I'm using python to download an application from a content distribution network. The application downloads as self extract file cabinet. When executed it creates a directory using a version naming format. For example app-version2/../app.exe, Thus I cannot rely on the folder name as it may change in the future. I'm trying to find the best way to work with the content inside the folder without depending on the actual folder name.
My idea was to rename the folder using os.listdir() and then os.rename(app-version2, myapp) This would work but is not automated. What would be the best automated method to find a folder name that contains version numbers and change that to something more static?
Assuming you want to find the path of the directory which begins with app, you can accomplish this using pathlib.Path:
from pathlib import Path
app_path = next(Path().glob('app*'))
This will give you the path to the first file or directory in your current directory whose name begins with "app".
Related
I am working on a spark java wrapper which uses third party libraries, which will read files from a hard coded directory name say "resdata" from where job executes. I know this is twisted but will try to explain.
when I execute the job it is trying to find the required files in the path something like this below,
/data/Hadoop/yarn/local//appcache/application_xxxxx_xxx/container_00_xxxxx_xxx/resdata
I am assuming it is looking for the files in the current data directory , under that looking for directory name "resdata". At this point I don't know how to configure the current directory to any path on hdfs or local.
So looking for options to create directory structure similar to what the third party libraries expecting and copying required files over there. This I need to do on each node. I am working on spark 2.2.0
Please help me in achieving this?
just now got the answer I need to put all the files under resdata directory and zip it say restdata.zip, pass the file using the options "--archives" . Then each node will have directory restdata.zip/restdata/file1 etc
I am extracting a zip at a location, in extracted folder directory there is a install.jar.I want to copy a file to the directory where install.jar is available. Now the zip I am extracting may have different folder structure every time, because of this I can not use
file{'EXTRACTED_PATH/ant.properties':
ensure: present
}
So I wrote a custom fact that will find out path of a install jar & I accessed value in manifest like
$install_jar_location=$::getinstallerpath
Now in facts file I have to give path to search, this path I want to pass as a parameter.
for that I have declared one variable in manifest file
$installer_location="/home/oracle/Installer"
how can I access it in my custom fact.I tried to do following but $installer_locationvariable value is coming up blank.
Facter.add(:getinstallerpath) do
setcode do
Facter::Util::Resolution.exec("echo \"$(dirname \"$(find $installer_location -name \*install*.jar)\")\" ")
end
end
I have created a package and am now creating my tests within the package. For one test my inputs are a set of files and my outputs will be a different set a files created within the test.
I am saving the input files in the test directory of my package and would like to save the output files there too. Since others may run this test, I do not want to specify the input/output file location using my own path eg /home/myname/.julia/v4.0/MyPackage/test/MyInputFile.txt
How do I specify that the input location is within the package's test folder?
So basically how do I tell Julia to look in the packages's folder under the test directory and not have to worry about specifying the entire path including user name etc?
For example currently I have to say
readtable(/home/myname/.julia/v4.0/MyPackage/test/MyInputFile.txt, separator = '\t', header = false)
But I'd like to just be able to say
readtable(/MyPackage/test/MyInputFile.txt, separator = '\t', header = false)
so that no matter who the user of the package is and where they may store the package, they can still run the test?
I know that LOAD_PATH gives the path Julia looks for packages but I can't find any information on where it looks when importing files.
joinpath(Pkg.dir("MyPackage"), "test") is what you need.
As #GnimucK mentioned in a comment, a better solution is
dirname(#__FILE__)
Why is this better? A package could be installed and used from somewhere else (not the standard package directory). Pkg.dir is "stupid" and does not know better. This is rare, of course, and in most cases it won't matter.
I don't want to specify the full directory of a folder or object within my program. I do not want to do this because if a user decides to change the installation folder, it will not function properly. I've seen in HTML you can do something like: ./folder/directory/name and it would work perfectly fine. Is there a way to do something like that within Python?
From https://docs.python.org/3/reference/datamodel.html
__file__ is the pathname of the file from which the module was loaded
You may find it helpful to apply os.path.abspath() to '.' or __file__.
In my linux python app for Fedora I want to save user's work to /home/user/Documents/MyCoolApp/TheirGreatWork.txt
But I am not sure how to find the "Documents" folder if the user is not using English as their default language.
What is the right way to determine the right path so that files go in their "Documents" folder.
EDIT
Here is a which comes up if you change locales... showing how paths can get easily changed.
I'd use the subprocess module to get the output of the command xdg-user-dir DOCUMENTS. For example:
import subprocess
documents_dir = subprocess.check_output(["xdg-user-dir", "DOCUMENTS"])
print documents_dir # This is what you're looking for.
There is no right way as the user may have changed their locale, which (fortunately) does not rename the directory. If you want a fixed place for files managed by your app, use ~user/.MyCoolApp or let the user specify the directory.