Encrypt file using AxCrypt in java - java

I want to encrypt the text file using AxCrypt. I did it through cmd but I want it from the Java program. The following 4 commands I've used in cmd.
1: location of axcrypt software directory
2: axcrypt command (which will be executed for encrypting the file)
3: import file location (file to encrypt)
4: export file location (encrypted file directory)
Here is my code:
public class TestCode {
String axcryptLocation = "C:\\Program Files\\Axantum\\AxCrypt";
String axcryptCommand = "AxCrypt.exe -e -k \"X2U4qPtdMTMZ K63D ABnS 3gO2 PHFL XKJ/ +UsZ /QuG yp5s X78k 2wH=\" -z";
String fileImportLocation = "E:\\ImportExport\\firstcheck.txt";
String fileExportLocation = "E:\\ImportExport\\";
public static void main(String[] args) {
}
}

You need the ProcessBuilder class. It's a bit tricky to use - for example, you should replace axcryptCommand with a list of arguments (because splitting on space but not on quotes - that's bash/cmd.exe stuff). The command itself should be an absolute path, as well.
Here is a tutorial on ProcessBuilder.
NB: Note that it is not particularly complicated to encrypt stuff in java code, without having to rely on windows-only executables. A search on the web should find you loads of tutorials on how to do this, though, as is usual with crypto, you'll probably mess something up. Such is the nature of doing a task where failure cannot be tested for.

Related

How to change the user.dir in java 8? [duplicate]

How can I change the current working directory from within a Java program? Everything I've been able to find about the issue claims that you simply can't do it, but I can't believe that that's really the case.
I have a piece of code that opens a file using a hard-coded relative file path from the directory it's normally started in, and I just want to be able to use that code from within a different Java program without having to start it from within a particular directory. It seems like you should just be able to call System.setProperty( "user.dir", "/path/to/dir" ), but as far as I can figure out, calling that line just silently fails and does nothing.
I would understand if Java didn't allow you to do this, if it weren't for the fact that it allows you to get the current working directory, and even allows you to open files using relative file paths....
There is no reliable way to do this in pure Java. Setting the user.dir property via System.setProperty() or java -Duser.dir=... does seem to affect subsequent creations of Files, but not e.g. FileOutputStreams.
The File(String parent, String child) constructor can help if you build up your directory path separately from your file path, allowing easier swapping.
An alternative is to set up a script to run Java from a different directory, or use JNI native code as suggested below.
The relevant OpenJDK bug was closed in 2008 as "will not fix".
If you run your legacy program with ProcessBuilder, you will be able to specify its working directory.
There is a way to do this using the system property "user.dir". The key part to understand is that getAbsoluteFile() must be called (as shown below) or else relative paths will be resolved against the default "user.dir" value.
import java.io.*;
public class FileUtils
{
public static boolean setCurrentDirectory(String directory_name)
{
boolean result = false; // Boolean indicating whether directory was set
File directory; // Desired current working directory
directory = new File(directory_name).getAbsoluteFile();
if (directory.exists() || directory.mkdirs())
{
result = (System.setProperty("user.dir", directory.getAbsolutePath()) != null);
}
return result;
}
public static PrintWriter openOutputFile(String file_name)
{
PrintWriter output = null; // File to open for writing
try
{
output = new PrintWriter(new File(file_name).getAbsoluteFile());
}
catch (Exception exception) {}
return output;
}
public static void main(String[] args) throws Exception
{
FileUtils.openOutputFile("DefaultDirectoryFile.txt");
FileUtils.setCurrentDirectory("NewCurrentDirectory");
FileUtils.openOutputFile("CurrentDirectoryFile.txt");
}
}
It is possible to change the PWD, using JNA/JNI to make calls to libc. The JRuby guys have a handy java library for making POSIX calls called jnr-posix. Here's the maven info
As mentioned you can't change the CWD of the JVM but if you were to launch another process using Runtime.exec() you can use the overloaded method that lets you specify the working directory. This is not really for running your Java program in another directory but for many cases when one needs to launch another program like a Perl script for example, you can specify the working directory of that script while leaving the working dir of the JVM unchanged.
