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Java run Python script Issue

ProcessBuilder pb = new ProcessBuilder("C:\\xxxxxxx\\python.exe", "C:\\xxxxxxxxxx\\1.py");
pb.redirectOutput(ProcessBuilder.Redirect.INHERIT);
Process p = pb.start();
I am trying to run Python script from my Java program. The problem is, java run doesn't give me any result. In Python script, there is an OCR operation and I am writing a txt file.
CMD - Manually - It runs OK
Python IDLE - Manually - It runs OK
.bat doesn't work. with administrative rights doesn't work.
Java run doesn't work.
I need help, I need to run the script from java program.

I solved the problem.
The problem is;
ProcessBuilder pb = new ProcessBuilder("C:\\xxxxxxx\\python.exe", "C:\\xxxxxxxxxx\\1.py");
pb.redirectOutput(ProcessBuilder.Redirect.INHERIT);
Process p = pb.start();
When you run this code, python script will process operations in that directory, I mean in your Java class' directory. No matter where py file is.
In my case, there were 'image.save("x.png")' lines in the python script. I hoped that images will be saved in the directory where the py file is. But it is not like that.

Shell script that sets LD_LIBRARY_PATH=`pwd` does not work from Java

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.

C program compilation from a java program

I am trying to compile a c program from a java program on Linux platform. My snippet is.
ProcessBuilder processBuilder = new ProcessBuilder("/usr/bin/gcc",
"-c","/hipad/UserProject/example.c");
Process proc = processBuilder.start();
There is no error during compilation of java program but I am not able to get .o file. I tried to find out solutions but no one is working.
Any suggestion.....

The default working directory of a child process is what ever directory the Java process has as a working directory, which usually is where it was launched from. And by default gcc writes output files to current working directory. That's where you should find example.o.
There are two simple ways to solve this. You can give gcc -o option and full path and name of desired output file, or you can set working directory of child process, like this:
ProcessBuilder processBuilder =
new ProcessBuilder("/usr/bin/gcc", "-c","example.c"); // source in working dir
processBuilder.directory(new File ("/hipad/UserProject")); // or whatever
Process proc = processBuilder.start();
See ProcessBuilder javadoc for more info.

How to execute unix commands through Windows/cygwin using Java

I am trying to accomplish two things:
I am running cygwin on Windows7 to execute my unix shell commands and I need to automate the process by writing a Java app. I already know how to use the windows shell through Java using the 'Process class' and Runtime.getRuntime().exec("cmd /c dir"). I need to be able to do the same with unix commands: i.e.: ls -la and so forth. What should I look into?
Is there a way to remember a shell's state?
explanation: when I use: Runtime.getRuntime().exec("cmd /c dir"), I always get a listing of my home directory. If I do Runtime.getRuntime().exec("cmd /c cd <some-folder>") and then do Runtime.getRuntime().exec("cmd /c dir") again, I will still get the listing of my home folder. Is there a way to tell the process to remember its state, like a regular shell would?
It seems that the bash command line proposed by Paŭlo does not work:
C:\cygwin\bin>bash -c ls -la
-la: ls: command not found
I am having trouble figuring out the technicalities.
This is my code:
p = Runtime.getRuntime().exec("C:\\cygwin\\bin\\bash.exe -c ls -la");
reader2 = new BufferedReader(new InputStreamReader(p.getInputStream()));
line = reader2.readLine();
line ends up having a null value.
I added this to my .bash_profile:
#BASH
export BASH_HOME=/cygdrive/c/cygwin
export PATH=$BASH_HOME/bin:$PATH
I added the following as well:
System Properties -> advanced -> Environment variables -> user variebales -> variable: BASH, value: c:\cygwin\bin
Still nothing...
However, if I execute this instead, it works!
p = Runtime.getRuntime().exec("c:\\cygwin\\bin\\ls -la ~/\"Eclipse_Workspace/RenameScript/files copy\"");

