Develop Reference

Cannot read resource from jar

I have the following prefix:
String prefix = TemplatesReader.class.getClassLoader().getResource("templates/").getPath();
and have method
public byte[] read(String pathToTemplate) {
return Files.readAllBytes(Paths.get(prefix + pathToTemplate));
}
in intellij idea works correctly, but when starting jar an error occurs:
java.nio.file.NoSuchFileException: file:/app.jar!/BOOT-INF/classes!/templates/request-orders/unmarked/RequestOrderUnmarked.pdf

You must not assume that a resource is a file. When the resource is inside a .jar file, it is a part of that .jar file; it is no longer a separate file at all.
You cannot use Files or Paths to read the resource.
You cannot use the getPath() method of URL. It does not return a file name. It only returns the path portion of the URL (that is, everything between the URL’s scheme/authority and its query portion), which is not a file path at all.
Instead, read the resource using getResourceAsStream:
private static final String RESOURCE_PREFIX = "/templates/";
public byte[] read(String pathToTemplate)
throws IOException {
try (InputStream stream = TemplatesReader.class.getResource(
RESOURCE_PREFIX + pathToTemplate)) {
return stream.readAllBytes();
}
}

Java - URI is not hierarchical error appears only in the exported JAR, not in Eclipse

I have a Java program that is able to change the wallpaper taking in input an image using WINAPI.
Everything works fine when I run it inside Eclipse IDE, but when I run the JAR I got the error:
Caused by: java.lang.IllegalArgumentException: URI is not hierarchical
public class Main {
//INIT USER32 for WINAPI
public static interface User32 extends Library {
User32 INSTANCE = (User32) Native.loadLibrary("user32",User32.class,W32APIOptions.DEFAULT_OPTIONS);
boolean SystemParametersInfo (int one, int two, String s ,int three);
}
public static void main(String[] args) throws IOException, URISyntaxException {
//Change wallpaper
System.out.println("Change wallpaper");
URL url = Main.class.getResource("/resources/img.jpg");
File f = new File(url.toURI());
String path = f.getPath();
User32.INSTANCE.SystemParametersInfo(0x0014, 0, path , 1);
}
}
The image is shipped within the JAR, so maybe the error is related to this since the program is not able to correctly read to URL inside the JAR.
Is there a way to solve this?

A jar file is just a compressed file when the resource is bundled as a jar java will be treated as a single file, which means it will not access to your resources.
try using this instead getResourceAsStream(...);

Jackson writeValue() to file not working with relative path

I'm trying to write an object to json file by jackson.
If i provide an absolute path "D:/Projects/quiz-red/src/main/resources/com/models/Quizzes.json" it's working and file appears in the directory
But if i provide a relative path - "/com/models/Quizzes.json" i'm just getting Process finished with exit code 0 in console and nothing happens. What am I doing wrong?
There is my code:
public static void writeEntityToJson(Object jsonDataObject, String path) throws IOException {
ObjectWriter writer = mapper.writer(new DefaultPrettyPrinter());
writer.writeValue(new File(path), jsonDataObject);
}
public static void main(String[] args) throws IOException {
Quiz quiz = new Quiz(5L, "Title", "Short desc");
writeEntityToJson(quiz, "/com/models/Quizzes2.json");
}
I want to save a file to resources from DataProvider using relative path
Exception:
Exception in thread "main" java.io.FileNotFoundException: com\models\Quizzes5.json (The system cannot find the path specified)
at java.base/java.io.FileOutputStream.open0(Native Method)
at java.base/java.io.FileOutputStream.open(FileOutputStream.java:298)
at java.base/java.io.FileOutputStream.<init>(FileOutputStream.java:237)
at java.base/java.io.FileOutputStream.<init>(FileOutputStream.java:187)
at com.fasterxml.jackson.core.JsonFactory.createGenerator(JsonFactory.java:1223)
at com.fasterxml.jackson.databind.ObjectWriter.writeValue(ObjectWriter.java:942)
at com.utils.DataProvider.writeEntityToJson(DataProvider.java:33)
at com.utils.DataProvider.main(DataProvider.java:50)

what is the correct file path in eclipse spring

I am working on an introduction to spring course, but I have an issue with my SpringConfig.xml file path. As show on the picture, I just added it everywhere as a hailmary but that's not working either:
Exception in thread "main" org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from class path resource [SpringConfig]; nested exception is java.io.FileNotFoundException: class path resource [SpringConfig] cannot be opened because it does not exist
My path:
Main method code:
public static void main(String[] args) {
ApplicationContext context = new ClassPathXmlApplicationContext("SpringConfig");
}

please don't forget to add the file extension ".xml", spring will read it from src/main/resource
public static void main(String[] args) {
ApplicationContext context = new ClassPathXmlApplicationContext("SpringConfig.xml");
}