See Runtime.exec javadocs
Specifically,
public Process exec(String[] cmdarray,String[] envp, File dir) throws IOException
where dir is the working directory to run the subprocess in
If I understand correctly, a Java program starts with a copy of the current environment variables. Any changes via System.setProperty(String, String) are modifying the copy, not the original environment variables. Not that this provides a thorough reason as to why Sun chose this behavior, but perhaps it sheds a little light...
The working directory is a operating system feature (set when the process starts).
Why don't you just pass your own System property (-Dsomeprop=/my/path) and use that in your code as the parent of your File:
File f = new File ( System.getProperty("someprop"), myFilename)
The smarter/easier thing to do here is to just change your code so that instead of opening the file assuming that it exists in the current working directory (I assume you are doing something like new File("blah.txt"), just build the path to the file yourself.
Let the user pass in the base directory, read it from a config file, fall back to user.dir if the other properties can't be found, etc. But it's a whole lot easier to improve the logic in your program than it is to change how environment variables work.
I have tried to invoke
String oldDir = System.setProperty("user.dir", currdir.getAbsolutePath());
It seems to work. But
File myFile = new File("localpath.ext");
InputStream openit = new FileInputStream(myFile);
throws a FileNotFoundException though
myFile.getAbsolutePath()
shows the correct path.
I have read this. I think the problem is:
Java knows the current directory with the new setting.
But the file handling is done by the operation system. It does not know the new set current directory, unfortunately.
The solution may be:
File myFile = new File(System.getPropety("user.dir"), "localpath.ext");
It creates a file Object as absolute one with the current directory which is known by the JVM. But that code should be existing in a used class, it needs changing of reused codes.
~~~~JcHartmut
You can use
new File("relative/path").getAbsoluteFile()
after
System.setProperty("user.dir", "/some/directory")
System.setProperty("user.dir", "C:/OtherProject");
File file = new File("data/data.csv").getAbsoluteFile();
System.out.println(file.getPath());
Will print
C:\OtherProject\data\data.csv
You can change the process's actual working directory using JNI or JNA.
With JNI, you can use native functions to set the directory. The POSIX method is chdir(). On Windows, you can use SetCurrentDirectory().
With JNA, you can wrap the native functions in Java binders.
For Windows:
private static interface MyKernel32 extends Library {
public MyKernel32 INSTANCE = (MyKernel32) Native.loadLibrary("Kernel32", MyKernel32.class);
/** BOOL SetCurrentDirectory( LPCTSTR lpPathName ); */
int SetCurrentDirectoryW(char[] pathName);
}
For POSIX systems:
private interface MyCLibrary extends Library {
MyCLibrary INSTANCE = (MyCLibrary) Native.loadLibrary("c", MyCLibrary.class);
/** int chdir(const char *path); */
int chdir( String path );
}
The other possible answer to this question may depend on the reason you are opening the file. Is this a property file or a file that has some configuration related to your application?
If this is the case you may consider trying to load the file through the classpath loader, this way you can load any file Java has access to.
If you run your commands in a shell you can write something like "java -cp" and add any directories you want separated by ":" if java doesnt find something in one directory it will go try and find them in the other directories, that is what I do.
Use FileSystemView
private FileSystemView fileSystemView;
fileSystemView = FileSystemView.getFileSystemView();
currentDirectory = new File(".");
//listing currentDirectory
File[] filesAndDirs = fileSystemView.getFiles(currentDirectory, false);
fileList = new ArrayList<File>();
dirList = new ArrayList<File>();
for (File file : filesAndDirs) {
if (file.isDirectory())
dirList.add(file);
else
fileList.add(file);
}
Collections.sort(dirList);
if (!fileSystemView.isFileSystemRoot(currentDirectory))
dirList.add(0, new File(".."));
Collections.sort(fileList);
//change
currentDirectory = fileSystemView.getParentDirectory(currentDirectory);

Java: Runtime.exec() and Unicode symbols on Windows: how to make it work with non-English letters?

Intro
I am using Runtime.exec() to execute some external command and I am using parameters that contain non-English characters. I simply want to run something like this:
python test.py шалом
It works correctly in cmd directly, but is incorrectly handled via Runtime.exec.getRuntime()("python test.py шалом")
On Windows my external program fails due to unknown symbols passed to it.
