I wat to save a file.I use this.
FileChooser fileChooser = new FileChooser();
File file = fileChooser.showSaveDialog(null);
But in the dialog I want to suggest a name for the file, so that the user only selects a directory for the given file.The name of the file is known already.So i want to suggest that filename.
ThankYou.
This is now fixed in Javafx 2.2.45 (bundled with java 7.0_45 now) and you can do what the OP is suggesing with the following property of fileChooser, setInitialFilename, used as such:
FileChooser myFile = new FileChooser();
myFile.setInitialFileName("Whatever_file_I_want.coolFile");
Now, I don't think there is anyway to STOP the user from choosing a different file, but at leas this will give them a default you want them to pick.
Initial file name providing - it is a thing, which requires to transfer your string (initial name) through native call, to the call of the native file chooser. It is a complex thing, and you can look at these issues about its implementing :
http://javafx-jira.kenai.com/browse/RT-16111 (main one)
http://javafx-jira.kenai.com/browse/RT-24588
http://javafx-jira.kenai.com/browse/RT-24612
They all have fix version lombard, so, they are fixed in JDK 8.
So, you can specify initial file name for a file, starting from JDK 8 (you can access it, downloading JDK early access).
Recently, I've tested this feature, and it is working.
There is a method setInitialName() or smth like that.
And, as I've mentioned, it is a complex thing, and you are not likely to be able to implement it by yourself (until you are able to build jfx).
So, the decision - to wait until JDK8 release, or to use early access builds. Or, to use your own implementation of file chooser.
Here's a workaround that worked for me:
you can use javafx.stage.DirectoryChooser to select a directory for the file you want to save and after saving create a new file in this directory with the default name and extension.
DirectoryChooser dc = new DirectoryChooser();
File file = dc.showDialog(null);
if (file != null) {
file = new File(file.getAbsolutePath() + "/dafaultFilename.extension");}
Related
I am new to kotlin.
I normally used val file = "Data.txt" that easily creates the file. But now the app is almost completed I noticed that code is not able to create file anymore. Whenever I tries to create file (execute it with a button and app crashes). With so many errors (one of them) "java.io.FileNotFoundException: /data/user/0/com.example.waterledger/files/Data.txt (No such file or directory)"
When I locate the folder and manually added the file there. The app works without any problem (it easily read, writes, modify the file "Data.txt") but when I build apk it didn't get that text file that is manually created via gradle so the app crashes when I press the button
I don't think that is the proper way to create a file in Kotlin. What your code is doing is simply creating a variable named file whose value is "data.txt".
However, Kotlin provides a number of methods to create files. One of them is the writeText() method. So you could create your file like this using the writeText() method
val fileName = "data.txt"
var file = File(fileName)
file.writeText("")
The "" will simply give an empty string, as argument to writeText(), to write data to file. You may provide the string you would like to write into this file.
I am using the latest version of pyRevit, v45.
I'm writing some info in temporary files with
myTempFile = script.get_instance_data_file("id")
This creates a file named pyRevit_2018_xxxx_id.tmp in which I store useful info. If I'm not mistaken, the "xxxx" part is changing every time I reload Revit. Now, I need to get access to this information from another pyRevit script.
How can I retrieve the name of the temp file I need to read? In other words, how do I access "myTempFile" from within the second script, which has no idea of the name of "myTempFile"?
I guess I can share somehow that variable between my script, but what's the proper way to do this? I know this must be a very basic programming question, but I'm indeed not a programmer ;)
Thanks a lot,
Arnaud.
Ok, I realise now that my variables in the 1st script cease to exist after its execution.
So for now I wrote the file name in another file, of which I know the name.. That works.
But if there's a cleaner way to do this, I'd be glad to learn ;)
Arnaud
pyrevit.script module provides 4 different methods for creating temporary files based on their use case:
get_instance_data_file:
for data files marked with Revit instance pid. This means that scripts running on another instance will not see this temp file.
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_instance_data_file
get_universal_data_file:
for temp files accessible to all Revit instances and versions
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_universal_data_file
get_data_file:
Base method to get a standard temp file for current revit version
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_data_file
get_document_data_file:
temp file marked with active document (so scripts working on another document will not see this)
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_document_data_file
Each method uses a pattern to create the temp file name. So as long as the call to the method is the same of different scripts, the method generates the same file name.
Example:
Script 1:
from pyrevit import script
tfile = script.get_data_file('mydata')
Script 2:
from pyrevit import script
tempfile = script.get_data_file('mydata')
In this example tempfile = tfile since the file id is the same.
There is documentation on each so make sure you take a look at those and pick the flavor that serves your purpose.
Is there a way to construct a filepath that links to the Documents folder of the active user. So instead of C:\Users\User\Documents\ something like C:\Active_User\Documents\
ps. I try to make use of this in KNIME.
The file chooser elements in KNIME understand a URL in the form of "knime://knime.workflow" which accesses the current workflow location regardless of higher directory path.
You could also use a Java Variable Edit to get the username in Java, which you use to create a string that can be used by the File Reader (or other node) as a flow variable.
It depends on what you're trying to achieve.
