Expanding files with 7zip - java

I am trying to expand a zip file using 7zip but I keep getting the 7zip Usage printout.
The zip exist in c:\temp
The same command succeed in batch window :
C:\TEMP>7z x "tryThis.zip"
I tried adding the workdir path to the file,And also without the working dir,
nothing help. - I can probably run this using CMD/c command but I prefer to keep the code clean
What am I doing wrong?
Thank you!
String pathTo7ZipExe = "c:\\program files\\7-zip\\7z.exe";
String fileName ="tryThis.zip";
String workingDir = "c:\\temp\\";
Process process = Runtime.getRuntime().exec(
new String[]{pathTo7ZipExe},
new String[]{" x \"" + fileName +"\""},
new File(workingDir));
BufferedReader in = new BufferedReader(new InputStreamReader(process.getInputStream()));
String line;
while ((line = in.readLine()) != null) {
System.out.println(line);
}
// wait for zip to end.
int exitVal = process.waitFor();

Please have a look at the documention for Runtime.exec
What you were actually trying to do is calling 7-zip without arguments and providing the arguments as your environment. Environment is something like Windows PATH etc.
so you would probably want to do something like:
Runtime.getRuntime().exec(new String[]{pathToZipExe, "x", fileName}, null, new File(workingDir));
On the other hand I would strongly advise to have a look on ZipInputStream which is included in java - using that you can also unpack zip files.
Cheers

You're invoking the overload of exec which accepts envp array as the second argument. This envp array is not for arguments at all, so actually you don't pass any arguments: that's why you get the usage printout.
Quotes and spaces aren't themselves part of arguments: they are used for separation into argv (with minor reservations, it's also true for Windows: that's how CommandLineToArgW works, even though full original command line with quotes and spaces is always available).
So it should be something like this:
Runtime.getRuntime().exec(new String[]{pathTo7ZipExe, "x", fileName},
new String[]{}, new File(workingDir));
(too bad I don't know Java, so the code might be unidiomatic, but it should work).

Related

Passing a path to exec

I am trying to use the exec function. The path to the executable contains spaces and this is giving me grief
My code looks like this
Runtime.getRuntime().exec("\"C:\\Program Files (x86)\\ASL\\_ASL Software Suite_installation.exe\"", null, new File("\"C:\\Program Files (x86)\\ASL\\_ASL Software Suite_installation\""));
When this is executed I get an exception -
Cannot run program ""c:\Program"
I would be grateful if someone can give me some help in solving this
Thanks in advance
From Runtime.exec(String command, String[] envp, File dir):
Executes the specified string command in a separate process with the specified environment and working directory.
This is a convenience method. An invocation of the form exec(command, envp, dir) behaves in exactly the same way as the invocation exec(cmdarray, envp, dir), where cmdarray is an array of all the tokens in command.
More precisely, the command string is broken into tokens using a StringTokenizer created by the call new StringTokenizer(command) with no further modification of the character categories. The tokens produced by the tokenizer are then placed in the new string array cmdarray, in the same order.
This means the first string is broken into tokens, regardless of the outer quotes. Use the Runtime.exec(String[] cmdarray, String[] envp, File dir) version to avoid the tokenization of the executable path.
Or, use ProcessBuilder:
File d = new File("C:/Program Files (x86)/ASL/_ASL Software Suite_installation");
ProcessBuilder pb = new ProcessBuilder(d.getAbsolutePath() + "/main.exe");
Process p = pb.directory(d)
.start();
See:
Why should avoid using Runtime.exec() in java?
ProcessBuilder vs Runtime.exec()
You don't need to quote the filenames again. Java will take care of it for you just give the proper filename as a string like so
Runtime.getRuntime().exec(
"C:\\Program Files (x86)\\ASL\\_ASL Software Suite_installation.exe",
null,
new File("C:\\Program Files (x86)\\ASL\\_ASL Software Suite_installation"));