I remember similar issue from early 2010s (!) - JDK-4947220, but I thought it is already fixed since Java core 1.6.
Environments:
OS: Name Microsoft Windows 10 Pro (Version 10.0.18362 Build 18362)
Java: jdk1.8.0_221
Code
To understand the question - the best way is to use code snippet listed below:
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class MainClass {
private static void foo(String filename) {
try {
BufferedReader input = new BufferedReader(
new InputStreamReader(
Runtime.getRuntime().exec(filename).getInputStream()));
String line;
while ((line = input.readLine()) != null) {
System.out.println(line);
}
input.close();
} catch (Exception e) { /* ... */ }
}
public static void main(String[] args) {
foo("你好.bat 你好"); // ??
foo("привет.bat привет"); // ??????
foo("hi.bat hi"); // hi
}
}
Where .bat file contains only simple #echo %1
The output will be:
??
??????
hi
PS
System.out.println("привет") - works fine and prints everything correctly
Questions are the following:
1) Is this issue related to Utf-8 utf-16 formats?
2) How to fix this issue? I do not like this answer as it looks like a very dangerous and ugly workaround.
3) Does anyone know why file names of batch file is not broken and this file can be found, but the argument gets broken? May be it is problem of #echo?
Yes, issue is related with UTF. Theoretically a setting 65001 codepage for cmd that executes the bat files should solve the issue (along with setting UTF-8 charset as default from the Java side)
Unfortunately there a bug in Windows mentioning here Java, Unicode, UTF-8, and Windows Command Prompt
So there's no simple and complete solution. What it's possible to do is to set the same default language-specific encoding, like cp1251 Cyrillic, for both java and cmd. Not all languages are well reflected in the windows encodings, for example Chinese is one of them.
If there's some non-technical restriction on the windows system to change default encoding to the language-specific one for all cmd processes, the java code will be more complicated. At beginning new cmd process have to be created and to its stdin/stdout streams should be attached reader with UTF-16LE (for `cmd /U' process) and writer with CP1251 from different threads. First command sending to stdin from java should be 'chcp 1251' and second is the name of bat-file with its parameters.
Complete solution still may use UTF-16LE for reading of cmd output but to pass a text in, other universal encoding should be used, for example base64, which again leads to increasing complexity

Call easy_install apps from java with ProcessBuilder/RunTime

The title doesn't really explain my question, but I don't know how to ask it in a better way. So, basicly, I'm writing a app that uses the program livestreamer. I installed it on my mac using: easy_install -U livestreamer . So far, so good, it works when I write livestream on my terminal. Now, my issue is that when I try to call it on java:
public static void runLiveStreamer(String channel, String quality) throws IOException{
String[] cmd = new String[]{"livestreamer", "twitch.tv/"+channel, quality};
Process proc = Runtime.getRuntime().exec(cmd);
InputStreamReader isr = new InputStreamReader(proc.getInputStream());
BufferedReader br = new BufferedReader(isr);
String line=null;
while ( (line = br.readLine()) != null)
System.out.println(line);
}
I get this error:
Exception in thread "main" java.io.IOException: Cannot run program "livestreamer": 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:591)
at java.base/java.lang.Runtime.exec(Runtime.java:450)
at livestream.runLiveStreamer(livestream.java:12)
I know the code works, because if I replace the String[] cmd = new String[]{"livestreamer", "twitch.tv/"+channel, quality}; with, for example, ls, it outputs without any problem. This is my first time messing with this kinds of stuff, so my error is probably a really newbie one. Thanks in advance for all the help!
livestreamer is not in your Java process’s path.
Every Windows and Unix operating system’s execution environment has a concept of a program path. The path is an environment variable (named PATH in all operating systems except Windows, which uses Path). It contains a list of directories, separated by colons :, except on Windows where they’re separated by semicolons (;).
As with any environment variable, each running process may have its own path defined, and child processes usually inherit it from their parent process.
Whenever you try to run a program without any directory separators (for instance, trying to run ls instead of /bin/ls), the system will look for that program in each directory in the path.
In your terminal, your PATH contains a directory which has livestream in it. When you run your Java process, you have a different PATH, one which does not include the directory which contains livestream.
