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Does SQLAlchemy have an equivalent of Django's get_or_create?
I want to get an object from the database if it already exists (based on provided parameters) or create it if it does not.
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How is the Linux kernel tested ?
...do the Linux kernel developers test their code locally and after they have it committed? Do they use some kind of unit testing, build automation? test plans?
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Why does my JavaScript code receive a “No 'Access-Control-Allow-Origin' header is present on the req
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If I understood it right you are doing an XMLHttpRequest to a different domain than your page is on. So the browser is blocking it as it usually allows a request in the same origin for security reasons. You need to do something different whe...
Rolling or sliding window iterator?
I need a rolling window (aka sliding window) iterable over a sequence/iterator/generator. Default Python iteration can be considered a special case, where the window length is 1. I'm currently using the following code. Does anyone have a more Pythonic, less verbose, or more efficient method for d...
Docker - how can I copy a file from an image to a host?
...and runs an executable. I also want to copy the build artifact (in my case it's a .zip produced by sbt dist in '../target/`, but I think this question also applies to jars, binaries, etc.
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Remove files from Git commit
I am using Git and I have committed few files using
29 Answers
29
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How to model type-safe enum types?
...follow
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edited Mar 14 '14 at 16:18
qtwo
45744 silver badges99 bronze badges
answered Aug...
What's the difference between $evalAsync and $timeout in AngularJS?
I've been using AngularJS for a little while now, and have found the need to use $timeout every once in a while (Seems to usually be to init a jQuery plugin).
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What is the Python equivalent of static variables inside a function?
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A bit reversed, but this should work:
def foo():
foo.counter += 1
print "Counter is %d" % foo.counter
foo.counter = 0
If you want the counter initialization code at the top instead of the bottom, you can create a deco...
Is it possible to determine whether ViewController is presented as Modal?
Is it possible to check inside ViewController class that it is presented as modal view controller?
14 Answers
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