From my application I have to execute an external jar of which I do not have the source.
Given an input file, it processes it, creates an "output" directory and puts in it an mxml output file. Problem is: it creates said directory in tomcat/bin instead of inside the directory of the original file.
Here's what I've tried so far.
Initially
Process p = new ProcessBuilder("java -jar "+newfile.getParent()+"\\converter.jar "+newfile.getPath()+" -mxml").start();
Then, seeing how from console the "output" directory was created in the directory the command was called from, I tried:
String startSim[] = {"cd "+newfile.getParent()+"\\" , "java -jar converter.jar "+newfile.getName()+" -mxml"};
try {
Runtime.getRuntime().exec(startSim).waitFor();
} catch (Exception e) {
e.printStackTrace();
System.out.println("Log non creato.");
}
But with this I get the "file not found" exception for the first instruction. Does anyone know how to possibly solve this problem? I'd like to avoid having to reach for my output file all the way in my tomcat/bin directory.
Thanks for any suggestion!
Paolo
P.s.: by the way, before trying all this I tried simply calling the method I need from the library, but had the same exact problem. So I resolved to execute the jar, instead. And here we are. :)
You can set working directory using ProcessBuilder.directory() method:
ProcessBuilder pb = new ProcessBuilder();
pb.directory(new File("mydirectory"));
pb.command(......);
etc
This does not work for you when you are using Runtime.exec() because cd command is a functionality of shell. You could solve it using this technique but you have to create platform specific command with prefix like cmd /c on windows or /bin/sh on Linux. This way is definitely not recommended.
But in your specific case you do not neither first nor second solution. Actually you are starting one java process from another. Why? you can easily invoke the main() method of the second process directly.
Take a look on META-INF/MANIFEST.mf file from converter.jar. Field Main-Class contains the fully qualified name of main class. Let's say it is com.converters.Main (just for example). In this case you can invoke
com.converters.Main.main(new String[] {newFile.getPath(), "-mxml"});
directly from your code. Just add the jar to your classpath.
Concerning to changing working directory in this case. Check again whether you really need this or your converters.jar supports parameter that does this.
A lazy approach to this may be going to the root directory and descending from there to your tomcat bin directory .
Related
I am using a .jar file, but unfortunatley as a black box, i.e. I do not know what exactly is in there nor how it all works.
I am sending commands to the Mac terminal from a Python script. I enter the following command:
java -jar jarfile.jar req_data /abs_path/to/required/data input path/to_my_/input/file.txt
This does what I need: analyses input using the 'black box' and creates and new file with analysis output. This new file is created in the folder where jarfile.jar is located.
I want to have this file put somewhere else upon creation.
I have tried using the > operator, specifying a path, e.g.:
java -jar jarfile.jar req_data /abs_path/to/required/data input path/to_my_/input/file.txt > /output/path/
this created a file in my desired location, but it was simply the message from Terminal, saying "The operation was carried out successfully" - the analysis results file was created in the same folder as before.
I tried %*> too, but it threw an error.
As a poor workaround I now have a function, which retrospectively finds and moves all the newly created files (analysis output) to my desired folder.
Is there a way to control the output files with the command line within the original command? Or is it something that is specified somewhere in my jar file? My problem is that editing it is not allowed.
I'm new to python. However, I may suggest to try few things, if they can work for you. Apology me, if does not work! I believe that you have already done the following step:
import subprocess
subprocess.call(['java', '-jar', 'Blender.jar'])
Like, if you have a properly configured jar path, then can run jar directly.
Secondly, look at the cwd parameter (is used for executable). Include a cwd param as x
def run_command(command, **x):
with subprocess.Popen(command,...., **x) as p:
for run_command specify the path of either the working directory (possibly it should be) or the full system path. I'm not sure, just try both.
for outputline in run_command(r'java -jar jarfilepath', cwd=r'workingdirpath', universal_newlines=True):
print(outputline, end='')
Alternatively, you can try to run command from the directory in which you wish to store output file. Try: run the popen as
subprocess.Popen(r'directory of running command', cwd=r'workingdir')
where workingdir could be your current directory path.
