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Can I use __init__.py to define global variables?
...package. But I don't know how to do this. Suppose I have a few subpackages and each with several modules. How can I access that variable from these modules?
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Dynamic constant assignment
..."blah"
If you really want to change the value of a constant in a method, and your constant is a String or an Array, you can 'cheat' and use the #replace method to cause the object to take on a new value without actually changing the object:
class MyClass
BAR = "blah"
def cheat(new_bar)
B...
How can I make my match non greedy in vim?
... answered Aug 20 '09 at 12:45
Randy MorrisRandy Morris
35.9k55 gold badges6262 silver badges7373 bronze badges
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Kill detached screen session [closed]
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@laffuste 's comment worked for me, but quit and :quit lead to command not found on my remote Linux server (perhaps differences between versions of the OS or screen are to blame)
– Hack-R
Mar 20 '15 at 15:05
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How to count total lines changed by a specific author in a Git repository?
Is there a command I can invoke which will count the lines changed by a specific author in a Git repository? I know that there must be ways to count the number of commits as Github does this for their Impact graph.
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Why does my 'git branch' have no master?
I'm a git newbie and I keep reading about a "master" branch. Is "master" just a conventional name that people used or does it have special meaning like HEAD ?
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How to create named and latest tag in Docker?
...n image containing Node.js 0.10.24). I built that image using a Dockerfile and executing docker build and by providing a tag using the -t parameter.
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Best way to get application folder path
...ple see @Vimvq1987's answer.
CodeBase is the place where a file was found and can be a URL beginning with http://. In which case Location will probably be the assembly download cache. CodeBase is not guaranteed to be set for assemblies in the GAC.
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Converting a List to a comma separated string
Is there a way to take a List and convert it into a comma separated string?
8 Answers
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Use a normal link to submit a form
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Two ways. Either create a button and style it so it looks like a link with css, or create a link and use onclick="this.closest('form').submit();return false;".
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