Spaces in file path in java

I have a folder which contains few jar files. I am referring to those jar files from another jar file which is in some other location.
My problem is, when I give the path of the jar folder like this C:\Trial Library\jar Folder\ ie. with space in the folder names (Trial Library) then it is unable to locate this folder.
If I give without space ie C:\Trial_Library\jar_Folder\ then it works fine.
Please help me to fix this issue ASAP.
Here is my Batch File
set CURRENT_DIRECTORY=%~dp0
set ANT_HOME=%"CURRENT_DIRECTORY"%ant\apache-ant-1.8.3
ECHO current directory is %CURRENT_DIRECTORY%
ECHO %ANT_HOME%
set Path=%ANT_HOME%\bin
set ADAPTER_LIBRAY_PATH=%1
set USER_JAR_PATH=%2
set CLASS_NAME=%3
set RESULTS_PATH=%4
set JUNIT_PATH=%"CURRENT_DIRECTORY"%ANT\test\junit-4.1.jar
set LIBRAIES_TO_INCLUDE="%JUNIT_PATH%";"%ADAPTER_LIBRAY_PATH%";"%USER_JAR_PATH%"
ECHO %LIBRAIES_TO_INCLUDE%
ECHO %ADAPTER_LIBRAY_PATH%
ECHO %JUNIT_PATH%
ECHO %USER_JAR_PATH%
ECHO %CLASS_NAME%
ECHO %RESULTS_PATH%
ant -lib "%LIBRAIES_TO_INCLUDE%" -Dlibraries="%ADAPTER_LIBRAY_PATH%" -Djunitlibrary="%JUNIT_PATH%" -Djartobeexec="%USER_JAR_PATH%" -Duserclass=%CLASS_NAME% -Dresultspath=%RESULTS_PATH% -buildfile build.xml test-html
Here is where i pass the values to my batch file
String[] commands=new String[5];
commands[0]="driver.bat";
commands[1]=finalLibraryPath;
commands[2]=executingJarLocation;
commands[3]=tempPackageName;
commands[4]=resultsFolderPath;
process = Runtime.getRuntime().exec(commands);
InputStream is = process.getInputStream();
InputStreamReader isr = new InputStreamReader(is);
BufferedReader br = new BufferedReader(isr);
StringBuffer errorStr = new StringBuffer();
String line;
while ((line = br.readLine()) != null) {
errorStr.append(line);
errorStr.append(System.getProperty("line.separator"));
}
Thanx in advance
Regards,
Prabhu
Okay, from what I understand I'm "guessing" that you are doing something like
Runtime.exec("myBatchFile.bat " + path);
This will end in tears. This is the equivalent of saying:
C:> myBatchFile.bat C:\Path to my jar files
This won't work. Basically, your batch file now thinks it has 5 parameters instead of one.
To fix the problem you need to pass each command/parameter seperatly...
Runtime.exec(new String[] {"mybatchFile.bat", path});
Or better still, use ProcessBuilder
ProcessBuilder pb = new ProcessBuilder("myBatchFile.bar", path);
Wrap the path in quotes. This means that the computer takes it literally. You can have a similar problem with notepad, where it adds a .txt extension on the end even if you supply the extension. Wrapping in quotes solves this problem.
Generally enclosing the path in double quotes ("path") works on platforms like Unix, Linux etc.
The problem only comes on WIN platform. The reason behind this is that as soon as the WIN sees a space in the path of a file to be executed, it reverts to 8.3 naming. In this naming, it takes the first 6 characters of the Sub directory as the param and searches for the pattern.
To solve the issue, you have to append the first 6 characters with a tilde(~) and a number representing the instance of that pattern.
For ex:
**Original PATH : C:/Program Files/Jdk1.6.0_07/bin
PATH to be used : C:/Progra~1/Jdk1.6.0_07/bin**
I have used the similar approach in lots of my Java applications and it has worked correctly all of the times.

How to run bat file from java with arguments (i.e file name with full path) having folder name with space

Am trying to execute the a bat file with some arguments through a JAVA programmes . the arguments are file name with full path, And this path had some folder name with space, which are creating issue and giving me the following error
Error: 'D:\Documents' is not recognized as an internal or external
command
the code is as below
String command = "D:\Documents and Settings\ A.bat" + " " D:\Documents and Settings\B.xml
1. process = Runtime.getRuntime().exec(new String[] {"cmd.exe","/c",command});
2. process.waitFor();
3. exitValue = process.exitValue();
You need to escape the \ in your string (i.e. doubling them: D:\\Documents), but that is not the problem. You can try to escape the spaces Documents\\ and\\ Settings or you use the exec method that does this for you. Just dont build the command line by yourself. Better use ProcessBuilder for starting processes.
String command = "\"D:\Documents and Settings\\" A.bat" + " \"D:\Documents and Settings\B.xml\""
Escape double quotes, so you can include double quotes in the literal, to give:
cmd.exe /x "D:\Documents and Settings\" A.bat "D:\Documents and Settings\B.xml"
I was trying to do the same thing. I googled whole day but didn't make it work. At Last I handled it in this way, I am sharing it if it comes to any use of anybody :
String command = "A.bat D:\\Documents and Settings\\B.xml";
File commandDir = new File ( "D:\\Documents and Settings ");
String[] cmdArray = { "cmd.exe", "/c", command };
1. Process process = Runtime.getRuntime().exec( cmdArray, null, cmdArray );
2. process.waitFor();
3. exitValue = process.exitValue();
I've spent a while searching on SO and the wider Internet and was about to post this as a new question when I came across this, which does seem identical to my issue...
I am trying to call a Windows batch file from Java. The batch file takes several arguments but just the first, which is a path to a data file, is of relevance to this problem. The cut-down command line that I have been experimenting with is essentially:
cmd /c c:\path\to\my\batchfile.bat c:\path\to\my\datafile.mdl
I'm using Apache Commons Exec which ultimately delegates to Runtime.getRuntime().exec(String[] cmdarray, String[] envp, File dir), the 'correct' version as opposed to the overloaded versions taking a single String command. Quoting of the arguments when they contain spaces is therefore taken care of.
Now, both the path to the batch file and/or the path to the data file can have spaces in them. If either the path to the batch file or the path to the data file have spaces in, then the batch file is executed. But if both have spaces in them then the path to the batch file is truncated at the first space.
This has to be a (Java or Windows?) bug, right? I've debugged right down to the native call to create() in java.lang.ProcessImpl and all seems ok. I'm on JDK1.6.

