I am trying to figure out how I may be able to setup my classpath file for java so that it could accept a jar and launch a class file not in the same directory as the one the user is currently in.
My code below shows java -cp .:jsoup-1.7.3.jar Class1 will allow me to launch the Class1 file without any problems if I am in the same directory as the Class1 file which also contains the jsoup file. However I am trying to launch it as if I were in another directory so I would try something like
java -cp .:jsoup-1.7.3.jar /pathtofile/Class1
However this doesn't work because you need to set the classpath correctly. How would I do this so I could do something like
java -cp {correct absolute class paths and arguments to include jsoup jar} Class1
Thank you in advanced.
"." means the current directory in Linux. so you'll need to substitute (or add) the path to where your .class file is.
Related
I have a jar that is reading a file using below code:
Thread.currentThread().getContextClassLoader().getResource(fileName);
I want to run this jar using java -jar .jar command but I want to keep this file outside my jar, so that I can edit the jar file later on without touching the jar. Can anyone help me, how to run this jar so that it will pick up the file from outside.
There are multiple approaches you can do that and it will depend on where would you like to place this external file. For the sake of this answer, I will refer to this file as config file
Not In The Same Directory
The first approach is where you will need to place this file outside the JAR and not necessarily next to the JAR file in the same directory. In that case, you can pass the file location of the config file using either an environment variable (if you are running the JAR in a shell for example) or a Java property.
To use an environment variable, assuming you are using some Linux distro, then you can use the export command to set the value; something like this:
$ export CONFIG_FILE_LOC=/etc/myapp/config.file
You can then read the value in your code using the System class by using the following code:
String fileLocationEnv = System.getenv("CONFIG_FILE_LOC");
Alternatively, you can set this as a property by adding the following segment to your launch command:
$ java -Dconfig.file.location=/etc/myapp/config.file -jar myapp.jar
You can then read the value in your code using the System class for properties using the following code:
String fileLocationProp = System.getProperty("config.file.location");
In The Same Directory
If you need the config file to co-exist in the same directory as your JAR file, then you can use the following code to get the JAR directory and then append the filename to it. Here's the code (assuming a class named MyApp)
try{
new File(MyApp.class.getProtectionDomain().getCodeSource().getLocation().toURI());
} catch(URISyntaxException exception){
System.out.println("Exception");
}
Hope that helps.
To open the file as a resoure, add the folder containing the file(s) you want to use, to your classpath:
java -classpath .;config -jar myjar.jar
This example adds the current directory and the config directory to your classpath.
Multiple folders can be specified by using a separator. On windows use ';' , on unix use ':' .
To open the file as a File, you can just use
new File("configfile")
which will look in the working directory (directory where you launched your java)
I am trying to compile and run some java files I have made in Eclipse. The full path to the .java file is C:\Users\MYNAME\Documents\Java\Introduction\src\tests\Test.java. tests is the package I created in Eclipse and src is a folder that Eclipse made under Introduction (which is the project name).
In my environment variables, I have the following relevant variable:
JAVA_HOME C:\Program Files (x86)\Java\jdk1.7.0_40\bin
Under system variables I have the following:
CLASSPATH %JAVA_HOME%
I go to my cmd and cd into the tests directory (cd C:\Users\MYNAME\Documents\Java\Introduction\src\tests). Then I compile using javac Test.java. This seems to work as I then have a Test.class file under the same directory. Now I want to run the file, I type java Test and I get the error, "could not find or load main class". I've tried a variety of things including appending .class and .java to the end but I keep getting the error. I looked at some answers and docs and I managed to get it to work if I cd into:
cd C:\Users\MYNAME\Documents\Java\Introduction\src (i.e, get out of the package)
and then run:
java -cp . tests.Test
So that seems to temporarily set the class path to the current directory, and run Test from the package tests. However, I want to simply be able to type java Test. I know it's possible as I used to be able to do it, but now for some reason I cannot (I must have changed something along the way...).
Any help is appreciated.
However, I want to simply be able to type java Test
That will only work if Test is in the default package - it's as simple as that. You need to pass the java executable the fully-qualified name of the class you want to launch. There's no way round that.
Of course, you could create your own launcher which looks in the current directory for class files, finds out the fully-qualified name of the classes within those files, and launches java providing the full name and probably specifying an appropriate classpath... but that seems like a lot of hassle compared with just including the package name in the command.
You could be making the same mistake I made. So, try the following.
Here is my code for your reference.
class A{
public static void main(String args[]) {
System.out.println("Hello world");
}
}
Once you saved this as "C:\JavaStudy\ClassA.java", try the following.
c:\JavaStudy>javac ClassA.java
c:\JavaStudy>java A.class
Error: Could not find or load main class A.class
c:\JavaStudy>java A
Hello world
c:\JavaStudy>
Note: You don't need to use " java.exe -cp . " if you have class file in the same directory from where you are executing.
