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Bundler: Command not found
I am hosting on a vps, ubuntu 10.04, rails 3, ruby and mysql installed correctly by following some tutorials. If I run bundle check or bundle install I get the error '-bash: bundle: command not found'. From gem list --local I see 'bundler (1.0.2, 1.0.0)' is installed.
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Most efficient way of making an if-elif-elif-else statement when the else is done the most?
...:
the_thing = 2
elif something == 'there':
the_thing = 3
else:
the_thing = 4
2.py
something = 'something'
options = {'this': 1, 'that': 2, 'there': 3}
for i in xrange(1000000):
the_thing = options.get(something, 4)
3.py
something = 'something'
options = {'t...
In Clojure, when should I use a vector over a list, and the other way around?
...: no, but they are sequential
[12:21] <rhickey> ,(sequential? [1 2 3])
[12:21] <clojurebot> true
[12:22] <Raynes> When would you want to use a list over a vector?
[12:22] <rhickey> when generating code, when generating back-to-front
[12:23] <rhickey> not too...
Removing numbers from string [closed]
... |
edited Oct 12 '12 at 3:54
answered Oct 12 '12 at 3:34
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The 'json' native gem requires installed build tools
I have ruby 1.9.2p180 (2011-02-18) [i386-mingw32] installed on my windows 7 machine. Now I tried to install the JSON gem using the command, "gem install json" and got the following error.
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not None test in Python [duplicate]
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1033
if val is not None:
# ...
is the Pythonic idiom for testing that a variable is not set to...
Fast permutation -> number -> permutation mapping algorithms
I have n elements. For the sake of an example, let's say, 7 elements, 1234567. I know there are 7! = 5040 permutations possible of these 7 elements.
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How to escape braces (curly brackets) in a format string in .NET
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For you to output foo {1, 2, 3} you have to do something like:
string t = "1, 2, 3";
string v = String.Format(" foo {{{0}}}", t);
To output a { you use {{ and to output a } you use }}.
or Now, you can also use c# string interpolation like this (featu...
How to split a string in shell and get the last field
Suppose I have the string 1:2:3:4:5 and I want to get its last field ( 5 in this case). How do I do that using Bash? I tried cut , but I don't know how to specify the last field with -f .
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Preferred way to create a Scala list
...end is O(1).
– Daniel C. Sobral
Aug 31 '15 at 19:33
@pgoggijr That is not true. First, there's no "change" anywhere, b...