Problem with copy files using cmd in Java

I'm trying to copy a bunch of files with a specific extension from one folder to another using the copy command, heres wat im doing,
String[] command = new String[3];
command[0] = "cmd";
command[1] = "/c";
command[2] = "copy C:\\output\\html\\*.txt C:\\output\\";
ProcessBuilder copyFiles = new ProcessBuilder(command);
p = copyFiles.start();
p.waitFor();
the thing is, this code works fine for files less than some 5 or so, but just stops responding wen the number of files are more (even for 15 files) !! and the files are not copied either!!
I dont know what the problem is, will be glad if someone could help! :)
You're not reading the output the copy command is generating.
When spawning a child process using ProcessBuilder, output generated by your child process gets written to a buffer. If this buffer isn't read from, it eventually fills up. When it fills up, the copy command can't write any more to it and so is blocked by the operating system. It is then forced to wait until space is made in the buffer by reading from it.
I ran your code with 20 files and I found that it did indeed hang.
One way to solve your problem is to redirect the output from copy to NUL. Most of the output from copy is a list of all the files it has copied, which you probably don't care too much for. To do this redirection, modify the line that assigns to command[2] to the following:
command[2] = "copy C:\\output\\html\\*.txt C:\\output\\ >NUL 2>NUL";
However, if there is a problem copying files, you might not know about it if you do this.
Alternatively, you can read the output that the copy command generates. The following code sends it to System.out, but you can easily send it elsewhere or completely ignore it if you wish:
String[] command = { "cmd", "/c", "copy C:\\output\\html\\*.txt C:\\output\\" };
ProcessBuilder copyFiles = new ProcessBuilder(command);
copyFiles.redirectErrorStream(true);
p = copyFiles.start();
// The InputStream we get from the Process reads from the standard output
// of the process (and also the standard error, by virtue of the line
// copyFiles.redirectErrorStream(true) ).
BufferedReader reader = new BufferedReader(new InputStreamReader(p.getInputStream()));
String line;
do {
line = reader.readLine();
if (line != null) { System.out.println(line); }
} while (line != null);
reader.close();
p.waitFor();
I gave each approach a quick test with the same 20 files and neither approach hung.
EDIT: You might also want to try a 'hybrid' approach, by throwing away what copy writes to standard output (e.g. the list of files it's copying) but using the second approach to read in what it writes to standard error (e.g. error messages). To do this, you'd add the >NUL, which redirects the standard output of copy to NUL, but you wouldn't add the 2>NUL, since that redirects standard error to NUL.

simple console Application using java

Im developing a simple console Application using java. The code is given below
` try {
File file = new File("writer.txt");
writer = new BufferedWriter(new FileWriter(file));
Process myProcess = Runtime.getRuntime().exec("jps -l");
BufferedReader stdout = new BufferedReader(new InputStreamReader(
myProcess.getInputStream()));
String line = stdout.readLine();
while (line != null) {
if (line.contains(".jar")) {
writer.write(line);
System.out.println(line);
}
line = stdout.readLine();
}
writer.close();
}
`
The code will display the currently running the jar in my windows. The output format is displayed 2356 Timeout.jar I want to display it only Timeout.jar How to remove that integer values. Thanks in advance.
Assuming you have "2356 Timeout.jar" in line, this will return just the jar name:
line.substring(line.indexOf(" ") + 1);
I think there must be an easier way to get the running jar though. I did a quick search and you may want to look at these questions:
How to get the path of a running JAR file?
http://arstechnica.com/civis/viewtopic.php?f=20&t=261563
You could:
Apply a regular expression to line before writing it out. (Think start-of-line, then integers, ending at the first whitespace)
Use ls (or dir) as your exec process instead of jps
Just grab the directory listing directly instead of via the external process as per below:
File dir = new File("directoryName");
String[] children = dir.list();
Doing what you have via JPS is probably not a good idea if this isn't a quick one-off app or a learning exercise because of the following note from the jps man page:
NOTE- You are advised not to write scripts to parse jps output since
the format may change in future releases. If you choose to write
scripts that parse jps output, expect to modify them for future
releases of this tool.
Tokenizing the result is one way.
if you are in unix, use awk to get the second field.
If you are using a Linux based OS,
Instead of
Process myProcess = Runtime.getRuntime().exec("jps -l");
try this one
Process myProcess = Runtime.getRuntime().exec("jps -l | cut -d \" \" -f2");

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