I am trying to compile and run simple Java program. This program basically prints out hello world phrase. I am not specifying -cp option and I don't have CLASSPATH environment variable. Hence the user classpath is limited only to current directory.
Now, compilation works beautifully.
rustam#rustam-laptop:~/temp/bird_test$ javac Sparrow.java
This command produces needed .class file. The weird stuff happens when I try to run .class file. The following command works good.
rustam#rustam-laptop:~/temp/bird_test$ java Sparrow
But when I try the following command
rustam#rustam-laptop:~/temp/bird_test$ java ./Sparrow
I receive the following error:
Error: Could not find or load main class ..Sparrow
WTF! i thought that symbol ./ refers to current directory.
java takes a class name as argument. It doesn't take a file path. The class name (Sparrow) is then resolved by the java class loader to a .class file based on the classpath, i.e. it looks for a Sparrow.class file in every directory and jar listed in the classpath.
Let's take an example that respects good practices, and thus doesn't use the default package:
package foo.bar;
public class Baz {
...
}
The class name of the above class is foo.bar.Baz. To execute it, you must use
java foo.bar.Baz
and java will look for a foo/bar/Baz.class in all the directories listed in the classpath. So if the classpath is set to /hello/world, it will look for the file /hello/world/foo/bar/Baz.class.
I am running the program as:
java -Dfirstarg=$1 -cp $APP_HOME/someFile.cfg;$APP_HOME/lib/*.jar :: Handler
With the intention to pass the argument and include someFile.cfg and all jars under $APP_HOME/lib/ to class path.
What am i doing wrong please?
You don't need to pass the someFile.cfg into the class-path. You just need to add the libraries, and current directory there.
I would do:
java -Dfirstarg=$1 -cp .:$APP_HOME/lib/*.jar YourMainClass
Assuming you are in a directory where your .class files are.
Also, do keep in mind that in linux you use : and not ; to separate your class path entries in the -cp arguments.
How can I create the class file and jar file for this coding, when I compile this program its not working because there is no main function in the program. And also I am trying in command prompt but I don't know how to set the classpath? please help me
My Coding is here
public class NewLogFields implements ILogNotify
{
public void onLog(Level level, String comment, IMediaStream stream, String category,String event, int status, String context) {
if (category.equals(WMSLoggerIDs.CAT_session) && event.equals(WMSLoggerIDs.EVT_destroy))
{
Long csBytes = (Long)WMSLoggerFactory.getGlobalLogValue(WMSLogger IDs.FD_cs_bytes);
Long scBytes = (Long)WMSLoggerFactory.getGlobalLogValue(WMSLogger IDs.FD_sc_bytes);
System.out.println("disconnect: csBytes:"+csBytes+" scBytes:"+scBytes);
}
}
}
Once you compiled the coding in wowza media Serever the jar file is automatically created in the library folder,see your Installation Library folder.Still you have problem Gothrough this link Wowza Quick Guide
What do you want to do?
Create a class and an jar file out of this Java code so that you can use this in another Java program?
Then you have to compile it:
java NewLogFields.java
Looks like you are unable to compile it at all. This could be because the interface ILogNotify (or the jar that contains this) is not in the classpath.
You can include the path/jar containing this interface in the classpath by using:
javac -cp .;path_to_jar_or_class NewLogFields.java
where path_to_jar_or_class is the path to the folder or jar file that contains ILogNotify.
For example, this may be something like ./logNotify.jar or ./log/
You can set use switch -cp or -classpath with javac command.
for example javac -cp path and name of jat file or class file yourjavafile.java
create the class file using the compiler: javac NewLogFileds.java
create the jar file using the jar command: jar cvf stuff.jar NewLogFileds.class
You are correct that the program needs a main() function in order to run.
add:
public static void main(String args[]) {
// code here
}
With that you could run the code with or without the jar:
java NewLogFields
or
java -cp stuff.jar NewLogFields
There are ways of creating a MANIFEST file that tells java which class to run from the jar making the last line more simple.
The link that you provided tells you how to do it:
Compile your class in the normal way.
Create a JAR file containing your class in the normal way.
Copy the JAR file into the wonza installation's lib as described in the javadoc.
Edit the startup script to add the -Dcom.wowza.wms.logging.LogNotify=... option to JAVA_OPTS ... as described in the javadoc.
The "full-path-to-your-ILogNotify-class" is actually supposed to be the fully qualified class name of your class; it is obvious from the examples.
Edit WowzaMediaServerPro-Service.conf and log4j.properties as described in the javadoc.
Restart the server.
If you put your JAR file in the directory like the instructions tell you to, you won't need to modify the classpath by changing -cp argument.
Your class doesn't need a main method because it is not a free-standing application. Rather, it is a "plugin" class that gets dynamically loaded and instantiated by the Wowza core as required. The "-D..." option and the config file change tell the Wonza core which class to try to load.