In Eclipse, my directory structure is this:
-src
-com.xxx.yyy
- MyClass.java
-assets
- car.txt
MyClass.java looks like this :
public class MyClass {
private static String FILE_PATH = "../assets/car.txt";
public static void main(String[] args) {
try {
//FileNotFoundException
FileInputStream fis = new FileInputStream(new File(FILE_PATH));
}
...
}
}
I this by default, the classpath is src/, so I point to my car.txt file by ../assets/car.txt. But I get :
java.io.FileNotFoundException: ../assets/car.txt (No such file or directory)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(FileInputStream.java:138)
Why ?
Relative file paths a relative to the execution, assuming that the program is executed in the same location as the src and assets directory, then the path should be assets/car.txt
You can check the current execution location using System.out.println(new File(".").getCanonicalPath());
Whe you run the program it is compiled to the parent folder of the src folder as far as I know. You should better add the assets folder to your build path and access the file using getClass().getResource[AsStream]().
To add the folder do the following:
Right click on your project
Click Build Path
Choose Configure Build Path
Select Source
Click Add Folder...
Select your assets folder
Inside your code you can either call it with MyClass.class.getResource[AsStream]() or getClass().getResource[AsStream]().
getResource() returns an URL and getResourceAsStream() an InputStream. Both methods expect a path as parameter. Check the docs for more information.
Your path is incorrect
Since you are inside com.xxx.yyy, you are inside three folders. You need to get out of all of them.
So you can use, the relative path as
FILE_PATH = "../../../assets/car.txt";
As src is a source folder, all the contents come directly in the classpasth, you can also use
FILE_PATH = "assets/car.txt";
You are in second level in directory structure.
Try
FILE_PATH = "../../assets/car.txt";
Hope, this will help you...
package com.xxx.yyy;
import java.io.File;
import java.io.FileInputStream;
public class MyClass {
private static String FILE_PATH = "src/assets/car.txt";
public static void main(String[] args) {
try {
//FileNotFoundException
FileInputStream fis = new FileInputStream(new File(FILE_PATH));
}
catch (Exception e){
e.printStackTrace();
}
}
}
Ankit Lamba 's suggestion works: that's using String FILE_PATH = "assets/car.txt";
Related
I have been happily using JUnit to run my tests and everything has been fine.
However, I now need to use Maven but for some reason it cannot find any of my resource files.
The files are in the expected place: src/main/resources
I am using the following code to try to read a file:
public Map<String, String> readCsv(String filename) {
Map<String, String> headersAsMap;
CSVDataManipulator csvDataManipulator = new CSVDataManipulator();
ClassLoader classLoader = getClass().getClassLoader();
String wrkbook = new File(classLoader.getResource(filename).getFile()).toString();
headersAsMap = csvDataManipulator.getAllRecordsAsMap(wrkbook);
return headersAsMap;
}
However, try as I might it cannot find the file.
I've tried lots of different code and tried moving the files to different locations but I cannot get Maven to find my resource files.
Any help would be greatly appreciated!
Thanks
To my understanding classLoader.getResource(..) expects the file to be in a folder structure matching the package of the class. So if the package of your class is com.matt.stuff, then you'll have to put the csv file in src/main/resources/com/matt/stuff.
Or you could just use this to grab your csv file:
private static String readFile(String fileName) throws IOException {
//filename can be src/main/resources/my-csv.csv
return new String(Files.readAllBytes(Paths.get(fileName)));
}
A File is indeed a file on the file system. For a resource which might be a file zipped in a jar, and has a path on the class path, you need something else.
Traditionally one would use a more general InputStream instead of a File.
