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.
Related
The code should be working on flutter2, with android and ios(if possible)
Searching for files
Do you want to simplify your life? Use this package:
import 'package:glob/glob.dart';
Stream<File> search(Directory dir) {
return Glob("**.mp3")
.list(root: dir.path)
.where((entity) => entity is File)
.cast<File>();
}
Do you want to avoid adding a new dependency in your project? Manipulate the Directory itself:
Stream<File> search(Directory dir) {
return dir
.list(recursive: true)
.where((entity) => entity is File && entity.path.endsWith(".mp3"))
.cast<File>();
}
Defining the search scope
You'll also need a Directory where you'll search for MP3 files.
Is it the root directory (i.e. search ALL the files in the device)? Use this answer.
Is it another directory? Pick one from this package.
Usage
final Directory dir = /* the directory you obtained in the last section */;
await search(dir).forEach((file) => print(file));
I am trying to copy a file from one dir to another using the copyFile(path, fileName, newPath, newFileName) function. It gives an error like {"code":13, "message":"input is not a directory"}. The documentation has only 12 error code and no 13th. I'd like to know what i did wrong please.
Here is a sample of my actual code.
this.path = "file:///storage/emulated/0/TheFolder/thefile.ext";
this.newPath = "file:///storage/emulated/0/NewFolder";
this.fileCtrl.copyFile(this.path, fileName, this.newPath, newFileName)
this.path must be a directory but your are showing some file name
change your code as follows
this.path = "file:///storage/emulated/0/TheFolder";
this.newPath = "file:///storage/emulated/0/NewFolder";
this.fileCtrl.copyFile(this.path, YOUR_EXISTING_FILE_NAME, this.newPath, NEW_FILE_NAME);
path -Base FileSystem
fileName - Name of file to copy
newPath - Base FileSystem of new location
newFileName - New name of file to copy to (leave blank to remain the same)
I'm writing a script to produce some artefacts from my build so I want to clean up unwanted files first. I'm using CleanDirectory(dirPath, predicate).
I'm finding it disturbingly hard to work out the directory for a file. If I use GetDirectoryName() that seems to just get me the immediate parent, not the full directory path.
Func<IFileSystemInfo, bool> predicate =
fileSystemInfo => {
// Dont filter out any directories
if (fileSystemInfo is IDirectory)
return false;
var path = fileSystemInfo.Path.FullPath;
var directory = ((DirectoryPath)path).GetDirectoryName();
...
}
Obviously I can use the .NET Framework System.IO classes to do this easily but then I get strings with the slashes in the wrong direction, and things do not smoothly inter-operate with Cake which uses POSIX paths.
OK I've worked out a solution. The key to IFileSystemInfo is to try and cast the Path to various derived types/interfaces, which then provide the functionality you are probably looking for. Example:
Func<IFileSystemInfo, bool> predicate =
fileSystemInfo => {
// Dont filter out any directories
if (fileSystemInfo is IDirectory)
return false;
// We can try and cast Path as an FilePath as know it's not a directory
var file = (FilePath) fileSystemInfo.Path;
if (file.FullPath.EndsWith("Help.xml", StringComparison.OrdinalIgnoreCase))
return false;
// GetDirectory() returns a Path of type DirectoryPath
var directory = file.GetDirectory().FullPath;
...
}
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
}
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.