I'm currently working on a program to zip and unzip files/directories and it behaves kind of weird when I call it over the git bash.
The program takes three arguments (zip/unzip, inputPath, outputPath).
For Example:
java -jar zip_unzip.jar --zip H:\\zip_test H:\\test5\zip_test_4.zip
Everything works fine when I call it over Eclipse or CMD. It creates the directory structure if it doesn't already exist and zips the input Folder into the newly created output folder. But when I call it over the git bash like in the example above it somehow "ignores" the backslash and instead of creating a folder called test5\, creates a zip-archive called test5zip_test_4.zip
Here's a snippet of the code that takes care of creating the directory structure, where zippedFolder is the outputPath-Parameter:
File directoryToZip = new File(inputFolder);
String targetZippedFolder = zippedFolder;
targetZippedFolder = targetZippedFolder.replace("\\", "/");
//create directory to store archive in, if it doesn't already exist
File destDir = new File(targetZippedFolder.substring(0, targetZippedFolder.lastIndexOf("/")));
if (!destDir.exists()) {
destDir.mkdirs();
}
In line 3 of the codesnippet I'm replacing every backslash with a forwardslash because I thought this would make it platform-independent.
Can someone explain the behavior of the git bash to me and maybe suggest a more platform-independent path handling method?
Related
Recently, I have switched to Ubuntu v20.04 from Windows 10 Pro v2004 because of performance purposes. When, I was on Windows I can freely compile a java project from another java program by writing:
String pathToCompiler = "\"C:/Program Files/Java/jdk-14/bin/javac\"";
Process compileProcess = Runtime.getRuntime().exec(pathToCompiler+" -d bin #.sources", null, new File("ProjectPath"))
Where the sources file is a file containing the list of classes of the project
The code above works successfully on Windows 10.
But On Linux(Ubuntu):
if I substitute the value of variable pathToCompiler as
pathToCompiler = "\"/usr/lib/jvm/java-11-openjdk-amd64/bin/javac\""
the below exception is raised up and the program executing the command exits:
"/usr/lib/jvm/java-11-openjdk-amd64/bin/javac" -d bin #.sources
java.io.IOException: Cannot run program ""/usr/lib/jvm/java-11-openjdk-amd64/bin/javac"" (in directory "/home/arham/Documents/Omega Projects/Project0"): error=2, No such file or directory
at java.base/java.lang.ProcessBuilder.start(ProcessBuilder.java:1128)
at java.base/java.lang.ProcessBuilder.start(ProcessBuilder.java:1071)
at java.base/java.lang.Runtime.exec(Runtime.java:592)
at java.base/java.lang.Runtime.exec(Runtime.java:416)
at ide.utils.systems.BuildView.lambda$3(BuildView.java:267)
at java.base/java.lang.Thread.run(Thread.java:834)
Caused by: java.io.IOException: error=2, No such file or directory
at java.base/java.lang.ProcessImpl.forkAndExec(Native Method)
at java.base/java.lang.ProcessImpl.<init>(ProcessImpl.java:340)
at java.base/java.lang.ProcessImpl.start(ProcessImpl.java:271)
at java.base/java.lang.ProcessBuilder.start(ProcessBuilder.java:1107)
The problem is that the file actually exists but it says No Such File or Directory
Actually, The program which is compiling the project is a Java IDE that I am creatiing.
Someone please tell if he/she knows how to fix this bug
The Runtime.exec method has several problems that make it difficult to use, and this is one of them. Use the newer ProcessBuilder class instead.
String pathToCompiler = "C:/Program Files/Java/jdk-14/bin/javac";
Process compileProcess = new ProcessBuilder(pathToCompiler, "-d", "bin", "#.sources")
.directory(new File("ProjectPath"))
.start();
The differences are:
Remove the extra quotes from around the path to the executable. If quoting is needed, the system takes care of it.
Pass the each command line arguments as a separate string. This way you don't have to worry about quoting.