You can use
C:\Users\%USERNAME%\Documents
which will use the environment variable %USERNAME% (= current user).
In C#/.NET you can use Environment.SpecialFolder.MyDocuments like this:
Console.WriteLine("GetFolderPath: {0}",
Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments));
In Java System.getProperty("user.home"); should give you the right base diretory to start with.
I am writing a small command line utility. It should hopefully be able to run on OSX, UNIX and Linux.
It needs to save a few preferences somewhere, like in a small YAML config file.
Where would one save such a file?
Language: Python 2.7
OS: *nix
Commonly, these files go somewhere like ~/.rc (eg: ~/.hgrc). This could be the path to a file, or to a directory if you need lots of configuration settings.
For a nice description see http://www.linuxtopia.org/online_books/programming_books/art_of_unix_programming/ch10s03.html
I would avoid putting the file in the ~ directory only because it has gotten totally flooded with crap. The recent trend, at least on ubuntu, is to use ~/.config/<appname>/ for whatever dot files you need. I really like that convention.
If your application is named "someapp" you save the configuration in a file such as $HOME/.someapp. You can give the config file an extension if you like. If you think your app may have more than one config file you can use the directory $HOME/.someapp and create regular-named (not hidden) files in there.
Many cross-platform tools use the same path on OS X as on linux (and other POSIX/non-Windows platforms). The main advantage of using the POSIX locations isn't saving a few lines of code, but saving the need for Mac-specific instructions, and allowing Mac users to get help from the linux users in the community (without any need to translate their suggestions).
The other alternative is to put them in the "Mac-friendly" locations under ~/Library instead. The main advantage of using the Mac locations is basically "Apple says so"—unless you plan to sandbox your code, in which case the main advantage is that you can do so.
If you choose to use the Library locations, you should read About the OS X File System and OS X Library Directory Details in the File System Programming Guide, but here's the short version:
Almost everything: Create a subdirectory with your app's name or bundle ID (unless you're going out of your way to set a bundle ID, you'll get org.python.python, which you don't want…) under ~/Library/Application Support. Ideally you should use APIs like -[NSFileManager URLForDirectory:inDomain:appropriateForURL:create:error:] to get the path; if not, you have to deal with things like localization, sandbox containers, etc. manually.
Anything that can be easily re-created (so it doesn't need to be backed up, migrated, etc.): An identically-named subdirectory of ~/Library/Caches.
Preferences: Use the NSUserDefaults or CFPreferences APIs instead. If you use your own format, the "old" way of doing things is to create a subdirectory under ~/Library/Preferences named with your app's name or bundle ID, and put your files in that. Apple no longer recommends that, but doesn't really recommend an alternative (short of "use CFPreferences, damnit!"); many apps (e.g., Aquamacs) still do it the old way, but others instead pretend they're not preferences and store them under Application Support.
In Python, this works as follows (leaving out the error handling, and assuming you're going by name instead of setting a bundle ID for yourself):
from Foundation import *
fm = NSFileManager.defaultManager()
appsupport = (fm.URLForDirectory_inDomain_appropriateForURL_create_error_(
NSApplicationSupportDirectory, NSUserDomainMask, None, True, None)[0].
URLByAppendingPathComponent_isDirectory_(
appname, True))
caches = (fm.URLForDirectory_inDomain_appropriateForURL_create_error_(
NSCachesDirectory, NSUserDomainMask, None, True, None)[0].
URLByAppendingPathComponent_isDirectory_(
appname, True))
prefs = NSUserDefaults.persistentDomainForName_(appname)
I want to know how to dynamically assign file name using Log4net .My application is such that 10 different files should be dynamically created based on user input ,and later based on the name the corresponding file name needs to be picked up and information written to it
For example in my application based on my buisness requirement for every xml file a corresponding log file with the same name as xml file should be created .Later whenever I do any modification to the xml file an entry needs to be in the corresponding log file
Please help . I having trouble to get control of the appropriate log to write it
Have not done this, but there are probably a number of ways of doing this, so this may not be the best way, but it should work
public OpenLogFile(string fileName)
{
log4net.Layout.ILayout layout = new log4net.Layout.PatternLayout("%d [%t]%-5p : - %m%n");;
log4net.Appender.FileAppender appender = new log4net.Appender.FileAppender(layout , filename);
appender.Threshold = log4net.Core.Level.Info;
log4net.Config.BasicConfigurator.Configure(appender);
}
Then just call OpenLogfile when you need to switch files.
You might need to tweak the layout or appender type.
A big disadvantage of this method is you losing the xml configuration and the ability to change settings at runtime. So a better way might be to configure your appender in the xml file to use a property
eg
file type="log4net.Util.PatternString" value="Logfiles\Log_For_%property{MyLogFileName}"
Then in your code you could change the property
log4net.GlobalContext.Properties["MyLogFileName"] = ...;
The tricky bit is to get log4net to reload itself. I haven't read the documentation of this, so I don't know if there is a way of forcing a reload. It might work if you just call log4net.Config.XmlConfigurator.ConfigureAndWatch again. Otherwise it should work if you opened the xml file and saved it again (without needing to change anything)
Hope this helps.