If it does not work, try without r'. If still does not work, try doubling slash in the path like (C:\\ abc\\def)
I am running windows 7 and have an exe file I am trying to run using the following command in Java:
File dir = new File("C:\\PATH\\TO\\DIR");
String[] cmdArray = {"file.exe"};
if(dir.exists()){
for(String s : dir.list()){
if(s.equals(cmdArray[0]))
System.out.println("File exists!");
}
}
Runtime.getRuntime().exec(cmdArray,null,dir);
The exec(...) command, however, gives me this error:
Cannot run program "file.exe" (in directory "C:\PATH\TO\DIR"): CreateProcess error=2, The system cannot find the file specified
The above code does print out "File Exists!" so the file is there and Java knows it is there. I dont know how this is possible.
Also, I need to use Runtime.getRuntime().exex(...). This call is actually inside another method that I am not suppose to change so alternative solutions will not work for me. Thanks in advance!
EDIT: I ran the file in a cmd window without admin privileges and got the following message:
Error in Opening Configuration File in Read Mode
So I assume I have permission issues. How would I change the permissions in order for me to be able to run it?
EDIT2:
I tried changing cmdArray to this:
String[] cmdArray = {"C:\\PATH\\TO\\DIR\\file.exe"};
and it worked like a charm. One question though. Why did this work but not the other way? I assumed that Runtime.exec() ran at a high level like so in a cmd wibndow:
pushd dir
cmdArray[0] cmdArray[1] cmdArray[2] ... cmdArray[length - 1]
popd dir
Is this the wrong assumption?
Perhaps, You will need to include your directory to your path. You may take a look at this question and this tutorial
You are having the permission issue in the C drive.Check by executing "file.exe" manually in the C drive.Better solution will be to keep "file.exe" in some location other than C drive and execute from there.
I'm trying to develop a website that takes user input and converts to a text file. The text file is then used as an input for a .jar file. (e.g. java -jar encoder.jar -i text.txt), the jar then outputs a .bin file for the user to download.
This jar is designed to be run from command line and I really don't know the best way to implement it within a .jsp page. I have created a few java test classes but nothing has worked so far.
Does anyone have any suggestions on possible methods?
An alternative to running it as an external process is to invoke its main class in the current JVM:
Extract/open META-INF/MANIFEST.MF of the jar
Identify the Main-Class:. Say it is called EncoderMainClass
Invoke its main method: EncoderMainClass.main("-i", "text.txt")
This aught to be faster because a new OS process does not need to be created, but there may be security considerations.
Have you tried somrthing like this,
Create a java file
use a ProcessBuilder and start a new JVM.
Here is something to get you started:
ProcessBuilder pb = new ProcessBuilder("/path/to/java", "-jar", "your.jar", "thetextfile.txt");
pb.directory(new File("preferred/working/directory"));
Process p = pb.start();
ps: do handle to destroy process else it will eat up all memory
You can put this jar to the web application on the classpath and use it's class and methods. Better if you have javadoc if you don't have sources. But even if not in classpath you can try the example or this example.
I have a specific need to unrar files found in different subdirectories during execution of my program on Os x. I do this by calling the freeware command line tool unrar, which works very well.
However, unrar will always unrar files in the current working directory, which means that I have to specify a working directory for the executed process or get every unpacked file in my .jar home folder.
This is easy enough to do using either the processBuilder.directory(dir) or runTime.getRuntime().exec(args,null,dir) for example where dir is a File.
This works excellent but not when the working directory contains a space. As a short example:
File dir=new File("/Users/chargedPeptide/rar 2");
String[] cmd = { "/Users/chargedPeptide/rar/unrar", "e", "-o+","/Users/chargedPeptide/rar", "2/weather.rar"};
Process pr = Runtime.getRuntime().exec(cmd,null,dir);
int exitValue=pr.waitFor();
Will not work, while using: Process pr = Runtime.getRuntime().exec(cmd); instead will launch the command successfully but leave me with all of the files in the jars working directory.
