Octokit allows me to easily retrieve the list of files in a given folder like so:
var directoryContent = await _githubClient.Repository.Content.GetAllContents("MyUsername", "MyProject", "TheFolderName").ConfigureAwait(false);
Unfortunately, the Content property for each file is null which means that I have to loop through the list of files and retrieve each one separately like so:
foreach (var file in directoryContent)
{
var fileWithContent = await _githubClient.Repository.Content.GetAllContents("MyUsername", "MyProject", file.Path).ConfigureAwait(false);
... do something with the content ...
}
Is there a more efficient way to get the content of all files in a given folder?
Just for reference: This is on purpose. If you read a whole directory and load all the content in one step it's huge load. So, the first call skips the content (Content property is null). The second step requests just this file and then the content is present.
Only thing I'm still missing is a method that can return just one result instead of a list if I pull a single file.
i am using in my unity project this method that saves the camera viewport to the file system:
Application.Capturescreenshot(fileName)
it is work great but i want it to be saved under a specific path.
for example : Application.Capturescreenshot(c:/screenshots/filename);
how can i managed this?
thnaks!
You could use the absolute file path easily by passing it as a string. In your case, using this statement should work just fine:
Application.CaptureScreenshot("C:/screenshots/filename");
Note that it will give you an error if screeshots folder does not exist. Hence, if you are uncertain of that, you could modify the code accordingly:
// File path
string folderPath = "C:/screenshots/";
string fileName = "filename";
// Create the folder beforehand if not exists
if(!System.IO.Directory.Exists(folderPath))
System.IO.Directory.CreateDirectory(folderPath);
// Capture and store the screenshot
Application.CaptureScreenshot(folderPath + fileName);
Lastly, if you want to use a relative path instead of an absolute one, Unity provides dataPath and persistentDataPath variables to access the data folder path of the project. So, if you want to store the screenshots inside the data folder, you could change the folderPath above accordingly:
string folderPath = Application.dataPath + "/screenshots/";
var nameScreenshot = "Screenshot_" + DateTime.Now.ToString("dd-mm-yyyy-hh-mm-ss") + ".png";
ScreenCapture.CaptureScreenshot(nameScreenshot);
Try this !
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);
My image upload doesn't work:
Controller:
if (Input::hasFile('image')) {
$bikecreate->image = Input::file('image');
$destinationPath = public_path().'/upload/';
$filename = str_random(6) . '_' . $bikecreate->users_id ;
Input::file('image')->move($destinationPath, $filename);
}
Form:
{{ Form::file('image', array('files' => true)) }}
After accepting form everything looks ok, but after the end of upload, filepath in database show .tmp/file at my server.
Without seeing the rest of your code it's hard to see exactly what's going on but my guess is that your line $bikecreate->image = Input::file('image') is where you're setting the file's path for the database. You've actually set the UploadedFile instance as the image property on $bikecreate there, which, presumably, when serialised to something to put into the database gets __toString() called on it.
__toString() called on a File instance (which itself inherits __toString from SPLFileInfo returns the path to that file. So you'd think you're get the uploaded filename, but actually because an uploaded file is actually a temporary file in PHP, you get the temporary name.
Try changing that line to the following:
$bikecreate->image = Input::file('image')->getClientOriginalName();
This retrieves the actual original name of the uploaded file, not the temporary path given to it by PHP.
It goes without saying that this is only pertinent to UploadedFiles, normal files should just be able to be __toStringed to get the path to the file, although you'll notice that it would be the full path and not the basename. To get that, use getBaseName().
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.