I've a WebGl Unity application with some scenes.
I need to do a build for each scene, suppose SceneA, SceneB, SceneC...
So, to do build I wrote:
"C:\Program Files\Unity\Editor\Unity.exe" -buildTarget WebGL ... SceneName ?
Questions are:
How to specify SceneName ?
And how to specify destination directory ?
Thanks
Are you asking how to specify which scenes are used when making a build through the command line? If so, here's how I would go about it.
For your first question,
How to specify SceneName ?
I am unaware if you can directly specify scene names in the command line, but you can run specific functions using the command line that can change which scenes are built for the project.
Here is a snippet you would need but can change for your use case
using UnityEditor;
class MyEditorScript
{
static void PerformBuild ()
{
string[] scenes = { "Assets/MyScene.unity" };
BuildPipeline.BuildPlayer(scenes, ...);
}
}
The above snippet is quite simple, you specify each scene directly then add it to your build settings. You can pass in parameters to get different results or make multiple functions, whichever is easiest for you. Here is how you would call this function in the command line
"C:\Program Files\Unity\Editor\Unity.exe" -batchmode -executeMethod MyEditorScript.PerformBuild
I use -batchmode here so if any prompts are brought up, Unity will silence them so the process can be much more automated.
As for your second question,
And how to specify destination directory ?
I am not sure if you want to tie the directory where you save and the scenes you build with, but it might make everything easier. There is a build setting called BuildPlayerOptions.locationPathName, which allows you to specify where the build is save
static void WebGLProductionBuild()
{
// Build the player.\
BuildPlayerOptions buildPlayerOptions = new BuildPlayerOptions();
buildPlayerOptions.scenes = new[] { "Assets/Scene1.unity", "Assets/Scene2.unity" };
buildPlayerOptions.locationPathName = "yourLocationNameHere";
buildPlayerOptions.target = BuildTarget.WebGL;
buildPlayerOptions.options = BuildOptions.None; // set whatever you want here
BuildPipeline.BuildPlayer(buildPlayerOptions); // apply the setting changes
}
To call the above method is exactly the same except for the function call as the above line
"C:\Program Files\Unity\Editor\Unity.exe" -batchmode -executeMethod MyEditorScript.WebGLProductionBuild
I do not believe what you want to achieve can completely be done from command line, but this answer is pretty close. I am also not exactly sure if you wanted the scenes and the new path bundled together. If you are looking to build each individual scene and specify a new path, you can do so by making a new function with a switch statement or just make multiple functions to call.
Unity already provides answer to a similar question here.
You can create a method to make build validations, change settings and build different players. These methods have to be inside a script file located in an Editor folder, for example Assets/Editor/Builders/Builder.cs. The class also doesn’t need to extend any Unity class - the only requirement is that static functions are used.
using UnityEditor;
using System;
class Builder
{
static void Build ()
{
string[] arguments = Environment.GetCommandLineArgs();
// ... your code here, validations, flag changes, etc.
// foreach(string arg in arguments)
// Filter...
string sceneName = argument[1];
string destinationDirectoryPath = arguments[2];
// Build the player.\
BuildPlayerOptions buildPlayerOptions = new BuildPlayerOptions();
buildPlayerOptions.scenes = new[] { sceneName };
buildPlayerOptions.locationPathName = destinationDirectoryPath;
//buildPlayerOptions.target = "Your Target";
BuildPipeline.BuildPlayer(buildPlayerOptions);
}
}
You can then call the builder function using these command:
/Path/To/Unity -batchmode -executeMethod Builder.Build SceneName
DestinationDirectoryPath
I got a tiny script that creates a text file in Resources folder before building that needs to be included in the build. So I wrote this script using Unity's IPreprocessBuildWithReport:
using UnityEngine;
using UnityEditor;
using UnityEditor.Build;
using System.IO;
class MyCustomBuildProcessor : IPreprocessBuildWithReport
{
public int callbackOrder { get { return 0; } }
public void OnPreprocessBuild(UnityEditor.Build.Reporting.BuildReport report)
{
File.WriteAllText(
Application.dataPath + "/Resources/version.txt",
string.Format("{0}", PlayerSettings.Android.bundleVersionCode));
AssetDatabase.Refresh();
}
}
Note the AssetDatabase.Refresh(); statement at the end. It makes sure Unity becomes aware of the change and includes the updated file in the build. Now, whenever time I change the bundleVersionCode and press build, Unity updates the file as expected but the build fails with this pretty generic error:
However, if I try to build the second time with the same bundleVersionCode than no file changes occur and the build succeeds.
So I guess the AssetDatabase.Refresh() doesn't really work in OnPreprocessBuild(), or am I doing something obviously stupid again? Can anyone suggest a workaround?
EDIT:
Please ignore the first two errors, just Unity doesn't like me when I'm excluding files from build by the ~ postfix.
Is there a way to exclude a subdirectory of "tests" from being watched by ember serve?
ex. I have a dir under "tests" called "selenium".
If 'ember serve' is running and I do "echo xyz > bam" in the selenium folder,
the ember serve console will output "file changed selenium/bam" and rebuild the app.
I would like to understand if there is a way to exclude the entire selenium dir without moving it out of the tests folder ( even though for now I am moving the dir)
either:
You can use broccoli-unwatched-tree.
Just add var seleniumTree = new UnwatchedTree('tests/selenium'); to your Brocfile.js before you add that tree to your application with module.exports = app.toTree(seleniumTree);
You can continue to use this tree just like any of the others, but due to the way it is implemented the standard Broccoli::Watcher class will not check it for file changes.
or:
You could also implement this functionality yourself by adding an object that exports the read() and cleanup() functions that are expected by the broccoli watcher, leaving the cleanup() function empty as is done for broccoli-bower.
// Almost the same as using a plain string for a tree; but unlike a plain
// string, Broccoli won't watch this
function UnwatchedTree (dir) { this.dir = dir }
UnwatchedTree.prototype.read = function (readTree) { return this.dir }
UnwatchedTree.prototype.cleanup = function () { }
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 have a Groovy script which is run like this:
File scriptFile = ...;
ScriptEngine engine = ...;
String script = FileUtils.readFileToString(scriptFile);
Object evalResult = engine.eval(script, bindings);
Unsurprisingly, breakpoint set in the script file doesn't trigger. What can I change to make it work? The script needs to be run in the context of the larger program (no separate launch configuration), and through a ScriptEngine, and the file is only known at runtime.
I'm using this hack: I've defined a Java class Debugger which looks like this:
public class Debugger {
private static final Logger log = LoggerFactory.getLogger( Debugger.class );
/**
* Invoke this method anywhere to end up in the Java debugger.
*
* <p>Sometimes, you have code that might eventually be executed or you have a place
* where you want to stop but no code line.
*
* <p>In these case, use this method.
*
* <p>Don't forget to set a breakpoint, first :-)
* */
public static void stopInDebugger() {
log.error( "Please set a breakpoint in Debugger.stopInDebugger() and try again whatever you just did!" );
}
}
I have a breakpoint in the log.error line in Ecipse.
Now, I can put this line into the script where I want a breakpoint:
Debugger.stopInDebugger();
Granted, it doesn't allow me to easily step through the script but it's better than nothing.
Is your script file in a source folder on the classpath (sounds like it's not)? If not, make it so. You can also change your preferences to ensure that the script file is never compiled by the compiler (and optionally not even copied to the output folder). Go to Preferences -> Groovy -> Compiler and look at script folders to make this happen.