InputStream in = getClass().getResourceAsStream("/.../x.csv"); // Path on the class path
With the new class Path, more general than File, you can deal with several (virtual) file systems:
URL url = getClass().getResource("/.../x.csv"); // Path on the class path
Path path = Paths.get(url.toURI());
Files.copy(path, Paths.get("..."));
With a bit of luck your CSVManipulator should besides being parametrized with a File, also with an InputStream or Reader (new InputStreamReader(in, "UTF-8"))
Here's my file structure of a Maven project built from the quick-start-archetype, with comons-io added as a dependency:
src
src/main
src/main/java
src/main/java/com
src/main/java/com/essexboy
src/main/java/com/essexboy/App.java
src/main/resources
src/main/resources/dir1
src/main/resources/dir1/test.txt
src/main/resources/dir2
src/main/resources/dir2/test.txt
src/main/resources/test.txt
src/test
src/test/java
src/test/java/com
src/test/java/com/essexboy
src/test/java/com/essexboy/AppTest.java
Here's my test
public class AppTest {
#Test
public void shouldAnswerWithTrue() throws Exception {
StringWriter writer = new StringWriter();
IOUtils.copy(getClass().getResourceAsStream("/test.txt"), writer, Charset.defaultCharset());
assertEquals("from root", writer.toString().trim());
writer = new StringWriter();
IOUtils.copy(getClass().getResourceAsStream("/dir1/test.txt"), writer, Charset.defaultCharset());
assertEquals("from dir1", writer.toString().trim());
writer = new StringWriter();
IOUtils.copy(getClass().getResourceAsStream("/dir2/test.txt"), writer, Charset.defaultCharset());
assertEquals("from dir2", writer.toString().trim());
}
}
I am trying to run the below sample code in java.
import java.util.Locale;
import java.util.ResourceBundle;
public class InternationalizationDemo {
public static void main(String[] args) {
ResourceBundle bundle = ResourceBundle.getBundle("MessageBundle", Locale.CANADA_FRENCH);
System.out.println("Message in "+Locale.CANADA_FRENCH +":"+bundle.getString("greeting"));
}
}
1.The above code executes properly when MessageBundle.properties is placed in the class path.
But I want to execute the above code successfully by removing the MessageBundle.properties from the classpath and placing it in some other location.
How can I do this?
Thanks in Advance.
You could use PropertyResourceBundle and get the path of your properties file from a System property for example something like that:
String configPath = System.getProperty("config.path");
ResourceBundle bundle = new PropertyResourceBundle(new FileReader(configPath));
Then in your launch command you will need to add -Dconfig.path=/path/to/my/config.properties
You can load properties file externally by this:
// Path to your file, if you have it on local, use something like C:\\MyFolder or \home\usr\etc
File file = new File("YOUR_PATH");
URL[] url = {file.toURI().toURL()};
ResourceBundle rb = ResourceBundle.getBundle("MessageBundle", Locale.CANADA_FRENCH, new URLClassLoader(url));
Example is from here: https://coderanch.com/t/432762/java/java/absolute-path-bundle-file
You are able to use remote Properties file or file which is saved on local.
public class unzipAll {
public static void main(final java.lang.String[] args) throws Exception{
TFile src = new TFile("C:/1/BULK.tar.gz");
File dest = new File("C:/Test/");
dest.mkdirs();
try {
src.cp_rp(dest);
TVFS.umount();
} catch (IOException e) {
e.printStackTrace();
}
}
}
I can use this code to unzip BULK.tar.gz. But I want to know the directory of the unzipped files.
Right now, all the files unzipped to C:/Test/. But it has a sub folder "AAAAA".
I want to get this sub folder name "AAAAA" How can I get it?
Try dest.listFiles(). It should give you an array of all files and directories in dest. There are also versions of listFiles that can filter out different kinds of files and/or directories which can be handy at times.
See java api for details: http://docs.oracle.com/javase/7/docs/api/java/io/File.html
I have a FileInputStream in a class in the package com.nishu.ld28.utilities, and I want to access sound files in the folder Sounds, which is not in the com.nishu.ld28 package. I specify the path for loading like so:
"sounds/merry_xmas.wav"
And then try to load it like this:
new BufferedInputStream(new FileInputStream(path))
When I export the jar, the command line prompt that I run it through says it can't find the file. I know how to access the files when I am running the program in Eclipse, but I can't figure out how to point the FileInputStream to the Sounds folder when I export it.
