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Get absolute path of resource file inside a jar

I want to access a file inside a resource folder of the current jar running.
The file is inside of My_app.jar where is located to /apps/dashboard/
I tried to access it like this
String customScriptPath = "script/template.sh";
public String getTemplatePath() {
Resource temp= new ClassPathResource(this.customScriptPath, this.getClass().getClassLoader());
try {
File templateFile = temp.getFile();
logger.info("Script template path = "+templateFile.getAbsolutePath());
return templateFile.getAbsolutePath();
} catch (IOException e) {
logger.error(e.getMessage());
e.printStackTrace();
}
return null;
}
and I got this error
class path resource [script/template.sh] cannot be resolved to absolute file path because it does not reside in the file system: jar:file:/apps/dashboard/My_app.jar!/BOOT-INF/classes!/script/template.sh

You can't use File to access the template.sh. File is used to reference files in the file system. In your case, you are trying to reference something inside a jar file.
If you want to read content of template.sh, take a stream using Resource.getInputStream(). If you want to log location of the file, use Resource.getURL().

Get file in the resources folder in Java

I want to read the file in my resource folder in my Java project. I used the following code for that
MyClass.class.getResource("/myFile.xsd").getPath();
And I wanted to check the path of the file. But it gives the following path
file:/home/malintha/.m2/repository/org/wso2/carbon/automation/org.wso2.carbon.automation.engine/4.2.0-SNAPSHOT/org.wso2.carbon.automation.engine-4.2.0-SNAPSHOT.jar!/myFile.xsd
I get the file path in the maven repository dependency and it is not getting the file. How can I do this?

You need to give the path of your res folder.
MyClass.class.getResource("/res/path/to/the/file/myFile.xsd").getPath();

Is your resource directory in the classpath?
You are not including resource directory in your path:
MyClass.class.getResource("/${YOUR_RES_DIR_HERE}/myFile.xsd").getPath();

A reliable way to construct a File instance from the resource folder is it to copy the resource as a stream into a temporary File (temp file will be deleted when the JVM exits):
public static File getResourceAsFile(String resourcePath) {
try {
InputStream in = ClassLoader.getSystemClassLoader().getResourceAsStream(resourcePath);
if (in == null) {
return null;
}
File tempFile = File.createTempFile(String.valueOf(in.hashCode()), ".tmp");
tempFile.deleteOnExit();
try (FileOutputStream out = new FileOutputStream(tempFile)) {
//copy stream
byte[] buffer = new byte[1024];
int bytesRead;
while ((bytesRead = in.read(buffer)) != -1) {
out.write(buffer, 0, bytesRead);
}
}
return tempFile;
} catch (IOException e) {
e.printStackTrace();
return null;
}
}

It is not possible to access resources of other maven modules. So you need to provide your resource myFile.xsd in your src/main/resources or src/test/resources folder.

The path is correct, though not on the file system, but inside the jar. That is, because the jar was running. A resource never is guaranteed to be a File.
However if you do not want to use resources, you can use a zip file system. However Files.copy would suffice to copy the file outside the jar. Modifying the file inside the jar is a bad idea. Better use the resource as "template" to make an initial copy in the user's home (sub-)directory (System.getProperty("user.home")).

In maven project, lets assume that, we have the file whose name is "config.cnf" and it's location is below.
/src
/main
/resources
/conf
config.cnf
In IDE (Eclipse), I access this file by using ClassLoader.getResource(..) method, but if I ran this application by using jar, I always across "File not found" exception. Finally, I wrote a method which accessing the file by looking at where app works.
public static File getResourceFile(String relativePath)
{
File file = null;
URL location = <Class>.class.getProtectionDomain().getCodeSource().getLocation();
String codeLoaction = location.toString();
try{
if (codeLocation.endsWith(".jar"){
//Call from jar
Path path = Paths.get(location.toURI()).resolve("../classes/" + relativePath).normalize();
file = path.toFile();
}else{
//Call from IDE
file = new File(<Class>.class.getClassLoader().getResource(relativePath).getPath());
}
}catch(URISyntaxException ex){
ex.printStackTrace();
}
return file;
}
If you call this method by sending "conf/config.conf" param, you access this file from both jar and IDE.

How can I access a non-class file inside a jar FROM 'within' the jar

This should be simple but it has cost me hours. Everything I find on this site indicates I am doing it right but the file still cannot be found.
Inside a jar file I have two files 'CDAkeystore.jks' and 'CDAtruststore.jks' at top level.
Yet when I call
securityProps.setProperty("javax.net.ssl.keyStore","CDAkeystore.jks");
I get a system cannot find the file requested error.
The class file calling this method is inside the same jar in the usual package arrangement.
The jar file is as follows:
com ..... (a lot more class files)
org ..... (lots of class files)
META-INF
CDAtruststore.jks
CDAkeystore.jks
How can this be SOOO difficult?!!
---------- Added INfo ------n
Since the object using the path is open source I found the routine they are using to load the file. It is:
InputStream keystoreInputStream = preBufferInputStream(new FileInputStream(keyStoreName));
which according to the documentation of FileInputStream(String name) is
Creates a FileInputStream by opening a connection to an actual file, the file named by the path name 'name' in the file system. So how should this path be expressed?

