FileNotFoundException when using FileInputStream

Question

I am trying to access a file I have contained in the jar.

The beginning of my code is as follows:

import java.io.*;
import java.net.*;

public class intro {
public static void main(String[] args) throws IOException{

    URL jarUrl = intro.class.getResource("myFile.jar");
    File myJar = new File(jarUrl.toString());

    FileInputStream fis = new FileInputStream(myjar);
}

I'm getting the following error:

Exception in thread "main" java.io.FileNotFoundException: file:\...myFile.jar (The filename, directory name, or volume label syntax is incorrect)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(Unknown Source)
at intro.main(intro.java:10)

It seems to have no problem finding the URL or initializing the file, but I can't get the FileInputStream to work. Does anybody know what I'm doing wrong? Help would be appreciated.

Hovercraft Full Of Eels · Accepted Answer · 2012-09-15 03:40:53Z

3

You can't access Jar'ed resources in this manner. A file in a Jar is not a File (ie as in a file system file), they are different concepts.

You should use the URL.openStream instead.

InputStream is = null;
try {
    is = jarURL.openStream();
    // ... Read from stream
} catch (IOException exp) {
} finally {
    try {
      if (is != null) {
        is.close();
      }
    } catch (Exception exp) {
    }    
}

edited Sep 15, 2012 at 3:40

Hovercraft Full Of Eels

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answered Sep 15, 2012 at 3:39

MadProgrammer

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2 Comments

MadProgrammer Over a year ago

@HovercraftFullOfEels (that's the other reason the close statement is in a try-catch ;))

MadProgrammer Over a year ago

@HovercraftFullOfEels no probs, your correction is more correct, I'm just lazy ;)

Community · Accepted Answer · 2017-05-23 12:27:15Z

1

If this is a known resource, you can just use:

InputStream is =intro.class.getResourceAsStream("myFile.jar");

The other answers are making some assumptions when they say you're doing this wrong. But it is suspicious that you are trying to get to a Jar first as a resource, and then as a file...

Also, see this: getResourceAsStream() vs FileInputStream

edited May 23, 2017 at 12:27

CommunityBot

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answered Sep 15, 2012 at 3:51

ykaganovich

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Comments

Stephen C · Accepted Answer · 2012-09-15 03:56:00Z

Your approach is incorrect on an number of levels:

This is wrong, because a URL is not a file name:
```
File myJar = new File(jarUrl.toString());
```
Assuming that jarUrl has type java.net.URL, it should be written as:
```
File myJar = new File(new URI(jarUrl));
```
This is wrong because it would (assuming that myJar was correct) open the JAR file as byte stream:
```
FileInputStream fis = new FileInputStream(myJar);
```
In fact, you need to open it using the JAR / ZIP file classes, and then use them to open a stream on the file contained by the JAR. It should be something like this:
```
ZipFile zip = new ZipFile(myJar);
ZipEntry entry = zip.getEntry(filePathInZip);
InputStream is = entry.getInputStream();
```
(I've left out all of the necessary exception handling and code to close things to prevent resource leaks.)
This looks wrong as a URL:
```
file:\...myFile.jar
```
What is with the backslash and the three dots? A well-formed URL would not use \.

rgettman · Accepted Answer · 2018-04-03 18:54:57Z

0

I won't beat around the bush, and I will straight away go to the point. You can't access the file inside the jar in this way. Instead try the below code.

URL jarUrl = intro.class.getResource("myFile.jar");
InputStream is = null;
String s = new String();

try {
       is = jarURL.openStream();
       Scanner scan = new Scanner(is);

       while(scan.hasNextLine){    
          s = scan.nextLine() + s;    
       }       
 } catch (IOException exp) {
 } finally {        
     is.close();
 }

edited Apr 3, 2018 at 18:54

rgettman

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answered Sep 15, 2012 at 5:16

Kumar Vivek Mitra

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