Edit: As requested, here's my code:
public void loadSound(String path) {
WaveData data = null;
data = WaveData.create(GameSound.class.getClassLoader().getResourceAsStream(path));
int buffer = alGenBuffers();
alBufferData(buffer, data.format, data.data, data.samplerate);
data.dispose();
source = alGenSources();
alSourcei(source, AL_BUFFER, buffer);
}
WaveData accepts an InputStream or other types of IO.
You don't need a FileInputStream, because you aren't reading from the filesystem. Use the InputStream returned by ClassLoader.getResourceAsStream(String res) or Class.getResourceAsStream(String res). So either
in = ClassLoader.getResourceAsStream("sounds/merry_xmas.wav");
or
in = getClass().getResourceAsStream("/sounds/merry_xmas.wav");
Note the leading slash in the second example.
I would put the com.nishu.ld28.utilities in the same package of your class , let's call it MyClass.
Your package:
Your code:
package com.nishu.ld28.utilities;
import java.io.InputStream;
public class MyClass {
public static void main(String[] args) {
InputStream is = MyClass.class.getResourceAsStream("sound/merry_xmas.wav");
System.out.format("is is null ? => %s", is==null);
}
}
Output
is is null ? => false
I have a properties file which is located under conf folder. conf folder is under the project root directory. I am using the following code.
public class PropertiesTest {
public static void main(String[] args) {
InputStream inputStream = PropertiesTest.class
.getResourceAsStream("/conf/sampleprop.conf");
Properties prop = new Properties();
try {
prop.load(inputStream);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println(prop.getProperty("TEST"));
}
}
But I get nullpointer exception.
I have tried using
InputStream inputStream = PropertiesTest.class
.getResourceAsStream("./conf/sampleprop.conf");
and
InputStream inputStream = PropertiesTest.class
.getResourceAsStream("conf/sampleprop.conf");
But all result in nullpointer exception.
Can anyone please help.
Thanks in advance
Try to recover your working directory first:
String workingDir = System.getProperty("user.dir");
System.out.println("Current working dir: " + workingDir);
and then is simple:
Properties propertiesFile = new Properties();
propertiesFile.load(new FileInputStream(workingDir+ "/yourFilePath"));
String first= propertiesFile.getProperty("myprop.first");
Regards, fabio
The getResourceAsStream() method tries to locate and load the resource using the ClassLoader of the class it is called on. Ideally it can locate the files only the class folders .. Rather you could use FileInputStream with relative path.
EDIT
if the conf folder is under src, then you still be able to access with getResourceAsStream()
InputStream inputStream = Test.class
.getResourceAsStream("../conf/sampleprop.conf");
the path would be relative to the class from you invoke getRes.. method.
If not
try {
FileInputStream fis = new FileInputStream("conf/sampleprop.conf");
Properties prop = new Properties();
prop.load(fis);
System.out.println(prop.getProperty("TEST"));
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
NOTE: this will only work if it is Stand alone application/in eclipse. This will not work if its web based (as the root will be Tomcat/bin, for eg)
I would suggest to copy the configuration file at designated place, then you can acess at ease. At certain extent 'System.getProperty("user.dir")' can be used if you are always copying the file 'tomcat` root or application root. But if the files to be used by external party, ideal to copy in a configurable folder (C:\appconf)
Your code works like a charm! But you might have to add the project root dir to your classpath.
If you work with Maven, place your configuration in src/main/resources/conf/sampleprop.conf
When invoking java directly add the project root dir with the java -classpath parameter. Something like:
java -classpath /my/classes/dir:/my/project/root/dir my.Main