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 created a framework which opens FXML in other jar files. I use the following to open them:
(fxml) is a string passed in from a DB query...
FXMLLoader loader = new FXMLLoader();
Parent node = loader.load(getClass().getClassLoader().getResource(fxml).openStream());
This works for all my FXML and I really don't want to change this.
I have one new window which will have a very similar implementation with another and I wanted to share the FXML between them with fx:include.
However this throws the error javafx.fxml.LoadException: Base location is undefined.
I found this link about linked files
Is there anyway around this - without changing my entire implementation? If not, likely will just duplicate the logic.
Thanks.
The problem is that if you provide an InputStream, the location (a URL) is undefined. Apparently your FXML is using the location somewhere (e.g. via location resolution) Try
FXMLLoader loader = new FXMLLoader(getClass().getClassLoader().getResource(fxml));
Parent node = loader.load();
I am using unity5.3.3, I would like to know how should I get the asset from an asset bundle, whose names are same but are kept in different folder. My AssetBundle Folder is set in the following manner:
MyAssets -> this Folder is packed as an AssetBundle
-ThemeOne(folder)
- Logo.png
-ThemerTwo(folder)
- Logo.Png
When I do AssetBundle.LoadAssetAsync("Logo"). I end getting the logo in the first(ThemeOne) folder. So how do I access the file in the other folder?
I have just created a sample project so that you can check it out. Check the Folder Assets\AssetBundleSample\SampleAssets\Theme and the script LoadAssets
You can put the ThemeOne(folder) and ThemeTwo(folder) in Resources Folder under assets and than use something like this
(AudioClip)(Resources.Load ("Sounds/" + "myaudioclip", typeof(AudioClip)) as AudioClip)
similarly load your png giving folder name first as i have given sounds and than name of file as i have given myaudioclip .
cast as texture or something as you want
As unity official docs provided
public string BundleURL;
public string AssetName;
IEnumerator Start() {
// Download the file from the URL. It will not be saved in the Cache
using (WWW www = new WWW(BundleURL)) {
yield return www;
if (www.error != null)
throw new Exception("WWW download had an error:" + www.error);
AssetBundle bundle = www.assetBundle;
if (AssetName == "")
Instantiate(bundle.mainAsset);
else
Instantiate(bundle.LoadAsset(AssetName));
// Unload the AssetBundles compressed contents to conserve memory
bundle.Unload(false);
} // memory is freed from the web stream (www.Dispose() gets called implicitly)
}
BundleURL will use the path where your actual assets is located. You can also use Application.dataPath in conjunction with ur specified path string.
Seems to be a bug. Still present in Unity version 5.3.3.
See: http://answers.unity3d.com/questions/1083400/asset-bundle-cant-have-multiple-files-with-the-sam.html
I'm not really sure if this is possible, you should read Explicit naming of assets which is FYI obsolete as on Unity 5.6. This allows you to explicitly name assets that you are including in the bundle. Alternatively, you could try,
bundle.LoadAsset("/Assets/ThemerTwo/Logo.png")
But generally, this is a bad practice to have assets of the same naming in the same bundle.
You can use the method LoadAssetAtPath() to specify the full path instead of just the asset name.
Example:
LoadAssetAtPath("Assets/Texture/Logo.jpg", typeof(Texture2D));
Documentation here: http://docs.unity3d.com/ScriptReference/AssetDatabase.LoadAssetAtPath.html
When I run my program this says:
Error: Could not find or load main class Hello world
Every setting I did several times but I get the same result.
Win 8.1 64 bit
Path settings ok
eclipse jre setting ok
public class newproject1
{
public static void main(String[] args) {
System.out.println("Hello World! ");
}
}
Error: Could not find or load main class newproject1
I would guess that your Hello world class does not have a main method defined. The format is:
public static void main(String[] args)
{
//in here you put what you want the program to do
System.out.println("Hello World");
}
Today I just got "Error: Could not find or load main class" in Eclipse indigo. I don't know if your problem is the same or not. In my case my project code was a backup of the original and I named the backup folder as "x86 31 mayıs yedek". "mayıs" is turkish word and there are generally problems associated with turkish characters and their language settings in software. "ı" in the this word is special character in turkish language.
Simply, I removed the project from the eclipse, I changed the backup folder's name by removing any extra character which is not in english. Then I imported the project again, rebuild it and It worked. Hope this helps.
I had the same problem and after trying some different ways, finally solve my problem by setting again the workspace!
File-> Switch workspace-> other
and set it again!
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);