I have a class that compiles a bunch of textures from a given folder into one big BufferedImage. It works perfectly when I run my program inside the IDE (Eclipse), but when I export it to a .jar, it throws a NullPointerException because it's not able to see the files inside the jar.
I think this has something to do with the File class, but I'm not sure.
The class's constructor takes a String path which represents the path to a folder inside the project's src folder. I use this line here to turn the partial path into a full path:
path = getFullPath(path);
The getFullPath method is just one line:
public String getFullPath(String path)
{
return Thread.currentThread().getContextClassLoader().getResource(path).getPath();
}
I then iterate through all files in the folder at path:
for(File f : new File(path).listFiles())
This works flawlessly in the IDE. For example, when I pass assets/textures/sky/ into the constructor as path, it successfully finds /C:/Users/MyName/eclipse-workspace/ProjectName/bin/assets/textures/sky/, then iterates through all the files in that folder.
However, when I export the project as a .jar and run it, it throws a NullPointerException when it gets to new File(path).lisFiles() because it can't find anything there. I've printed path to the console and the path is correct save for an !exclamation mark after the name of the .jar, which makes me think that the File class can't read .jars.
I'm aware of classes like ZipFile or JarFile that work differently, but I'm not familiar with them, and Google has not been helpful. Would anyone like to lend me a hand with this?
Related
I am trying to execute a windows command inside java code using Runtime.exec() command. It is working fine when put all the necessary batch file and properties file on the root directory. But when i am exporting this is as jar, the java program is throwing error, which is becuase it is not able to find all those dependent .bat and .properties files. Can some one please tell me, where should i keep all the .bat and .properties files in side the folder. Thanks in Advance.
You can do so
Something like
Runtime.getRuntime().exec("cmd /c start yourFile.bat");
You should be able to keep it in the root of your jar if you want
EDIT :
On second thought I don't think you can run bat files inside a JAR
you would have to extract it and then run it
Please give more information on what it is you want the bat file to be doing and I can update this answer maybe there is another way?
Your problem can be divided into two parts: get the bat from the jarand run it.
To get the bat from the jar, you will have to use the ClassLoader to get a resource. you can achieve this by using the method Class.getResource to get the URL or Class.getResourceAsStream to get an InputStream
Anyway, i dont' think you can run the bat from inside the jar. If you try and fail, my advice is to create a temp file, copy your bat into your temp file and run that file.
P.S: Class.getResource finds file in the classpath. If your file is not in your classpath, you won't be able to find it this way.
EDIT: i add the code i'm using to get resources from a general relative path, given the path exist both starting from you working directory and from the home of your jar. It works, i've been able to just pack every folder i need into the jar and ship the jar to another compute rwhere eveything worked fine.
public static URL getResource(String name) {
if ("jar".equals(Main.class.getResource("Main.class").getProtocol())) {
return Main.class.getResource(("\\" + name).replace('\\', '/'));
} else {
try {
return (new File(System.getProperty("user.dir") + "\\" + name)).toURI().toURL();
} catch (MalformedURLException ex) {
return null;
}
}
}
Main is a known class, in this case the class where this static method is. I first use it to get a known url, and see if i am executing from a jar. If i am, i use the getResource, otherwise i use the File api.
the structure i use is this
main_folder\
res\
src\
package\
and, in the jar
file.jar\
package\
res\
and i need to use both File api and getResource since in the rist case the res folder is not in my classpath. with a different structure probably only the getResource method is fine.
This should solve your problem of getting the bat file, you still need to see if you are able to run it, and if you are not, copy everything into a temp file and run the temp file instead.
I am trying to make a mess management application in Java using NetBeans. I want to save images of Members in a specified folder inside my src directory. I just created folder named EmpImgs for storing employees images. Here is my code:
File srcDir = new File(file); // current path of image
File dstDir = new File("src\\J_Mess_Mgnt\\EmpImgs\\"+Txt_C_G_M_M_ID.getText());
objm.copyFile(srcDir, dstDir);` // copy image from srcDir to dstDir
Here I use another class for copying images to predefined folders and renaming the images based on their ID.
