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Convert bytes to a string
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You need to decode the bytes object to produce a string:
>>> b"abcde"
b'abcde'
# utf-8 is used here because it is a very common encoding, but you
# need to use the encoding your data is actually in.
>>> b"abcde".decode("utf-8") ...
Test whether a glob has any matches in bash
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Bash specific solution:
compgen -G "<glob-pattern>"
Escape the pattern or it'll get pre-expanded into matches.
Exit status is:
1 for no-match,
0 for 'one or more matches'
stdout is a list of files matching t...
Do I need to explicitly call the base virtual destructor?
...I am implementing the destructor again as virtual on the inheriting class, but do I need to call the base destructor?
7 Ans...
C library function to perform sort
Is there any library function available in C standard library to do sort?
7 Answers
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LINQ - Left Join, Group By, and Count
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from p in context.ParentTable
join c in context.ChildTable on p.ParentId equals c.ChildParentId into j1
from j2 in j1.DefaultIfEmpty()
group j2 by p.ParentId into grouped
select new { ParentId = grouped.Key, Count = grouped.Count(t=>t.ChildId != nu...
jQuery SVG, why can't I addClass?
I am using jQuery SVG. I can't add or remove a class to an object. Anyone know my mistake?
15 Answers
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First-time database design: am I overengineering? [closed]
I'm a first year CS student and I work part time for my dad's small business. I don't have any experience in real world application development. I have written scripts in Python, some coursework in C, but nothing like this.
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Nested defaultdict of defaultdict
Is there a way to make a defaultdict also be the default for the defaultdict? (i.e. infinite-level recursive defaultdict?)
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How to convert a string to an integer in JavaScript?
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The simplest way would be to use the native Number function:
var x = Number("1000")
If that doesn't work for you, then there are the parseInt, unary plus, parseFloat with floor, and Math.round methods.
parseInt:
var x = parseInt("1000", 10); /...
How to generate a random string in Ruby
...'..'z').to_a[rand(26)] }.join
And a last one that's even more confusing, but more flexible and wastes fewer cycles:
o = [('a'..'z'), ('A'..'Z')].map(&:to_a).flatten
string = (0...50).map { o[rand(o.length)] }.join
sh...