I have created a jar file , it reads from standard input and print them out to the standard output.
Below is how you can use the app:
cat /ephemeral/test/input | java -jar /ephemeral/test/Hello.jar >> /ephemeral/test/output
I am wondering is it possible to add my Hello.jar to be a Linux Command, let's say -hello-:
So next time I could just do something like:
hello /ephemeral/test/input /ephemeral/test/output
I am not that much familiar with Linux Soft Link and Java, so any help or information would be appricated!
UPDATE:
thanks for reno's inspiring answer, I put the cat ... | .. >> .. into a shell script hello.sh.
Now I can do something like:
./hello.sh /inputpath/input outputpath/output
(Note: you should not use $0, $0 is actually the name the file itself instead of input arguments)
Now, the last question is how could you access the shell script EVERYWHERE, like how you call basic shell cmds: ls..cat..echo..
I have read a tutorial described exactly what I want(reach to the end of the tutorial directly):
I did add the path of the directory where the hello.sh exists into the path variable.
I created a bin folder which is like mkdir /home/ubuntu/bin and put my shell script there.
However, I still have problem do what the author described:
hello input output (in my case)
Does anyone know what is going wrong here?
you can write a shell script say test.sh and copy your code to test.sh
!/bin/bash
cat $0 | java -jar /ephemeral/test/Hello.jar >> $1
and execute the script by passing your input and output file name something like this:
test.sh /ephemeral/test/input /ephemeral/test/output
The way to do this is to write a shell script (usually hello.sh) that contains the necessary java -jar command.
Related
I'm using Apple's automator to create a Shell Script. I can get it to run if I change directory specifically to where the jar file is. But what if I want to change to directory to wherever the Shell script happens to be running?
Right now I have the following, which works:
cd desktop/CommonDenom/
java -XstartOnFirstThread -jar CommonDenom.jar
I know there's a way to target whatever directory the Shell script is launched from, but I can't seem to get anything to work. Please be specific with instructions as I havent been using Automator very long. Unless someone can specify how ot writ ethe script without Automator. Thanks in advance.
A standard idiom for this in shell scripts is dirname $0. The $0 variable is the path to the script that was executed, and the dirname command takes a path and strips off the last component to leave the path to the containing directory
cd "`dirname $0`"
Just wanted to weigh in here. I've seen some people with this problem. If you are JUST on OSX and making your own scripts. This will do the trick :) kind of a hack, but works like a charm.
#! /bin/bash
sudo /Applications/XAMPP/xamppfiles/xampp startapache
open /Applications/XAMPP/htdocs
#!/bin/bash
DIR=$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )
cd "${DIR}"
...the rest of your script...
Credits to Ian C. on AskDifferent: https://apple.stackexchange.com/a/179064
I am not sure you mean "you want to change directory to wherever the script is run from" because you will be in the directory the script is run from when you start anyway, so there will be no need to change directory there. I think you mean you want the script to stay wherever it starts, yet still be able to find the jar file.
In which case, you probably jus need the following without any changing directory:
java -XstartOnFirstThread -jar ~/Desktop/CommonDemon/CommonDemon.jar
I have created one java program on my Linux system which indents and formats the given file. I want to make that program work like a command in Linux which will take file names and other options as arguments and then will produce the output. I can do this with a C program by just copying the compiled executable in /bin folder but I don't know how to do it with java.
Sample script that can might further help-
#!/bin/bash
#Set whatever number of arguments you expect for the Java jar you have
ARGS_EXPECTED=3
if [ $# -ne $ARGS_EXPECTED ]
then
echo "[$HOSTNAME]: Usage: `basename $0` filename arg1 arg2"
exit 1
fi
java -cp yourfile.jar com.yourpkg.Driver $1 $2 $3
Save the above content to a file, say test.sh
and use the command to give an executable permission chmod +x test.sh
run like ./test.sh filename arg1 arg2 from current directory where test.sh is
I thing this can be useful for your case: http://git.kernel.org/cgit/linux/kernel/git/torvalds/linux.git/tree/Documentation/binfmt_misc.txt?id=HEAD
You can simply run a Java class file or jar file with "java" command from command line. Example:
java -jar yourprogram.jar argument1 argument2
If you save this line in a text file saved for example as "script.sh" and then give it the permission for execution you can run it double clicking or from terminal typing ./script.sh in the same folder containing the file script.sh.
You can also produce scripts that use arguments with $1 $2 etc. avoiding the need of editing file.
http://www.linuxquestions.org/questions/linux-newbie-8/how-to-pass-command-line-parameter-to-shell-script-254396/
You can use named parameters, too.
You can also produce a C program for a new command like you suggested that run the "java" command. In this case you can introduce arguments directly from terminal and pass them to java command in the C source.
As others have pointed out it is probably best to use a small shell script to run the Java application. There are several open source products that will help you wrap your Java code to produce a runnable (set of) .jar(s).
If you have correctly separated your business logic from your interface (as you should) then it is probably best if your Java application parses the parameters given on the command line interface. To do this create a separate class for parsing such parameters and calling the classes making up the business logic. Of course this will lead quickly - if not immediately - in writing a parser for Linux like CLI parameters. When this happens you may wish to consider the Apache Commons CLI project.
If you don't want to use any wrapping application/runtime, my method is generally pointing to all the class file containers in the classpath and directly pointing to the class containing the static main method:
java -cp "path_to_jar;path_to_class_folder;etc" "nl.owlstead.stackoverflow.LinuxMain"
I have an application that reads a text file.
