I'm surprised that IntelliJ 13.1.6 can't do this.
I've tried the solution in this answer but it shows the pictured dialog without the ability to specify a file name. So choosing the same directory causes an error saying we can't create the file because it already exists (obviously -- I want to be able to rename it!).
It only applies to files with both an class and object (or multiple of either).
Is there anything I can do to duplicate a file or must I switch to explorer and do it there?
Related
I was trying to get yapf configured for a Python project, and so I created a .style.yapf file, which VSCode did not recognize by its extension. I thought it was in TOML format, but when I changed the file association for .style.yapf to TOML, the parser shows an error (because values aren't quoted). So I tried changing the name of the file (not really a solution, because the tool is looking for a file with that name), and found that if I change it to a name ending in .cfg, VSCode seems to parse and highlight it appropriately. Unfortunately, there doesn't seem to be anything in the list of file associations corresponding to this file mode.
I expected to find a cfg mode in the file associations list, but there wasn't an entry with that name, nor was there an entry that used the same gear icon displayed next to the open file name in its tab (which I have been assuming is an indicator of the mode that VSCode is using to display and format the file).
Is there any way for me to get VSCode to tell me what the current file mode is for an open file? Or give me a list of the default associations?
I'm more interested in understanding the tools in general than solving the one minor annoyance that sparked the question, but I'll settle for a solution to that problem, too.
Edit: Here's the content of the .style.yapf file in question:
[style]
based_on_style = google
If I rename the file to .yapf.cfg (or any other name ending in .cfg, I believe), it gets handled properly, but all of the file associations I've tried (including ini, Properties, and TOML) indicate an error because there are no quotes around google.
Hi im new to programming, how do i import a .txt file? My code cant find the file, is there any specific directory it has to be put into?
My code:
object Zettel01 extends App {
import scala.io.Source
object Suchtest {
val gesch = Source.fromFile("DieUnendlicheGeschichte.txt").getLines()
for (w <- gesch) println(w)
}
}
I have tried different code but the problem is always the same, i cant find the .txt file...
Thanks in advance for any help
Flurry1337
Every Scala program that you run on your computer is ultimately a java process. This process will have a "working directory", just as every process on your computer does. By default, the working directory is the working directory of the process that started it, that is, the current directory of the shell or command-line interpreter at the time when you started your program.
Now, that means it is important to know how exactly you start your program. If you are using a command line and start your program in the fashion of java MyCoolProgram, then the current directory of the shell will become the working directory of the program. If you use an IDE like Eclipse or IntelliJ IDEA, those typically use the project folder of your IDE project as the working directory of the process that they start.
There is a simple way to find it out quickly: You can always print out the result of new java.io.File(".").getAbsolutePath(). This will print the full path to the working directory. For example, you can write a little Scala program like this:
object PrintWorkingDirectory extends App {
println(new java.io.File(".").getAbsolutePath())
}
and start it. On the console, you should find the full path of the program's working directory. If you put a file named "DieUnendlicheGeschichte.txt" in this directory, your program will find that file under exactly that file name.
Of course, you don't have to dump all your files into that one directory. You can make subdirectories in order to organize your files better. For example, you might put your file in a path like "resources/text/DieUnendlicheGeschichte.txt".
Finally, I would like to point out that there is also a different way to associate resource files with your program, and to load them. The idea is that you put the code (class files) as well as resources like texts, images, CSV files, XML files and the like into one big file. This would be a JAR file. You can then use ClassLoader to access resources inside the JAR file by a URL.
Explaining that process in detail is out of scope for this question; this is just dropping a couple of buzzwords that you (or other readers) can search for in case they want to look up a more elaborated process.
System.getProperty("user.dir") also tells you the working directory.
I have a small perl script that I use to search archives for members matching a name. I'd like to enhance this so that if it finds any members in the archive that are also archives (zip, jar, etc) it will then recursively scan those, looking for the original desired pattern.
I've looked through the "Archive::Zip" documentation, and I thought I saw how to do this. I noticed the "fh()" and "readFromFileHandle()" methods. However, in my testing, it appears that the "fh()" call on an archive member returns the file handle for the containing archive, not the member. Perhaps I'm doing it wrong, but I would appreciate an example of how to do this.
You can't read the contents of any sort of archive member (whether it is text, picture, or another archive) without extracting it from the archive file.
Once you have identified a member that you want to view, you must call extractMember (or, more likely, extractMemberWithoutPaths if the file is to be temporary) to extract it to a disk file. Then you can create a new Archive::Zip object and read the new file while keeping the old one open.
You will presumably want to unlink the archive file once you have catalogued its contents.
Edit
I hadn't come across the Archive::Zip::MemberRead module before. It appears you were on the right track with readFromFileHandle. I would guess that it should work like this, but it would be awkward for me to test it at present.
my $zip = Archive::Zip->new;
$zip->read('myfile.zip');
my $zipfh = Archive::Zip::MemberRead->new($zip, 'archive/path/to/member.zip');
my $newzip = Archive::Zip->new;
$newzip->readFromFileHandle($zipfh)
I know there's a way to make the enter-key on the keyboard confirm the inputdlg() dialog, see here:
http://www.mathworks.de/support/solutions/en/data/1-39UWQT/index.html?product=ML&solution=1-39UWQT
Okay it's a workaround, the problem is, I have to store it in the matlab directory (because as said on the page: "Since it is dependent on several private functions, newid.m will only work if stored in this location.").. The problem now is, I want to make a code which doesn#t rely on user changes in the Matlab directory because of missing privileges... Is there still another (perhaps dirty) way to achieve this behaviour?
Thanks!
Solution:
Copy the newid.m from http://www.mathworks.de/support/solutions/en/data/1-39UWQT/index.html?product=ML&solution=1-39UWQT to a arbitrary folder, e.g. into your projects folder
Go into the folder where the original inputdlg() is stored; you can find this out via which inputdlg (e.g. C:\Program Files\MATLAB\R2011b\toolbox\matlab\uitools\inputdlg.m)
Go into the private-subdirectory and copy the two files getnicedialoglocation.m and setdefaultbutton.m and paste them into your project's folder where the newid.m is located. Now call all your input dialogs by using newid() instead of inputdlg()
On some of the controls im using locatisation (language property).
Besides default (that is in english) im also using additional language (slovenian).
Other localisation is done with application resource file.
If i debug the app on emulator there is a directory (si-sl), but if i build a CAB and install the CAB on emulator there is no such directory there.
Does any one know why is that so?
I'd have to see your CAB's INF, but I have a theory. CAB files have a limitation that they can only have one instance of a source file name. If you have more than one instance, CABWIZ will not complain or give an error, it simply will use (IIRC) the last one referenced.
This means is you have source files like this:
./en/resource.dll
./es/resource.dll
./jp/resource.dll
If you build your CAB file, all instances of the resource.dll file in the target will be identical, not the separate file you think you're pushing. To get around this you have to rename them at the source side going into the CAB.
You can use the CAB to rename them coming back out to all have the same name (i.e. resource.dll), so hand-modifying the INF can still get you the desired output structure. You just have to modify the CAB build process to rename your duplicates.