Use YourClass.class.getResourceAsStream() or this.getClass().getResourceAsStream(). You can also use class loader if you are in multiple class loaders environment.

The answer is, in short, that you can't. At least in this situation. I am stuck with passing a path to a file to a library implementation that I have no control over. So the library method accesses the file on the assumption that the file exists in unzipped form in the OS's file system. It is getting the path from a Property.setProperty(stringKey, stringPath)
So the only solution I found was an ugly hack. I need to take the resource in my jar and copy it to a file on the system. Then I would pass the path to that file in the above setProperty() method. The ugly hack is implemented as follows (if anyone else can come up with a nicer solution I would be happy). It does solve the problem. The library routine is able to find my newly created file.
/* This evil reads a file as a resource inside a jar and dumps the file where ever
* the loader of this jar/application defines as the current directory. This pain is done
* so the SecurityDomian class can load the file; it cannot access the file from
* the jar. This means the 'name' passed in contains the file name inside the jar
* prefixed with "/" so it is not read with an assumed package extension.
*/
private boolean createFileFromResource(String name)
{
// Dont bother if the file already exists
if(new File(name.replace("/", "")).exists())
{
return true;
}
byte[] keystoreFile = new byte[2048];
ByteArrayOutputStream byteArrayOut = new ByteArrayOutputStream(2048);
// Get the resource
InputStream inputStream = this.getClass().getResourceAsStream(name);
try
{
int bytesRead = 0;
while(true)
{
// Read the resource into the buffer keystoreFile in 2048 byte chunks
bytesRead = inputStream.read(keystoreFile);
if(bytesRead < 0)
{
break;
}
// Copy and append the chunks to the ByteArrayOutputStream (this class
// does the auto-extending of the output array as more chunks are
// added so you don't have to do it.
byteArrayOut.write(keystoreFile, 0, bytesRead);
}
inputStream.close();
// Now create a file at the root of where ever the loader happens to think
// the root is. So remove the "/" in front of the file name
FileOutputStream outputStream = new FileOutputStream(name.replace("/", ""));
// Write out the file. Note you will be left with garbage at that location.
byteArrayOut.writeTo(outputStream);
outputStream.flush();
outputStream.close();
}
catch (IOException e)
{
e.printStackTrace();
return false;
}
return true;
}

Create Directory along with file using file stream in java

I have a String like this "D:/Data/files/store/file.txt" now I want to check ,is directory is already exist or not, if not I want to create directory along with text file. I have tried mkdirs() but its creating directory like this data->files->store->file.txt. means its creates file.txt as folder, not a file. can any one kindly help me to do this. thanks in advance.

You need to run mkdirs() on parent directory, not the file itself
File file = new File("D:/Data/files/store/file.txt");
file.getParentFile().mkdirs();
try {
file.createNewFile();
} catch (IOException e) {
e.printStackTrace();
}

Here you go...
boolean b = (new File("D:/Data/files/store/file.txt").getParentFile()).mkdirs();

java - reading xml file from jar

I am trying to read the an xml configuration from a jar file deployed to dm-server
here is the code
Reader fileReader = null;
try {
fileReader = new FileReader("test.xml");
} catch (FileNotFoundException fnfex) {
fnfex.printStackTrace();
} catch (IOException ioex) {
ioex.printStackTrace();
}
i was able to read it if i just write a junit test w/o the jar and w/o the dm-server.
the test is packed into the jar and is at the root of the jar file.
please help!!!!
thanks,
A

You need to use the Class's getResource or getResourceAsStream methods to read files from within the jar.
This can be done like this:
Reader fileReader = null;
InputStream is = this.getClass().getResourceAsStream("/test.xml");
if (null != is) {
fileReader = new InputStreamReader(is);
}
Note that getResourceAsStream returns null if it can't find the file.
Edit:
Corrected test.xml to /test.xml
Also note that the Class<T> version of getResourceAsStream defers to the ClassLoader's getResourceAsStream or getSystemResourceAsStream as appropriate.

ClassLoader.getResourceAsStream() will allow you to read resources from within a .jar file. If your file is at the root of the .jar file then:
this.getClass().getClassLoader().getResourceAsStream("/test.xml");
will give you an InputStream from that file.