1. Calling unix commands:
You simply need to call your unix shell (e.g. the bash delivered with cygwin) instead of cmd.
bash -c "ls -la"
should do. Of course, if your command is an external program, you could simply call it directly:
ls -la
When starting this from Java, it is best to use the variant which takes a string array, as then
you don't have Java let it parse to see where the arguments start and stop:
Process p =
Runtime.getRuntime().exec(new String[]{"C:\\cygwin\\bin\\bash.exe",
"-c", "ls -la"},
new String[]{"PATH=/cygdrive/c/cygwin/bin"});
The error message in your example (ls: command not found) seems to show that your bash can't find the ls command. Maybe you need to put it into the PATH variable (see above for a way to do this from Java).
Maybe instead of /cygdrive/c/cygwin/bin, the right directory name would be /usr/bin.
(Everything is a bit complicated here by having to bridge between Unix and Windows
conventions everywhere.)
The simple ls command can be called like this:
Process p = Runtime.getRuntime().exec(new String[]{"C:\\cygwin\\bin\\ls.exe", "-la"});
2. Invoking multiple commands:
There are basically two ways of invoking multiple commands in one shell:
passing them all at once to the shell; or
passing them interactively to the shell.
For the first way, simply give multiple commands as argument to the -c option, separated by ; or \n (a newline), like this:
bash -c "cd /bin/ ; ls -la"
or from Java (adapting the example above):
Process p =
Runtime.getRuntime().exec(new String[]{"C:\\cygwin\\bin\\bash.exe",
"-c", "cd /bin/; ls -la"},
new String[]{"PATH=/cygdrive/c/cygwin/bin"});
Here the shell will parse the command line as, and execute it as a script. If it contains multiple commands, they will all be executed, if the shell does not somehow exit before for some reason (like an exit command). (I'm not sure if the Windows cmd does work in a similar way. Please test and report.)
Instead of passing the bash (or cmd or whatever shell you are using) the commands on the
command line, you can pass them via the Process' input stream.
A shell started in "input mode" (e.g. one which got neither the -c option nor a shell script file argument) will read input from the stream, and interpret the first line as a command (or several ones).
Then it will execute this command. The command itself might read more input from the stream, if it wants.
Then the shell will read the next line, interpret it as a command, and execute.
(In some cases the shell has to read more than one line, for example for long strings or composed commands like if or loops.)
This will go on until either the end of the stream (e.g. stream.close() at your side) or executing an explicit exit command (or some other reasons to exit).
Here would be an example for this:
Process p = Runtime.getRuntime().exec(new String[]{"C:\\cygwin\\bin\\bash.exe", "-s"});
InputStream outStream = p.getInputStream(); // normal output of the shell
InputStream errStream = p.getInputStream(); // error output of the shell
// TODO: start separate threads to read these streams
PrintStream ps = new PrintStream(p.getOutputStream());
ps.println("cd /bin/");
ps.println("ls -la");
ps.println("exit");
ps.close();

You do not need cygwin here. There are several pure Java libraries implementing SSH protocol. Use them. BTW they will solve your second problem. You will open session and execute command withing the same session, so the shell state will be preserved automatically.
One example would be JSch.

How to run .reg file in Java

I need to install a .reg file (INTRANET) by using Java. How do i get my goal ?
Cheers,

You could use System.exec to launch regedit yourfile.reg
Here is how to do it :
String[] cmd = {"regedit", "yourfile.reg"};
Process p = Runtime.exec(cmd);
p.waitFor();
Last line is optional, it only allows you to wait until the operation is over.

If you're already on Java 1.6, just grab java.awt.Desktop:
Desktop.getDesktop().open(new File("c:/yourfile.reg"));
It will launch the file using the default application associated with it, as if you're doubleclicking the particular file in Windows explorer.

This can achieved through Process Builder in JAVA. Please consider the following example for this:
ProcessBuilder processBuilder = new ProcessBuilder("regedit", "reg_file_to_run.reg");
Process processToExecute = processBuilder.start();
And then you can optionally wait for the completion of process execution with this line:
processToExecute.waitFor();
Note: If command in your registry file asks for confirmation prompts while making changes in registry entries, you can perform it silently as well with '/s' option. Like this:
ProcessBuilder processBuilder = new ProcessBuilder("regedit", "/s", "reg_file_to_run.reg");
With this command would be executed silently without any confirmation prompt.