I'm having the following situation.
I have a java programm packed in a jar file. If I call java -jar myProgramm.jar everything is working fine. The file is reading some values from build.xml (ant file). This file is in the same directory where myProgramm.jar is located.
In our company we wrap everything in shell scripts to have a uninfied way to call our scripts.
So my shell script myProgrammWrapper.sh looks like this:
#!bin/bash
java -jar $(cygpath -w ~/path/to/tools/myProgramm.jar) "$#"
The cygpath command is there because the sh is executed within cygwin and otherwise the path would not be found.
The"$#" passes the arguments to the program.
Following is the problem:
Our cygwin environment has been setup in a way that I can call myProgrammWrapper.sh from every directory. But of course when I call it from any random location, the build.xml is not found.
Is there a way to reference the build.xml in the shell script. It is located in ~/path/to/tools/ ?
First I was thinking about copying the build file to the current directory and deleting it afterwards. This is working fine but has one fundamental flaw. We are working a lot with ant and have build.xml files. So if someone would execute myProgrammWrapper.sh in a directory where there is already a build.xml file. It would be overwritten.
Maybe the problem can be tackled from the Java side. Any ideas and input is appreciated.
Why not make the Bash script change the working directory to where the XML file is contained?
Try:
(cd $(cygpath -w ~/path/to/tools/) && java -jar $(cygpath -w ~/path/to/tools/myProgramm.jar) "$#")
Related
shell script file directory: /some/location/myShellScript.sh
Properties-Type: shell script (application/x-shellscript)
EDIT
content of shell script:
#!/bin/bash
export LD_LIBRARY_PATH=`pwd`
echo `pwd`
./someExecutable ../input/cfg/test1.ini
The test1.ini is generated one step before in the java code,
it provides settings for some testing, which is done in the background. Then the shell script ends up with the file I need for further processing.
/EDIT
When I am running this shell script on linux terminal in its own directory just with "./myShellScript.sh" it works perfectly fine...
The part my shell script shall be executed:
//Do something before
//Shell scripts creates a file
String cmd = /some/location/myShellScript.sh;
ProcessBuilder pb = new ProcessBuilder(cmd);
Process process = pb.start();
int exitValue = process.waitFor();
System.out.println(exitValue);
//Afterwards I am processing the generated file
When running my java program as an executable .jar file, this process gets not executed and the exitValue is 127, but I don't know why...
I tried many things like:
using the Runtime to exec
adding #!/bin/bash or #!/bin/sh on top of the shell script
adding a "sh" parameter to the process command in form of String[]
In my execution directory, I changed the permission with chmod 755 -R * recursively so every associated library used by the shell script is indeed available (also due to the fact, that I can just execute it on the terminal).
I really tried to find a proper answer on the internet but I wasn't successful.
And no, I cannot just do everything in java, the shell script is mandatory and cannot be replaced in this case.
Thanks in advance for helpful suggestions!
The script you are executing is highly sensitive to its working directory. It uses pwd to set the LD_LIBRARY_PATH and it attempts to execute another program via a relative path to that program, providing a relative path as a command-line argument, as well.
The working directory for an execution of the script has no essential relationship with the directory in which the script resides -- it completely depends on how and in what context the script is launched. For example, you report that the script works as you expect "When I am running this shell script [...] in its own directory." But when you run the script from Java you very likely are not running it with its own directory as the working directory, and that will strongly affect this script's behavior.
One solution would be to hardcode the script's installation path into the script itself, and to express all your paths relative to that:
#!/bin/bash
installation_dir=/path/to/the/script/dir
export LD_LIBRARY_PATH=$installation_dir
"$installation_dir"/someExecutable "$installation_dir"/../input/cfg/test1.ini
It's a bit kludgy to hardcode the path, though. You could further improve it by having the script identify its own directory at runtime:
#!/bin/bash
installation_dir="$( cd "$( dirname "${BASH_SOURCE[0]}" )" >/dev/null && pwd )"
export LD_LIBRARY_PATH=$installation_dir
"$installation_dir"/someExecutable "$installation_dir"/../input/cfg/test1.ini
That's obviously Bash-specific, but you were using bash anyway. Alternatively, if the executable your script launches is also sensitive to its working directory, then perhaps you just want the script to change directory (which will be effective only for the script and processes downstream from it, not for its parent process):
#!/bin/bash
cd "$( dirname "${BASH_SOURCE[0]}" )"
export LD_LIBRARY_PATH=`pwd`
./someExecutable ../input/cfg/test1.ini
The 127 exit status means that a command used in the script is not found.
