This may be a silly question but I really don't know where to look.
I'm creating a browser testing environment for a pretty large-scale API written in typescript. This API uses esbuild to build the typescript files into a /dist/ folder with a single index.js entry-point and its appropriate d.ts file.
I've created a /tests/ folder to hold some browser files that includes an index.html file with Mocha and Chai imported. It also imports /dist/index.js which is set globally to a window.myAPI variable.
In /tests/index.html:
import * as myAPI from "./dist/index.js"
Alongside index.html in the tests folder, there are separate JS files included for different tests that run things on window.myAPI... to do assertion tests.
search.test.js
book.test.js
navigate.test.js
I then run a server to host at the root. These separate tests are then imported from /tests/index.html. The separate tests look like this inside:
const { chai, mocha } = window;
const { assert } = chai;
describe("Search", function() {
describe("Setup", function() {
it("Setting URL should work", function() {
const call = myAPI.someCall()
assert.ok(call);
});
});
});
mocha.run();
Everything works, but I have no code hinting for myAPI. I'd like to be able to see what functions are available when I type myAPI, and what parameters they take, and what they should return - along with all my comments on each function.
In typescript you can do things like ambient declarations, but I don't want to make my tests typescript because then I add an unnecessary build step to the tests. But it would be as easy as:
/// <reference path = "/dist/index.d.ts" />
How can I tell VSCode that window.myAPI is an import of /dist/index.js and should import the types as well so I can see them ?
I'm open to different solutions to this, but I feel like this should be pretty simple. I don't know if ESLint is capable of doing something like this, but I tagged it because I feel it's relevant.
Thanks!
I have a file inside my Flutter-project. A simple .dart file which looks like this:
class EnLanguage implements BaseLanguage {
#override
Map<String, String> get language => {'test': 'test'};
}
Now my goal is that I write I script which I by executing goes through all my Project-files, searches for specific Strings ( the ones with a .tr ending) and adds it to the map in the class above (key and value are the same).
I couldn't find any way to achieve this. How does a simple script looks like that can write inside my project files? Im not asking for the whole logic, I just need a start. I couldn't find anything..
Have a look at the package dcli and specifically the pack command. It does a chunk of what you need.
Not quite certain what you mean by strings ending with a .tr.
But to process each script.
var project = DartProject.self.pathToPackage;
find('*.dart', workingDirectory: project)
.forEach((script) {
read(script). forEach((line) {
If (line. contains('.tr'))
{
Extract line...
Write to generated file..
}
I am developing a custom module for a payment method in magento 2. Currently I am using cc-form.html from the vendor directory and the module is working fine. see below path.
vendor/magento/module-payment/view/frontend/web/template/payment/cc-form.html
Is there any way to override the HTML file?
Yes, there is. You can look in pub static to see how path to static asset constructed.
How it works
Every asset is accessible from the page by itenter code heres "RequireJS ID". It similar to real path, but varied.
For example file http://magento.vg/static/adminhtml/Magento/backend/en_US/Magento_Theme/favicon.ico.
It's real path is /app/code/Magento/Theme/view/adminhtml/web/favicon.ico.
It's RequireJS ID is Magento_Theme/favicon.ico. This means that file could be accessible via require("text!Magento_Theme/favicon.ico") or similar command.
You can find that RequireJS ID consist with module name and useful part of path (after folder web).
How can I replace a file
So you have file
vendor/magento/module-payment/view/frontend/web/template/payment/cc-form.html
On the page it loaded with src as
http://magento.vg/static/frontend/Magento/luma/en_US/Magento_Payment/template/payment/cc-form.html
So its RequireJS ID is
Magento_Payment/template/payment/cc-form.html
Side note: Inside UI components stuff it equals to Magento_Payment/payment/cc-form. Words "template" and ".html" are added automatically.
And now you can replace this file for application via RequireJS config
var config = {
"map": {
"*": {
"Magento_Payment/template/payment/cc-form.html":
"<OwnBrand>_<OwnModule>/template/payment/cc-form.html"
}
}
};
This code snippet you place in requirejs-config.js file in your module. That is all.
In general, I want to know how to do code-generation/fabrication in a Webpack plugin on demand. I want to generate contents for files that do not exist when they are "required."
Specifically, I want a plugin that, when I require a directory, automatically requires all files in that directory (recursively).
For example, suppose we have the directory structure:
foo
bar.js
baz.js
main.js
And main.js has:
var foo = require("./foo");
// ...
I want webpack to automatically generate foo/index.js:
module.exports = {
bar: require("./bar"),
baz: require("./baz")
};
I've read most of the webpack docs. github.com/webpack/docs/wiki/How-to-write-a-plugin has an example of generating assets. However, I can't find an example of how to generate an asset on demand. It seems this should be a Resolver, but resolvers seem to only output file paths, not file contents.
Actually for your use case:
Specifically, I want a plugin that, when I require a directory, automatically requires all files in that directory (recursively).
you don't need a plugin. See How to load all files in a subdirectories using webpack without require statements
Doing code-generation/fabrication on demand can be done in JavaScript quite easily, why would you restrict your code generation specifically to only applied, when "required" by WebPack?
As NodeJS itself will look for an index.js, if you require a directory, you can quite easily generate arbitrary exports:
//index.js generating dynamic exports
var time = new Date();
var dynamicExport = {
staticFn : function() {
console.log('Time is:', time);
}
}
//dynamically create a function as a property in dynamicExport
//here you could add some file processing logic that is requiring stuff on demand and export it accordingly
dynamicExport['dyn' + time.getDay()] = function() {
console.log('Take this Java!');
}
module.exports = dynamicExport;
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);