Here I have two files:
file.rb
def method
puts "This won't be outputted."
end
puts "This will be outputted."
main.rb
require "./file"
When running main.rb it will load all the code inside file.rb so I will get "This will be outputted." on the screen.
Is it possible to load a file without having it to run the code?
Cause I want to load all the methods (in modules and classes too) without having to execute code outside these scopes.
Is it possible to load a file without having it to run the code?
No, everything in a ruby file is executable code, including class and method definitions (you can see this when you try to define a method inside an if-statement for example, which works just fine). So if you wouldn't execute anything in the file, nothing would be defined.
You can however tell ruby that certain code shall only execute if the file is run directly - not if it is required. For this simply put the code in question inside an if __FILE__ == $0 block. So for your example, this would work:
file.rb
def method
puts "This won't be outputted."
end
if __FILE__ == $0
puts "This will not be outputted."
end
main.rb
require "./file"
the if __FILE__ == $0 is nice, but a way more in keeping with ruby's Object Oriented approach is to put all the methods you want access to in a class (as class methods), and then call them from main.rb.
e.g.
file.rb
class MyUtils
def self.method
puts "this won't be outputted"
end
end
and then in main.rb
require "/.file.rb"
and when you want to use your utility methods:
MyUtils.method
I don't think modifying file is good idea - there are could be a lot of files like this one or these files belong to customer and a ton of another reasons.
Ruby is good at metaprogramming so why don't use this feature?
It could be like this.
Create file with fake module and put here the file.
File.open("mfile.rb","w") do |f|
f.write "module FakeModule
"
f.write File.open("file.rb").read
f.write "
end"
end
Then load this file:
require "/.mfile.rb
and accessing to the method:
FakeModule::method
Related
If I want to run a bunch of ruby scripts (super similar, with maybe a number changed as a commandline argument) and still have them output to stdout, is there a way to do this?
i.e a script to run these:
ruby program1.rb input_1.txt
ruby program1.rb input_2.txt
ruby program1.rb input_3.txt
like
(1..3).each do |i|
ruby program1.rb input_#{i}'
end
in another script, so I can just run that script and see the output in a terminal from all 3 runs?
EDIT:
I'm struggling to implement the second highest voted suggested answer.
I don't have a main function within my program1.rb, whereas the suggested answer has one.
I've tried this, for script.rb:
require "program1.rb"
(1..6).each do |i|
driver("cmd_line_arg_#{i}","cmd_line_arg2")
end
but no luck. Is that right?
You can use load to run the script you need (the difference between load and require is that require will not run the script again if it has already been loaded).
To make each run have different arguments (given that they are read from the ARGV variable), you need to override the ARGV variable:
(1..6).each do |i|
ARGV = ["cmd_line_arg_#{i}","cmd_line_arg2"]
load 'program1.rb'
end
# script_runner.rb
require_relative 'program_1'
module ScriptRunner
class << self
def run
ARGV.each do | file |
SomeClass.new(file).process
end
end
end
end
ScriptRunner.run
.
# programe_1.rb
class SomeClass
attr_reader :file_path
def initialize(file_path)
#file_path = file_path
end
def process
puts File.open(file_path).read
end
end
Doing something like the code shown above would allow you to run:
ruby script_runner.rb input_1.txt input_2.txt input_3.txt
from the command line - useful if your input files change. Or even:
ruby script_runner.rb *.txt
if you want to run it on all text files. Or:
ruby script_runner.rb inputs/*
if you want to run it on all files in a specific dir (in this case called 'inputs').
This assumes whatever is in program_1.rb is not just a block of procedural code and instead provides some object (class) that encapsulates the logic you want to use on each file,meaning you can require program_1.rb once and then use the object it provides as many times as you like, otherwise you'll need to use #load:
# script_runner.rb
module ScriptRunner
class << self
def run
ARGV.each do | file |
load('program_1.rb', file)
end
end
end
end
ScriptRunner.run
.
# program_1.rb
puts File.open(ARGV[0]).read
In my Ruby without Rails application, I have a bunch of methods saved in various files, like methods.rb, other_methods.rb etc, that I want to access from my main file, main.rb.
Rather than spell out each 'require_relative' individually, my main.rb file currently reads:
main.rb
Dir["./*.rb"].each {|file| require_relative file }
puts "executing"
The only problem is this outputs:
"executing"
"executing"
My current hacky approach is:
Dir["./*.rb"].each {|file| next if file == "./main.rb"; require file }
Is there a better way?
Should I be wrapping my methods in a Module, or a Mixin instead and then using 'include'?
I'm new to Ruby, so apologies if this sounds really silly.
I can't seem to figure out how to write a "main" code and have methods in the same file (similar to C). I end up with a "main" file which loads a seperate file that has all the methods. I appreciate any guidance on this.
I spotted the following SO post but I don't understand it:
Should I define a main method in my ruby scripts?
While it's not a big deal, it's just easier being able to see all the relevant code in the same file. Thank you.
[-EDIT-]
Thanks to everyone who responded - turns out you just need to define all the methods above the code. An example is below:
def callTest1
puts "in test 1"
end
def callTest2
puts "in test 2"
end
callTest1
callTest2
I think this makes sense as Ruby needs to know all methods beforehand. This is unlike C where there is a header file which clearly list the available functions and therefore, can define them beneath the main() function
Again, thanks to everyone who responded.
