputStream("prova.txt");
BufferedReader br = new BufferedReader(new InputStreamReader(fstream));
String strLine = br.readLine(); // step 1
if (strLine != null) {
String[] delims = strLine.split(","); // step 2
// step 3
int[] a = new int[Integer.parseInt(delims[0])];
int[] b = new int[Integer.parseInt(delims[1])];
// step 4
for (int i=0; i < a.length; i++)
a[i] = Integer.parseInt(br.readLine());
// step 5
for (int i=0; i < b.length; i++)
b[i] = Integer.parseInt(br.readLine());
br.close(); }// step 6
}catch (Exception e){//Catch exception if any
System.err.println("Error: " + e.getMessage());
}
// step 7
System.out.println(Arrays.toString(a));
System.out.println(Arrays.toString(b));
}
}
But now at the end i have this error:
但现在最后我有这个错误：
a cannot be resolved to a variable
a 不能解析为变量
and the same for b.
But i import java.util.Arrays;
b 也一样。但是我import java.util.Arrays;
回答by sonichy
void addRule(String content){
FileOutputStream outStream = null;
try {
outStream = this.openFileOutput("BlockList", Context.MODE_APPEND);
outStream.write(("\n"+content).getBytes());
outStream.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
}catch (IOException e) {
e.printStackTrace();
}
}