Everything is working properly in Java IDE.
But unfortunately after making an executable .jar file, this code will not work. I cannot save or access any image file in that directory.
I just went through this site, but I didn't find a suitable answer.
All I need is saving and editing images inside jar folder
Hehe hi mate you need some help. This is a duplicate but I will cut you some slack and maybe you should delete this later. So back to basics, the jvm runs byte code, which you get from compiling java source code to .class files. Now this is different to C and C++ were you just get a .exe. You don't want to give your users a bunch of .class files in all these folders which they can edit and must run a command on the command line, but instead give them what is known as an 'archive' which is just an imutable file structure so they can't screw up the application, known as a jar in java. They can just double click on the archive (which is a jar), and the jvm will call the main method specified in the MetaInf directory (just some information about the jar, same as a manifest in other programming languages).
Now remember your application is now a jar! It is immutable! for the resasons I explained. You can't save anymore data there! Your program will still work on the command line and in IDEs because it is working as if you used your application is distrubuted as bunch of folders with the .class files, and you can write to this location.
If you want to package resources with your application you need to use streams (google it). BUT REMEMBER! you cant then save more resources into the jar! You need to write somewhere else! Maybe use a user.home directory! or a location specified from the class path and the photos will be right next to the jar! Sometimes you might need an installer for your java application, but usually you don't want to create the extra work if you don't need to.
At last I find an answer suit for my question.It is not possible to copy images or files to a executive jar folder.So I used a different Idea.Create some folders(as per our requirement),Where my executable jar folder is located(No matter which drive or where the location is).The code is..
String PRJT_PATH=""; //variable to store path of working directory.
private void getdire() throws IOException{
File f=new File(".");
File[] f1=f.listFiles();
PRJT_PATH=f.getCanonicalPath(); //get path details.for eg:-E:/java/dist
}
private void new_Doc_folder(){ //function for creating new folders
try{
String strManyDirectories="Docs"+File.separator+"Bil_Img"; //i need to create 2 folders,1st a folder namedDocs and In Docs folder another folder named Bil_Img
String SubDirectories="Docs"+File.separator+"EmpImgs"; //same as above but keep in mind that It will not create a Same folder again if already exists,
// Create one directory
boolean success = (new File(strManyDirectories)).mkdirs(); //create more than one directory
boolean success1 = (new File(SubDirectories)).mkdir(); //Creates a single directory
if (success && success1) {
}
}catch (Exception e){//Catch exception if any
System.err.println("Error: " + e.getMessage());
}
It works Successfully.
Regds
I have a code where i create Java Actions and try to associate Icons with them. One snapshot of code is
FileOpenCommand fileOpen = new FileOpenCommand(this);
fileOpen.putValue("ImageOnly", false);
fileOpen.putValue(Action.NAME, "Open");
fileOpen.putValue(Action.SMALL_ICON, new ImageIcon(getClass().getResource("../resources/File-Open-icon24x24.png")));
fileOpen.putValue(Action.SHORT_DESCRIPTION, "Opens the existing file.");
fileOpen.putValue("Group", "File");
fileOpen.putValue(Action.ACCELERATOR_KEY, KeyStroke.getKeyStroke(KeyEvent.VK_O, ActionEvent.CTRL_MASK));
this.commands.put("FileOpen", fileOpen);
The accent is on the line where I try to set the Action.SMALL_ICON property to the action. This works when executed in NetBeans environment either in debug or release mode. But when I've tried to execute jar file from the command line, it fails with exception.
Any idea? Anything to do with classpath? Resources folder is put as the package inside the main package.
Thanks in Advance!