Java opening File Streams in one class and closing/deletion of file in another class

I want to delete the file which is opened and done writing but not closed. Please refer to code below:
Class A (can't be changed):
import java.io.FileOutputStream;
public class A {
public void run(String file) throws Exception {
FileOutputStream s = new FileOutputStream(file);
}
}
Class B:
import java.io.File;
import java.nio.file.Files;
import java.nio.file.Paths;
public class B {
public static void main(String[] args) throws Exception {
String path = "D:\\CONFLUX_HOME\\TestClient\\Maps\\test\\newTest.txt";
A a = new A();
a.run(path);
File f = new File(path);
Files.delete(Paths.get(f.getAbsolutePath()));
}
}
In Class A , just open the stream without closing the file.
In class B , calling A's run method and then try to delete the file.
Since the file is still opened. I'm unable to delete the file.
Error is :
The process cannot access the file because it is being used by another process.
Actual Scenario is :
We are loading the jars dynamically. Classes inside jar are creating the file. When there is an exception, a file gets created whose size will be 0 bytes. We need to delete this file. Since the file is not closed during the exception, we can't delete the file.
We could fix the issue if we could close the streams in the jar classes, but we can't modify the jars that create the files as they are client specific jars.
Please suggest how to delete the opened file, without modifying the code in class A.

Make sure you close the file, even if there was an Exception when writing to it.
E.g.
public void run(String file) throws Exception {
FileOutputStream s = null;
try {
s = new FileOutputStream(file);
} finally {
try {
s.close();
} catch(Exception e) {
// log this exception
}
}
}

You have to close the file before any delete operation as firstly its a bad practice and second is it will lead to memory leaks.

If you are using Tomcat, it is possible to set AntiLockingOption and antiJARLocking in $CATALINA_HOME/conf/context.xml for Windows:
<Context antiJARLocking="true" antiResourceLocking="true" >
Important note:
The antiResourceLocking option can stop JSPs from redeploying when they are edited requiring a redeploy.
Read more about this option:
http://tomcat.apache.org/tomcat-7.0-doc/config/context.html
antiResourceLocking:
If true, Tomcat will prevent any file locking. This will significantly impact startup time of applications, but allows full webapp hot deploy and undeploy on platforms or configurations where file locking can occur. If not specified, the default value is false.

Pass the resource as a parameter and it becomes the caller's responsibility to clear up the resources
public void run(FileOutputStream stream) throws Exception {
...
}
caller:
try(FileStream stream = new FileStream(path)){
A a = new A();
a.run(stream);
}catch(Exception e){
.. exception handling
}

Updated according to OPs comment.
Another approach could be to subclass A and override run().
public static void main(String[] args) throws Exception {
String path = "D:\\CONFLUX_HOME\\TestClient\\Maps\\test\\newTest.txt";
A a = new A() {
#Override
public void run(String file) throws Exception {
FileOutputStream s = new FileOutputStream(file);
s.close();
}
};
a.run(path);
File f = new File(path);
Files.delete(Paths.get(f.getAbsolutePath()));
System.out.println("foo");
}

I don't think you'll find a pure java solution to this problem. One option is to install Unlocker (being careful to hit "Skip" on all the junkware) and invoke it from your code.
If you have UAC enabled, you'll also need to be running your java in an elevated process (e.g. start command prompt as Administrator). Then, assuming unlocker is in C:\Program Files\Unlocker:
Process p = new ProcessBuilder("c:\\Program Files\\Unlocker\\Unlocker.exe",path,"-s").start();
p.waitFor();
And after that you can delete the file as before. Or you could use "-d" instead of "-s" and Unlocker will delete the file for you.