The easiest solution is to refer to livestream by its absolute file name, thus making the system execution path irrelevant:
String[] cmd = { "/usr/bin/livestreamer", "twitch.tv/" + channel, quality };
/usr/bin/livestreamer is just an example. I don’t know where livestreamer was actually installed on your system.
To find it, do which livestreamer in your terminal. That should tell you the absolute location of it. (I think in Windows, the command would be where livestreamer.)

Using files as arguments for a Java program in the terminal

java someJavaProgram fsa.fsa <test.txt
That, apparently, is a legitimate command to take with two files as arguments for a Java program in the terminal - one to read in, and then the other (and I think the idea is that it prints the output to the terminal directly). someJavaProgram, fsa.fsa and test.txt are all files in the same directory (being someProject/src, and someJavaProgram in the default package).
However, the response I am given in the terminal just says:
FSA file not found - please scan in the appropriate file.
Testing file not found, please scan in the new relevant file.
My question is two-fold:
What is this command and what is it for?
Does it need refining or modifying or is it the program that needs improvement?
I should note that I wrote the code in Eclipse, where I simply hardcoded filepaths into the program. I'm not sure if that affects anything but it's related.
EDIT: The filepaths and related code are as follows:
private static final String FILE_PATH = "src/test.txt";
private static final String FSA_PATH = "src/fsa1.fsa";
...
public static void main(String[] args) throws FileNotFoundException {
interpretAutomaton();
testAutomaton();
}
...
interpretAutomaton() {
...
Scanner fsaScanner = new Scanner(new BufferedReader(new FileReader(FSA_PATH)));
...
testAutomaton() {
...
Scanner fileScanner = new Scanner(new BufferedReader(new FileReader(FILE_PATH)));
*Both are surrounded by try/catch blocks - which clearly work!
Thanks to anyone who can help clarify on the matter!
Based on the comments so far, to answer your actual questions:
1) The command has four elements:
java - execute the java program
someJavaProgram - the name of the Java class to execute
fsa.fsa - the first argument to the java program, accessible via argv[]
<test.txt - standard input redirection, the contents of the file will be available on the program's standard input, ie. System.in
The net effect is to run your Java program with one argument and one file's contents on the standard input.
2) Both the command line and the program look like they need to change:
change the command line to:
java someJavaProgram fsa.fsa test.txt
That is, remove the <. You will also need to check the paths to the files are correct. This command line assume you are in the same directory as the files when you execute it.
Change your code to use the filenames on the command line rather than the hard-coded names.

Compressing and Archiving the files in the folder using Java Runtime

I am trying to Compress and Archive all the files in a folder, using Java Runtime class. My code snippet looks as this :
public static void compressFileRuntime() throws IOException, InterruptedException {
String date = Util.getDateAsString("yyyy-MM-dd");
Runtime rt = Runtime.getRuntime();
String archivedFile = "myuserData"+date+".tar.bz2";
String command = "tar --remove-files -cjvf "+archivedFile+" marketData*";
File f = new File("/home/amit/Documents/");
Process pr = rt.exec(command, null, f);
System.out.println("Exit value: "+pr.exitValue());
}
The above code doesn't archive and compress the file as expected, though it creates a file myuserData2009-11-18.tar.bz2 in the folder "/home/amit/Documents/".
Also the output is
Exit value: 2.
While if I execute the same command from command line, it gives the expected result.
Please tell me what I am missing.
Thanks
Amit
The problem lies in this part:
" marketData*"
you expect the filenames to be compressed to be globbed from the * wildcard. Globbing is done by the shell, not by the tools themselves. your choices are to either:
numerate the files to be archived yourself
start the shell to perform the command ("/bin/sh -c")
start tar on the folder containing the files to be archived
Edit:
For the shell option, your command would look like:
String command = "sh -c \"tar --remove-files -cjvf "+archivedFile+" marketData*\"";
(mind the \"s that delimit the command to be executed by the shell, don't use single quotes ot the shell won't interpret the glob.)
If really you want to create a bzip2 archive, I'd use a Java implementation instead of a native command which is good for portability, for example the one available at http://www.kohsuke.org/bzip2/ (it is not really optimized though, compression seems to be slower than with Java LZMA).

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