I have a zip file in hdfs and i need to add a file to the zip and save in the same HDFS location. Any examples would be appreciated.
I have the following code.
val filePattern = s"${hdfsFolderPath}/${filePath}.txt"
val zipFilePath = hdfsWrapper.getFileNameFromPattern(s"${targetFilePath}/*.zip")
if (hdfsWrapper.filter(filePattern).size() > 0)
{
Try
{
val zipEntry = new ZipEntry(filePattern)
val zos: ZipOutputStream = new ZipOutputStream(new FileOutputStream(zipFilePath))
zos.putNextEntry(zipEntry)
zos.closeEntry()
zos.close()
}
}
Would like to know if above code is right?
I believe that your code would result in the zip file being replaced with a new one containing just the new file. See Appending files to a zip file with Java for an example of adding a file to a zip archive.
I'm not extremely familiar with HDFS, but I suspect that you can't write to that directly either, you probably have to create the new zip file and then replace it in HDFS
I just want to have 2 Objects from Type "Media". Therefore i have 2 .mp3 files.
The first is no problem, i can play it.
But if i enable the Code for the second file, the System throws an Error
"InvocationtargetException" it says "System can not find the file"
I checked both paths of the .mp3 files, here are the results:
file:/C:/Users/Lennie-Admin/Documents/NetBeansProjects/Lost/Rise.mp3
file:/C:/Users/Lennie-Admin/Documents/NetBeansProjects/Lost/TrueStrength.mp3
I checked my working directory with System.out.println( new File("").getAbsolutePath()); the Result is
C:\Users\Lennie-Admin\Documents\NetBeansProjects\Lost
And finally here is the java Code, which should create the 2 Objects:
String uriString1 = new File("Rise.mp3").toURI().toString();
Media hit1 = new Media(uriString1);
String uriString2 = new File("TrueStrength.mp3").toURI().toString();
Media hit2 = new Media(uriString2);
Hope you can help me :)
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);
I have a FileWriter to write a text to a file. The file is not getting created at all. I've checked the current working directory. Moreover, when I debug, the code for writing the text into the file is not getting executed at all...Have anyone experienced anything similar?
File ff = new File("ffile.txt");
FileWriter frr = new FileWriter(ff);
frr.write("hello");
frr.close();
to create new file use createNewFile();
File ff = new File("ffile.txt");
ff.createNewFile();
FileWriter frr = new FileWriter(ff);
frr.write("hello");
frr.close();
I have a file reader channel picking up an xml document. By default, a file reader channel populates the 'originalFilename' in the channel map, which ony gives me the name of the file, not the full path. Is there any way to get the full path, withouth having to hard code something?
You can get any of the Source reader properties like this:
var sourceFolder = Packages.com.mirth.connect.server.controllers.ChannelController.getInstance().getDeployedChannelById(channelId).getSourceConnector().getProperties().getProperty('host');
I put it up in the Mirth forums with a list of the other properties you can access
http://www.mirthcorp.com/community/forums/showthread.php?t=2210
You could put the directory in a channel deploy script:
globalChannelMap.put("pickupDirectory", "/Mirth/inbox");
then use that map in both your source connector:
${pickupDirectory}
and in another channel script:
function getFileLastModified(fileName) {
var directory = globalChannelMap.get("pickupDirectory").toString();
var fullPath = directory + "/" + fileName;
var file = Packages.java.io.File(fullPath);
var formatter = new Packages.java.text.SimpleDateFormat("yyyyMMddhhmmss");
formatter.setTimeZone(Packages.java.util.TimeZone.getTimeZone("UTC"));
return formatter.format(file.lastModified());
};
Unfortunately, there is no variable or method for retrieving the file's full path. Of course, you probably already know the path, since you would have had to provide it in the Directory field. I experimented with using the preprocessor to store the path in a channel variable, but the Directory field is unable to reference variables. Thus, you're stuck having to hard code the full path everywhere you need it.