Update the path to the following:
String pathToCompiler = "/usr/lib/jvm/java-11-openjdk-amd64/bin/javac/";
I am trying to run a yarn job from a java wrapper program. The mapreduce jar takes two inputs:
A header file: I dont know the name of the file but the location and file extension and there's only one file at that location
A Input files directory
Apart from these I have an Output directory.
the processbuilder code looks like:
HEADER_PATH = INPUT_DIRECTORY+"/HEADER/*.tsv";
INPUT_FILES = INPUT_DIRECTORY+"/DATA/";
OUTPUT_DIRECTORY = OUTPUT_DIRECTORY+"/";
ProcessBuilder mapRProcessBuilder = new ProcessBuilder("yarn","jar",JAR_LOCATION,"-Dmapred.job.queue.name=name","-Dmapred.reduce.tasks=500",HEADER_PATH,INPUT_DIRECTORY,OUTPUT_DIRECTORY);
System.out.println(mapRProcessBuilder.command().toString());
Process mapRProcess = mapRProcessBuilder.start();
On run, I get the following error:
Exception in thread "main" java.io.FileNotFoundException: Requested
file /input/path/dir1/HEADER/*.tsv does not exist.
But when I run the same command as :
yarn jar jarfile.jar -Dmapred.job.queue.name=name -Dmapred.reduce.tasks=500 /input/path/dir1/HEADER/*.tsv /input/Dir /output/Dir/
It works all fine.
what can be the issue when running the command from java is causing this issue?
The * is being treated as part of the literal string in this case rather than a wildcard. Therefore globbing isn't expanding to your desired path name.
If there is only one file in the directory, why don't you find what its path is and pass that as the argument instead
eg.
File dir = new File(INPUT_DIRECTORY+"/HEADER);
if (dir.list().length > 0)
String HEADER_PATH = dir.list()[0].getAbsolutePath();
I have a DIY cartridge. My project structure looks like
MY_PROJECT
-diy
-myProgram.jar
-resources
-file1, file2...
-.openshift
-action_hooks
-start
The myProgram.jar uses files from the folder 'resources'.
The code looks like
File imageFolder = new File("resources");
System.out.println("Image Folder read:"+imageFolder.canRead()); //canRead returns false
File[] listOfFiles = imageFolder.listFiles(); // here I get null
The program runs by action hook 'start':
nohup java -jar $OPENSHIFT_REPO_DIR/diy/*.jar --server.port=${OPENSHIFT_DIY_PORT} --server.address=${OPENSHIFT_DIY_IP} &
The problem is that I'm not able to work with files.
As described in code comments I get null on listFiles().
If I run the program on Openshift manually(ssh to server/$OPENSHIFT_REPO_DIR/diy/ and run java -jar ...) it works, but it doesn't work via action_hooks.
Thank you in advance!
I resolved the issue with Openshift env variable 'OPENSHIFT_REPO_DIR'.
Instead of using relative path
new File("resources");
I use absolute
String absolutePath = System.getenv("OPENSHIFT_REPO_DIR");
new File(absolutePath + "diy/resources")
I use the following code in a views.py
def berkeleyParser(infile,outfile):
cmd="java -Xmx1024m -jar nlptools/BerkeleyParser/berkeleyParser-1.7.jar -gr nlptools/BerkeleyParser/chn_sm5.gr < "+infile+" > "+outfile
os.system(cmd)
and then call this function to use berkeley parser.
I think the file path is ok, because the jar can successfully create the output file.
Meanwhile, I use a independent .py code to run the code above (with path modified), and got correct result in output file.
So, I don't know what's wrong with it.
In my war, I have file exe in WEB-INF\classes\
How can I execute this file in Java code (How can I specify path to this file) ?
command = " ? ";
Process x = p.exec(command);
Te following approach could work:
1) Prepare full path of your executable:
ServletContext context = getContext();
String fullPath = context.getRealPath("/WEB-INF/classes/executable");
2) Execute like you would normally do it:
String[] cmd = { fullPath /*[...] arguments */};
Process p = Runtime.getRuntime().exec(cmd);
p.waitFor();
This is a simplified example; you may also want to read more about ProcessBuilder.
This is bad idea. Imagine simply fact that your .war packege should run on almost any server (".war is platform independend") and your .exe file is compiled just for one architecture.
Better should be execute your .exe as external program just for separate platform independent and platform dependent part. Then in java you can test operating system and on this basis run desired externel programm.
Read this link with similar question.
The best way to do find a file's real location inside a web app is to use the ServletContext.getRealPath (see http://docs.oracle.com/javaee/6/api/javax/servlet/ServletContext.html#getRealPath(java.lang.String))
You can access that object from the session...