How to exclude UnityEditor reference from asmdef?
Why I need it:
I have an asmdef file. For example, it is MyAssembly/MyAssembly.asmdef. The MyAssembly contains a lot of features and each feature staff is placed in its own folder. And some of these features has a code that is needed only in editor, and it refers to UnityEditor namespace. Such editor code is placed into an Editor folder.
But as you know, Editor folder name means nothing in terms of asmdef usage. So I add AssemblyDefenitionReference in each folder and refer it to the MyAssemblyEditor.asmdef assembly definition. So the paths looks like this:
MyAssembly/MyAssembly.asmdef
MyAssembly/Editor/MyAssemblyEditor.asmdef - this folder contains no code. It's needed just to place asmdef, because it's not allowed to place two asmdefs in a single folder.
MyAssembly/SomeFeature/Editor/*feature editor staff*
MyAssembly/SomeFeature/Editor/Editor.asmref - refers to MyAssemblyEditor.asmdef
MyAssembly/SomeFeature/*feature staff*
All this works good. But the problem is that, when some developer adds a new feature, he can forget to add a reference to the MyAssemblyEditor.asmdef in the editor folder. And there are no any errors will be shown in this case. This mistake will be revealed only when the build will be cooked. But I'd like that using of UnityEditor in MyAssembly will be instantly marked as an error.
Feel free to suggest other solution for this problem.
This thread got me thinking I can use CsprojPostprocessor to remove all references to UnityEditor from my csproj file. I wrote such class:
using System.Text.RegularExpressions;
using UnityEditor;
// ReSharper disable once CheckNamespace
public class CsprojPostprocessor : AssetPostprocessor
{
public static string OnGeneratedCSProject(string path, string content)
{
if (!path.EndsWith("Editor.csproj") && !path.EndsWith("Tests.csproj"))
{
var newContent =
Regex.Replace(content, "<Reference Include=\"UnityEditor(.|\n)*?</Reference>", "");
return newContent;
}
return content;
}
}
It also can be done with an xml parser or something.
The only thing, that confuse me is that this mechanism is badly documented and doesn't look like something simple users should use. So I use it at my own risk, but looks like there is no guarantee it will be strongly supported in future.
I made a word game app for android in Unity, where the player has to find a word from a category previously loaded to the game.
The way I load the categories is:
There is a folder named Categories, inside Assets, I run through the folder and read each text file as a category.
The categories are stored in a dictionary where the key is name the of the file and the value is every line of the file as an array element.
It worked well on the PC however no luck on android. Tried changing the path to
"public string categoriesDirectoryPath = Application.persistentDataPath +"Categories";" still does not work.
Original path was "Assets/Categories"
Code for initiating the dictionary with the file values is (Happens on GameManager's Awake()):
private Dictionary<string, string[]> createCategories(string directoryPath)
{
Dictionary<string, string[]> categories = new Dictionary<string, string[]>();
string[] categoryPaths = Directory.GetFiles(directoryPath);
foreach (string path in categoryPaths)
{
if (!path.EndsWith("meta")) {
Debug.Log(path);
string categoryName = Path.GetFileNameWithoutExtension(path);
Debug.Log(categoryName);
string[] categoryData = File.ReadAllLines(path).ToArray();
categories.Add(categoryName, categoryData);
}
}
return categories;
}
Is there a way of iterating the folder and reading the text files that were in Assets/Categories after building the APK?
Is there a way of iterating the folder and reading the text files that
were in Assets/Categories after building the APK?
No.
If you want to access from the project, in a build, you have two options:
1.Put the file a folder named "Resources" then use the Resources to read the file and copy it to the Application.persistentDataPath path. By copying it to Application.persistentDataPath, you'll be able to modify it. Anything in the "Resources" is read only.
2.Put the file in the StreamingAssets folder then use UnityWebRequest, WWW or the System.IO.File API to read it. Atfer this, you can copy it to the Application.persistentDataPath.
Here is a post with code examples on how to do both of these.
3.AssetBundle(Recommended due to performance and loading reaons).
You can build the file as AssetBundle then put them in the StreamingAssets folder and use the AssetBundle API to read it.
Here is a complete example for building and reading AssetBundle data.
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 am trying to generate external modules rather than a type definition file. I believe I need to do the following:
Change the extension of the file to .ts instead of .d.ts.
Generate one file per module.
Add the key word "Export" in front of each interface and enum.
I was easily able to change the extension of the file by changing the "output extension" setting in the tt file.
I cannot figure out how to split the modules into separate files.
I cannot figure out how to add the Export key word to each interface.
TypeLITE doesn't support generating multiple files. This feature has been requested by several users, but I am not aware of a simple way to generate multiple files from the single tt file.
export keyword can't be added without changing source code of the library (TsGenerator.cs). This is very specific requirement, so I probably won't add it to the library.
TypeLite is a good project but lacking in Documentation and examples, it's open source so anyone can contribute and make it better.
As for creating a file per class i solved it using the code below.
private static void GenerateTypeScriptContracts(string assemblyFile, string outputPath)
{
// Clean TS Folder
System.IO.DirectoryInfo di = new DirectoryInfo(outputPath);
foreach (FileInfo file in di.GetFiles())
{
file.Delete();
}
// --
var assembly = Assembly.LoadFrom(assemblyFile);
// If you want a subset of classes from this assembly, filter them here
var models = assembly.GetTypes();
foreach (var model in models)
{
var generator = new TypeScriptFluent()
.WithConvertor<Guid>(c => "string")
.WithMemberFormatter((identifier) => Char.ToLower(identifier.Name[0]) + identifier.Name.Substring(1));
generator.ModelBuilder.Add(model);
// Generate TS interface definitions
var tsClassDefinitions = generator.Generate(TsGeneratorOutput.Properties | TsGeneratorOutput.Fields);
File.WriteAllText(Path.Combine(outputPath, "I" + model.FullName.Replace("ProjectName.DtoModels.", "") + ".ts"), tsClassDefinitions);
}
}
I want to write an eclipse plugin to show the actual value of the message code. The values are to be loaded from the given resource bundle. Only classes of the resource bundle will be available. So I require to load the resource bundle class which is declared in the current file. These class files will be in the classes folder or in a jar file in the lib folder. Is there any way to load the classes dynamically from the eclipse plugin? Thanks in advance
An Example how to get a externalized string from a message.properties file
private static final String BUNDLE_NAME = "de.stackoverflow.package.messages"; //$NON-NLS-1$
private static final ResourceBundle RESOURCE_BUNDLE = ResourceBundle.getBundle(BUNDLE_NAME);
RESOURCE_BUNDLE.getString(key);
Every plugin does have a bundle object where you can load the files of a plugin. The Bundle object should contain every information you want to use.
Have a look at the Bundle class. There is the following method:
public Class loadClass(String name) throws ClassNotFoundException;
To get the bundle from your plugin:
Activator.getDefault().getBundle()
Hope this help
Manu