Using processbuilder instead and using processbuilder.directory(dir); to set the directory
exits with the message:
Exception: Cannot run program "/Users/chargedPeptide/rar/unrar" (in directory "/Users/chargedPeptide/rar 2"): error=2, No such file or directory
Help? How do I handle the spaces? I've tried adding backslashes before the spaces to make them literal no help since the File object treats them like actual part of the path.
Edit: To make the whole thing a bit more clear:
1. I have a separate method that feeds the execute method a command and a directory to processbuilder, all directories are found by the previous method and exist. This works except when the dir contains a space.
2.I need to set the working dir or the command will execute in the wrong place.
3.Parsing the file object by:
dir=new File(dir.tostring.replace(" ","\ ");
to put a backslash in front of all spaces does not work since the java File object then looks for a path containing actual backslashes. No luck.
4.Using rt.exec instead of processbuilder dosen't help either.
Any ideas most welcome.
How about:
dir.mkdirs();
before launching the process.
This creates the missing directory.
Edit:
This looks strange.
String[] cmd = { "/Users/chargedPeptide/rar/unrar", "e", "-o+","/Users/chargedPeptide/rar", "2/weather.rar"};
If this was a shell you'd write:
unrar e -o+ "/Users/chargedPeptide/rar 2/weather.rar"
You actually have to put the .rar file in quotes, since otherwise it will be interpreted as 2 arguments to the process.
The way you've split 'cmd' will do exactly that, break the rar argument in two. Try:
String[] cmd = { "/Users/chargedPeptide/rar/unrar", "e", "-o+","/Users/chargedPeptide/rar 2/weather.rar"};
Not sure if it will work, but can you try putting a / at the end of the path.
i.e.,
File dir=new File("/Users/chargedPeptide/rar 2/");
I'm making an editor-like program. If the user chooses File->Open in the main window I want to start a new copy of the editor process with the chosen filename as an argument. However, for that I need to know what command was used to start the first process:
java -jar myapp.jar blabalsomearguments // --- need this information
> Open File (fileUrl)
> exec("java -jar myapp.jar blabalsomearguments fileUrl");
I'm not looking for an in-process solution, I've already implemented that. I'd like to have the benefits that seperate processes bring.
Since you are launching Java -> Java, you can use the existing classpath to set the classpath on the command line. This type of thing works really nice in the dev environment too.
ProcessBuilder selfLauncher = new ProcessBuilder(
"java", "-cp", System.getProperty("java.class.path"),
"com.my.mainClass" );
selfLauncher.start();
Update:
For executable jar files, you will have a classpath which is simply the relative path to the jar file itself. If you want the command line arguments, you will have to save them from main, and re-apply them when launching.
You can see this by packing the following program into a jar. I'm not actually sure what happens if you have jars inside the executable jar file. They probably show up in the classpath.
public class TestJarPath {
public static void main(String args[]) throws Exception {
for (String s : args)
System.out.print("[" + s + "] ");
System.out.println();
String cp = System.getProperty("java.class.path");
for (String s : cp.split(";"))
System.out.println(s);
}
}
For java -jar ..\tst.jar X, you get output like:
[X]
..\tst.jar
If all else fails, try writing a batch/shell script to launch your app. In windows you can pass %CmdCmdLine% to Java to get the entire command line.
See http://www.robvanderwoude.com/parameters.php
As far as I know is there no portable way to get this info. I found a property in the gcj runtime but I doubt this will cover a large percentage of the users.
I think the accepted practice is "Try and Pray" :
Hope it is on the path, (the path IS available, so that can be checked)
if not, check if JAVA_HOME is defined, and use that to find java.
if not check in the most likely places on all OS's you have received bug reports for.
Well, it is messy... porbably best to check for JAVA_HOME and the path and ask the user to configure a JVL explicitely if that fails.