EDIT
Debug the script, when bash is used, add the line below on the second line:
exec > /tmp/debug.txt 2>&1 ; set -x
After the next attempt, analyze the traces generated into the /tmp/debug.txt file.
OLD INTRO
(the script content was not yet provided)
The Java program which executes the myShellScript.sh script has probably not the same PATH environment variable than the one which is set in your environment when you execute the script manually from a terminal.
I have a simple .sh script, which runs a Java program:
#!/bin/bash
java -jar Test.jar
However, the Test.jar is referencing a .jar file in the folder. I would like to know, whether it's possible to execute a file, that has been previously uploaded to a website. Reason: The script is sent onto (and then ran from) a server, to which I do not have access, which is why I cannot upload the said .jar file onto it.
I've tried with wget, but that command only downloads the file, rather than uses the said file in the script.
What about
#!/bin/bash
wget http://server/Test.jar
java -jar Test.jar
I'm trying to run a script from an Amazon Linux machine. The script invokes checkstyle like this (in a script called eval.sh):
CHECKSTYLE="java -jar /home/ec2-user/grader/ext/checkstyle-6.15-all.jar"
CHECKSTYLE_RULES="/home/ec2-user/grader/config/checks.xml"
CHECKSTYLE_OUT="quality.log"
"${CHECKSTYLE}" -c "${CHECKSTYLE_RULES}" -f xml -o "${CHECKSTYLE_OUT}" $(find "${_toCheck}" -name "*.java") 2>"quality.err"
When I run this, I get the following error in quality.err:
./grader/eval.sh: line 10: java -jar /home/ec2-user/grader/ext/checkstyle-6.15-all.jar: No such file or directory
I have tried to run the same command directly in the terminal and it is working. Both checkstyle-6.15-all.jar and checks.xml are where they should be.
What could cause this problem?
Change "${CHECKSTYLE}" to ${CHECKSTYLE} (without the quotes).
You are passing the entire value of the CHECKSTYLE variable as a single word (that's what the quotes do), so the shell is looking for a relative directory named java -jar, and is trying to find a file under that (nonexistent) directory with the path home/ec2-user/grader/ext/checkstyle-6.15-all.jar.
When you envoke "${CHECKSTYLE}" the shell thinks that is the command you are running. There is no such file name with the spaces and options have you have included there. If you envoke it simply as ${CHECKSTYLE} (drop the quotes) the shell will process it for whitespace as normal and split it into the appropriate pieces for creating the process.
Essentially what I'd like to achieve is add a .jar file to the startup of my Linux computer. I created a script that would run the file and then tried adding that script to run when the system boots up.
I have a .java file on my desktop named Box.java. The file contains no errors and I manually compiled it and it was working fine. I then created a script on my desktop called start.sh and it's contents are
#!/bin/bash
javac /home/maple/Desktop/Box.java
So what this should do is compile the java class and the result would be a class file on my desktop called Box.class
I then created a file in /etc/init.d/ and it is called **start_java* it's contents are
#!/bin/sh
home/maple/Desktop/start.sh
I then opened up terminal and did
chmod +x /etc/init.d/start_java
I know the sh file will compile a java file and not run a file, in the completed copy I will be doing it with a jar file. How can I add that jar to startup without using a 3rd party software or any pre-installed programs?
EDIT: The current way I have it is not working.
You just want to do the same thing you already did, but changing the content of start.sh to:
#!/bin/bash
java -jar <filename>
In a bash shell script I tried these two versions:
java -jar abc.jar&
and
CMD="java -jar abc.jar&"
$CMD
The first verison works, and the second version complains that abc.jar cannot be found. Why?
Commands do run from current directory in a shell script.
This is why the first command in your test script worked.
The second command may not work because either java isn't in your ${PATH} or abc.jar isn't in your ${CLASSPATH}. You could echo these environment variables or set +x to debug your bash script.
Bash (and others) won't let you do backgrounding (&) within the value of a variable (nor will they let you do redirection that way or pipelines). You should avoid putting commands into variables. See BashFAQ/050 for some additional information.
What is the actual error message you're getting? I bet it's something like "abc.jar& not found" (note the ampersand) because the ampersand is seen as a character in the filename.
Also, the current directory for the script is the directory that it is run from - not the directory in which it resides. You should be explicit about the directory that you want to have your file in.
java -jar /path/to/abc.jar&