I'm not entirely sure what exception is being thrown in your case, although assuming it is a NullPointerException, IOException, or IllegalArgumentException deriving from
fileOpen.putValue(Action.SMALL_ICON, new ImageIcon(getClass().getResource("../resources/File-Open-icon24x24.png")));
Your issue should be resolved simply by adding getClassLoader() between the getClass() and getResource(), like so:
fileOpen.putValue(Action.SMALL_ICON, new ImageIcon(getClass().getClassLoader().getResource("../resources/File-Open-icon24x24.png")));
Additionally, you must be exact in your filenames, specifically the extension. In this case, you are accessing File-Open-icon24x24.png, which will work perfectly fine regardless of whether the actual file is extended by png or PNG within Netbeans, but once exported the extension case matters.
Lastly, if neither of those changes resolve your problem, I would check your filepath, as there is most likely a logical error somewhere down the line.
When using embedded resources in Netbeans, you should have a resources folder containing additional folder or whatever data you need, which you seem to have, but this folder should be located inside the Netbeans project's src folder. getClass().getResource() returns the directory at the top of the package line, meaning if your class package is com.example.code, then the compiler will look for files/folder on the same level as com. Opening the Netbeans src folder you should see the initial com folder. Your resource folder should be placed directly next to that folder, as then it will be properly embedded in the jar file export.
In your code your path is ../resources/File-Open-icon24x24.png, which confuses me as to why you begin with ... I cannot see your folder structure so I cannot give a precise answer on this note, but you may be accessing the wrong location, although I feel like you are not as you said your project runs correctly within Netbeans. However, your resource files may not be correctly encoding into the jar file due to placement as mentioned. To test what your jar file actually contains, make a copy of it (for safety reasons) and change the file extension from jar to zip. You can then look through its contents in Windows Explorer, and see its directory structure. Another debugging trick for folder structures is to create a text file at the URL you are trying to access to see where it is placed.
I need to read a text file when I start my program. I'm using eclipse and started a new java project. In my project folder I got the "src" folder and the standard "JRE System Library" + staedteliste.txt... I just don't know where to put the text file. I literally tried every folder I could think off....I cannot use a "hard coded" path because the text file needs to be included with my app...
I use the following code to read the file, but I get this error:
Error:java.io.FileNotFoundException:staedteliste.txt(No such file or directory)
public class Test {
ArrayList<String[]> values;
public static void main(String[] args) {
// TODO Auto-generated method stub
URL url = Test.class.getClassLoader().getResource("src/mjb/staedteliste.txt");
System.out.println(url.getPath()); // I get a nullpointerexception here!
loadList();
}
public static void loadList() {
BufferedReader reader;
String zeile = null;
try {
reader = new BufferedReader(new FileReader("src/mjb/staedteliste.txt"));
zeile = reader.readLine();
ArrayList<String[]> values = new ArrayList<String[]>();
while (zeile != null) {
values.add(zeile.split(";"));
zeile = reader.readLine();
}
System.out.println(values.size());
System.out.println(zeile);
} catch (IOException e) {
System.err.println("Error :"+e);
}
}
}
Ask first yourself: Is your file an internal component of your application?
(That usually implies that it's packed inside your JAR, or WAR if it is a web-app; typically, it's some configuration file or static resource, read-only).
If the answer is yes, you don't want to specify an absolute path for the file. But you neither want to access it with a relative path (as your example), because Java assumes that path is relative to the "current directory". Usually the preferred way for this scenario is to load it relatively from the classpath.
Java provides you the classLoader.getResource() method for doing this. And Eclipse (in the normal setup) assumes src/ is to be in the root of your classpath, so that, after compiling, it copies everything to your output directory ( bin/ ), the java files in compiled form ( .class ), the rest as is.
So, for example, if you place your file in src/Files/myfile.txt, it will be copied at compile time to bin/Files/myfile.txt ; and, at runtime, bin/ will be in (the root of) your classpath. So, by calling getResource("/Files/myfile.txt") (in some of its variants) you will be able to read it.