If the application reads it from stdin, then I could forward the input / pipe it
$ app < input.txt
$ cat input.txt | app
Unfortunately, the application expects a filename as an argument..
$ app --input input.txt
and I can't easily change the source of input (eg. swap local file [as in example] for a result of a wget).
Is there an option to somehow trick this app?
If a solution requires a temporary file, it is necessary that this file is not left on the FS..
The app is actually an executable JAR; I'd like the trick to run on Win/Lin..
I am not sure if it could work for you, but I have already solve a similar problem with a fifo.
http://linux.die.net/man/3/mkfifo
This could give you the indirection between your app and your source.
mkfifo fifo.input
app --input fifo.input
Now you can fill the pipe with different sources.
cat input > fifo.input
ctrl-c
cat otherInput > fifo.unput
Hope it solve your problem on Linux. So far I know it does not exist on Windows.
You want to do something like this.
java -jar myjar.jar arg1 arg2
where myjar.jar is the executable jar and arg1 arg2 etc are parameters the program expects.
So suppose you want to pass input.txt to your application and its the only commandline parameter required then command will be
java -jar myjar.jar input.txt
I was trying to execute my java program through shell script so I wrote:
java -jar $(pwd)"/test.jar"
It worked flawlessly but when I turned to the code below:
PATH=$(pwd)"/test.jar"'
java -jar $PATH
Then I got an error: "Run.sh: 3: java: not found"
(Running on Ubuntu)
I have very little experience in shell script so please let me know what's wrong with it. Thanks.
PATH is a special environment variable which the shell uses to find executables. You've changed PATH to point at test.jar, so now the shell can't find java.
Call your variable something else.
Example:
LIB_PATH="$(pwd)/test.jar"
java -jar ${LIB_PATH}
The value in $(PWD) depends on the directory the script is called from (print working directory). If you call the script from another directory, than the one your jar-files resides in, you'll get the wrong path. And you changed the search path of the SHELL, that will prevent the shell from finding any other binary e.g. java.
PATH is system-reserved variable, that define the way where your system should look to find the executable (in your case java). Therefore you shouldn't use it in your code as variable to your test.jar .
In my opinion, your code should be something like:
#!/bin/sh
PROGPATH='/path/to/your/test.jar'
JAVAEXEC=`which java`
JAVAPARAMS='-j'
GLOBALPATH="$JAVAEXEC $JAVAPARAMS $PROGPATH"
echo $GLOBALPATH
I am trying to run a Jar file in the backend of my php code.But I am not getting the desired output to it.There is a jar file which runs in the background and returns the Page Rank of any of the keyword and Domain given to it.
I am attaching the code,please suggest me any solution to it,because when I run it on the terminal,it is giving correct output.
Here is the Code :
<?php
set_time_limit(0);
function returnJarPath()
{
$jarPath = $_SERVER['DOCUMENT_ROOT'] . "myFolder/tools_new/includes/Rank.jar";
return $jarPath;
}
$jar = returnJarPath();
$command = "java -jar $jar aspdotnet/microsoft.com";//Passing the Argument to the Jar file.
$shellOutput = shell_exec($command);
print "The Shell Output is : " ; var_dump($shellOutput);print "<br />";
exec($command,$executeCommmand);
print "The Exec returns the value : " ; var_dump($executeCommmand);print "<br />";
passthru($command,$passthruCommand);
print "The Passthru returns the value : " . $passthruCommand. "<br />";
?>
I just checked apache's error log and the last error I found was :
sh: java: command not found
But as I have already said,I have been using the same command through SSH to run the Java command.So there's no such possibility of not having JAVA installed on the server.Please help me out of this mess...
If the jar file writes to standard output you can use exec.
Here is an example how I use it:
may be first: exec("cd jar dir"); // if jar fine needs to be executed from the same dir
$output = exec("/usr/bin/java -jar $jar aspdotnet/microsoft.com");
But as you say:
sh: java: command not found
It means the there is no path alias to java from php. Just use the full java path to the executable /usr/bin/java.
Given you are calling java. My bet is the output is being displayed in the Java Console, and not in the shell, where PHP could pull the text information.
How to solve this dilemma?
Well you could write the results to a file, if you have the java source to modify, and then read that file through php to get the results. The possibility of a collision here would be pretty good. The other option is to have Java connect to your MySQL database (if you had one) and then run the java then query the database for the response. Of course, you would need to pass Java a way for it to input the data to insert a record you could identify (a hash of some sort), I have never done that in Java, just a theory of how you might be able to do it.
Update
You may want to try the standard output as suggested by darko petreski as another option as well.
If the PHP code is to be executed in a server (and not via command line) the user that runs the java executable is www-data, not you. In that case make sure that www-data has the permissions to read the jar file and to execute the java executable
The first thing that I would check/change is the line in the function where you are building the $jarPath variable from this:
$jarPath = $_SERVER['DOCUMENT_ROOT'] . "myFolder/tools_new/includes/Rank.jar";
to this:
$jarPath = $_SERVER['DOCUMENT_ROOT'] . "/myFolder/tools_new/includes/Rank.jar";
The trailing slash may not be present in $_SERVER['DOCUMENT_ROOT'] which could cause issues.
I am assuming that when you say it runs from the console, you are running the java command like so:
$ java -jar /rest/of/path/myFolder/tools_new/includes/Rank.jar aspdotnet/microsoft.com
I would ensure that you include the path to the java binary in the $command variable like so...
$command = "/path/to/java -jar $jar aspdotnet/microsoft.com";
The user that owns the web server process may not have a $PATH variable that includes the path to the Java binary.