I am using Apache commons ftps client to connect to an ftps server. I have the remote file path which is a directory. This directory contains a tree of sub-directories and files. I want to get the path for each file or folder. Is there any way I can get this property? Or if there is any way I could get the parent folder path, I could concatenate the file name to it.
I am currently using below function to get path and size of all files under a certain directory. It gets all the file in current directory and check if it is a directory or file. If it is a directory call recursively until end of the tree, if it is a file save the path and size. You may not need these "map" things you can edit according to your needs.
Usage:
getServerFiles(ftp,"") // start from root
or
getServerFiles(ftp,"directory_name") // start from given directory
Implementation:
def getServerFiles(ftp: FTPClient, dir: String): Map[String, Double] = {
var fileList = Array[FTPFile]()
var base = ""
if (dir == "") {
fileList = ftp.listFiles
} else {
base = dir + "/"
fileList = ftp.listFiles(dir)
}
fileList.flatMap {
x => if (x.isDirectory) {
getServerFiles(ftp, base + x.getName)
} else {
Map[String, Double](base + x.getName -> x.getSize)
}
}.toMap
}
Related
How do we download directory content from SFTP server recursively in Scala? Can someone please help me with an example?
def recursiveDirectoryDownload(sourcePath: String, destinationPath: String) : Unit = {
val fileAndFolderList = channelSftp.ls(sourcePath)
for(item <- fileAndFolderList)
{
if(! item.getAttrs.isDir)
{
ChannelSftp.get(sourcePath + “\” + item.getFilename,destinationPath +”\” + item.getFilename)
}
}
}
You have to call back your recursiveDirectoryDownload function, when you encounter a subfolder.
See this question for implementation in Java:
Transfer folder and subfolders using channelsftp in JSch?
It should be trivial to translate to Scala.
Is there a way to get a gzip archive file to unzip to a different destination when a user double clicks on the archive? Currently, my compression code looks something like this in Scala:
val filename = SetFilename.getOrElse {
val path = files.head.getAbsolutePath
val baseUrl = FilenameUtils.getFullPathNoEndSeparator(path)
...
}
val output = new File(filename)
val fos = new FileOutputStream(output)
val gzos = new GZIPOutputStream(new BufferedOutputStream(fos))
try {
files.foreach { input =>
val fis = new FileInputStream(input)
try {
ioStream(fis, gzos)
gzos.flush()
}
finally {
fis.close()
}
}
}
finally {
gzos.close()
fos.close()
}
IS there any way to tell the compressed files to decompress in a different destination when a user double clicks on the archive?
It is not the gzip archive that decides it will be unzipped in the same location, it's something the operating system you're unzipping it on decides.
If you need to unzip into a specific place, you should look for a packaging solution like deb for Ubuntu or Debian systems; or dmg for OSX.
I'm attempting to save a screenshot using a generic method in protractor. Two features, it creates the folder if it does not exist and it saves the file (with certain conditions).
export function WriteScreenShot(data: string, filename: string) {
let datetime = moment().format('YYYYMMDD-hhmmss');
filename = `../../../test-reports/${filename}.${datetime}.png`;
let path =filename.substring(0, filename.lastIndexOf('/'));
if (!fs.existsSync(path)) {
fs.mkdirSync(path);
}
let stream = fs.createWriteStream(filename);
stream.write(new Buffer(data, 'base64'));
stream.end();
}
This can be used by calling browser.takeScreenshot().then(png => WriteScreenShot(png, 'login/login-page')); Using this example call, a file will be created, I assumed, in the path relative where my WriteScreenShot method's file resides. But that does not appear to be the case.
For example, when I run my spec test in the spec's folder, the image gets saved in the correct place. But if I run it at the project root, an error is capture. Obviously, this has to do with my relative path reference. How do I capture the project's root directory and build from that so that I can run the test from any directory?
This is a classical directory access error. Let me just explain what is happening to your code -
let path =filename.substring(0, filename.lastIndexOf('/'));
The above line outputs to ../../../test-reports
fs.existsSync checks whether thispath exists or not -
case 1 :(postive flow) Your spec folder is in the same current working directory in which you are trying to create reports folder. When you run your test, the path exists, it generates the test-reports directory & screenshots and your code works fine.
case 2:(negative flow) When you try to run it from the root directory which is the current working directory now, fs.existsSync tries to check the path & the reports folder inside it. If it doesn't exist , fs.mkdirSync tries to create your directories but it would fail as it cannot create multiple directories.
You should be using native path module of nodejs to extract the path instead of using file substring and the mkdirp external module for creating multiple directories.
import * as path from 'path';
let {mkdirp} = require('mkdirp'); // npm i -D mkdirp
export function WriteScreenShot(data: string, filename: string) {
let datetime = moment().format('YYYYMMDD-hhmmss');
filename = `../../../test-reports/${filename}.${datetime}.png`;
let filePath = path.dirname(filename); // output: '../../..' (relative path)
// or
let filePath = path.resolve(__dirname); // output: 'your_root_dir_path' (absolute path)
// or
let filePath = path.resolve('.'); // output: 'your_root_dir_path' (absolute path)
if (!fs.existsSync(filePath )) {
mkdirp.sync(filePath); // creates multiple folders if they don't exist
}
let stream = fs.createWriteStream(filename);
stream.write(new Buffer(data, 'base64'));
stream.end();
}
If you are curious to know the difference btw mkdir & mkdir-p please read this SO thread.
I'm trying to read some files from my Scala project, and if I use: java.io.File(".").getCanonicalPath() I find that my current directory is far away from them (exactly where I have installed Scala Eclipse). So how can I change the current directory to the root of my project, or get the path to my project? I really don't want to have an absolute path to my input files.
val PATH = raw"E:\lang\scala\progfun\src\examples\"
def printFileContents(filename: String) {
try {
println("\n" + PATH + filename)
io.Source.fromFile(PATH + filename).getLines.foreach(println)
} catch {
case _:Throwable => println("filename " + filename + " not found")
}
}
val filenames = List("random.txt", "a.txt", "b.txt", "c.txt")
filenames foreach printFileContents
Add your files to src/main/resources/<packageName> where <packageName> is your class package.
Change the line val PATH = getClass.getResource("").getPath
new File(".").getCanonicalPath
will give you the base-path you need
Another workaround is to put the path you need in an user environmental variable, and call it with sys.env (returns exception if failure) or System.getenv (returns null if failure), for example val PATH = sys.env("ScalaProjectPath") but the problem is that if you move the project you have to update the variable, which I didn't want.
I'm developing an eclipse plug-in and I need to traverse a directory and whole content of the directory. I found the method which reads a file in plug-in (bundleresource) as InputStream.
InputStream stream = Activator.class.getResourceAsStream("/dir1/dir2/file.ext");
this method works for files only. I need a way to read directories, list subdirectories and files like File.io.
Thanks.
Do you want to read a resource directory of your plugin? Otherwise you have to traverse a directory and open one stream per file:
String path = "c:\\temp\\";
File directory = new File(path);
if (directory.isDirectory()) {
String[] list = directory.list();
for (String entry : list) {
String absolutePath = path + entry;
System.out.println("processing " + absolutePath);
File file = new File(absolutePath);
if (file.isFile()) {
FileInputStream stream = new FileInputStream(file);
// use stream
stream.close();
}
}
}
If you want to traverse subdirectories as well you should wrap this into a recursive method, check if file is a directory and call the recursive method in this case.