Edited: Further, if your file is conceptually tied to a java class (eg, some com.example.MyClass has a MyClass.cfg associated configuration file), you can use the getResource() method from the class and use a (resource) relative path: MyClass.getResource("MyClass.cfg"). The file then will be searched in the classpath, but with the class package pre-appended. So that, in this scenario, you'll typically place your MyClass.cfg and MyClass.java files in the same directory.
One path to take is to
Add the file you're working with to the classpath
Use the resource loader to locate the file:
URL url = Test.class.getClassLoader().getResource("myfile.txt");
System.out.println(url.getPath());
...
Open it
Suppose you have a project called "TestProject" on Eclipse and your workspace folder is located at E:/eclipse/workspace. When you build an Eclipse project, your classpath is then e:/eclipse/workspace/TestProject. When you try to read "staedteliste.txt", you're trying to access the file at e:/eclipse/workspace/TestProject/staedteliste.txt.
If you want to have a separate folder for your project, then create the Files folder under TestProject and then access the file with (the relative path) /Files/staedteliste.txt. If you put the file under the src folder, then you have to access it using /src/staedteliste.txt. A Files folder inside the src folder would be /src/Files/staedteliste.txt
Instead of using the the relative path you can use the absolute one by adding e:/eclipse/workspace/ at the beginning, but using the relative path is better because you can move the project without worrying about refactoring as long as the project folder structure is the same.
Just create a folder Files under src and put your file there.
This will look like src/Files/myFile.txt
Note:
In your code you need to specify like this Files/myFile.txt
e.g.
getResource("Files/myFile.txt");
So when you build your project and run the .jar file this should be able to work.
Depending on your Java class package name, you're probably 4 or 5 levels down the directory structure.
If your Java class package is, for example, com.stackoverflow.project, then your class is located at src/com/stackoverflow/project.
You can either move up the directory structure with multiple ../, or you can move the text file to the same package as your class. It would be easier to move the text file.
MJB
Please try this
In eclipse "Right click" on the text file u wanna use,
see and copy the complete path stored in HDD like (if in UNIX "/home/sjaisawal/Space-11.4-template/provisioning/devenv/Test/src/testpath/testfile.txt")
put this complete path and try.
if it works then class-path issue else GOK :)
If this is a simple project, you should be able to drag the txt file right into the project folder. Specifically, the "project folder" would be the highest level folder. I tried to do this (for a homework project that I'm doing) by putting the txt file in the src folder, but that didn't work. But finally I figured out to put it in the project file.
A good tutorial for this is http://www.vogella.com/articles/JavaIO/article.html. I used this as an intro to i/o and it helped.
Take a look at this video
All what you have to do is to select your file (assuming it's same simple form of txt file), then drag it to the project in Eclipse and then drop it there. Choose Copy instead of Link as it's more flexible. That's it - I just tried that.
You should probably take a look at the various flavours of getResource in the ClassLoader class: https://docs.oracle.com/javase/1.5.0/docs/api/java/lang/ClassLoader.html.
I'm not able to give this question an apt title so apology for that.
I am making a modularised application. I load various jar files at runtime and invoke particular method of a particular class (of the jar file) at run time.
The jar file has some supported file. Now my jar file uses another application , lets say abc which is located in the same directory where i have kept the jar file. When i run the jar file then
new File(".").getAbsolutePath()
gives the correct path (this is where abc is located) and program runs fine. But when i load this jar file dynamically and invoke method using reflection above code gives the path of the parent program and abc is not found at that path.
Now my question is how do i find the path in which my jar file exists when i'm running my jar file's code using reflection.
Please let me know if you need more explanation.
Try something like this:
public static void main(String[] args) {
System.out.println(StringUtils.class.getResource("StringUtils.class"));
}
(Note: StringUtils is present on my classpath as a Maven dependency at the time) This gives me:
jar:file:/home/******/.m2/repository/org/apache/commons/commons-lang3/3.4/commons-lang3-3.4.jar!/org/apache/commons/lang3/StringUtils.class
Since the class is in a JAR file, it also gives me the location of the class file within the JAR.