#Hauleth's answer is correct: there is no main method or structure in Ruby. I just want to provide a slightly different view here along with some explanation.
When you execute ruby somefile.rb, Ruby executes all of the code in somefile.rb. So if you have a very small project and want it to be self-contained in a single file, there's absolutely nothing wrong with doing something like this:
# somefile.rb
class MyClass
def say_hello
puts "Hello World"
end
end
def another_hello
puts "Hello World (from a method)"
end
c = MyClass.new
c.say_hello
another_hello
It's not that the first two blocks aren't executed, it's just that you don't see the effects until you actually use the corresponding class/method.
The if __FILE__ == $0 bit is just a way to block off code that you only want to run if this file is being run directly from the command line. __FILE__
is the name of the current file, $0 is the command that was executed by the shell (though it's smart enough to drop the ruby), so comparing the two tells you precisely that: is this the file that was executed from the command line? This is sometimes done by coders who want to define a class/module in a file and also provide a command-line utility that uses it. IMHO that's not very good project structure, but just like anything there are use cases where doing it makes perfect sense.
If you want to be able to execute your code directly, you can add a shebang line
#!/usr/bin/env ruby
# rest of somefile.rb
and make it executable with chmod +x somefile.rb (optionally rename it without the .rb extension). This doesn't really change your situation. The if __FILE__ == $0 still works and still probably isn't necessary.
Edit
As #steenslag correctly points out, the top-level scope in Ruby is an Object called main. It has slightly funky behavior, though:
irb
>> self
=> main
>> self.class
=> Object
>> main
NameError: undefined local variable or method `main' for main:Object
from (irb):8
Don't worry about this until you start to dig much deeper into the language. If you do want to learn lots more about this kind of stuff, Metaprogramming Ruby is a great read :)
No there isn't such structure. Of course you can define main function but it won't be called until you do so. Ruby execute line by line so if you want to print 'Hello World' you simply write:
puts 'Hello World'
The question that you mentioned is about using one file as module and executable, so if you write
if __FILE__ == $0
# your code
end
It will be called only if you run this file. If you only require it in other file then this code will never run. But IMHO it's bad idea, better option is using RubyGems and there add executables.
Actually there is a main, but it is not a method; it's the top-level object that is the initial execution context of a Ruby program.
class Foo
p self
end
#=> Foo
p self
#=> main
def foo
p self
end
foo
#=> main
There is no magic main function in Ruby. See http://en.wikipedia.org/wiki/Main_function#Ruby
If you wish to run Ruby scripts like C compiled files, do the following:
#!/usr/bin/env ruby
puts "Hello"
and then chmod a+x file_name.rb. Everything that is below the first line will be run, as if it was contents of main in C. Of course class and function definitions won't give you any results until they are instantiated/invoked (although the code inside class definitions is actually evaluated, so you could get some output but this is not expected in normal circumstances).
Another way to write main() method is:
class HelloWorld
def initialize(name)
#name = name
end
def sayHello()
print "Hello ##name!"
end
end
def main()
helloWorld = HelloWorld.new("Alice")
helloWorld.sayHello
end
main
My simple attempt to redefine instance methods are not working
class File
alias_method :old_atime, :atime
def atime(*args)
puts "helllllo"
old_atime(*args)
end
end
f = File.new("C:\\abc.txt","w")
puts f.atime
Any idea why?
I'm attempting to print "helllllo" everytime File#atime is called. Even alias old_atime atime is not working.
Is there something I'm doing wrong here?
Above code works perfectly as it should be. Puts "helllllo" writes "helllllo" in to your opened file. Puts inside the file instance meant for writing.
Just call f.close and open your file in text editor. You can see the content.
Yep, Ramesh is right. Try this:
class File
alias_method :old_atime, :atime
def atime(*args)
Kernel.puts "helllllo" # <---- Kernel method
old_atime(*args)
end
end
f = File.new("C:\\abc.txt","w")
puts f.atime
The issue is that 'puts' is defined in File for writing to files. You want the Kernel one which is used unless defined in a more specific scope.
This should work fine, but IO#puts writes to the IO object itself, not STDOUT. In other words, it's writing to the file.
Call f.atime a few times and then f.close within irb and you should see it printing helllllo to the file for each call to atime.
To print to STDOUT, you could use $stdout.puts or Kernel.puts.
Is there a shorter way to require a file located in the same directory (as the script being executed)?
require File.expand_path(File.dirname(__FILE__) + '/some_other_script')
I read that require "my_script" and require "./my_script" will actually load the script twice (ruby will not recognize that it is actually the same script), and this is the reason why File.expand_path is recommended: if it is used every time the script is required, then it will only be loaded once.
It seems weird to me that a concise language like Ruby does not seem to have a shorter solution. For example, python simply has this:
import .some_other_module_in_the_same_directory
I guess I could monkey-patch require... but that's just evil! ;-)
Since ruby 1.9 you can use require_relative.
Check the latest doc for require_relative or another version of the Core API.
Just require filename.
Yes, it will import it twice if you specify it as filename and ./filename, so don't do that. You're not specifying the .rb, so don't specify the path. I usually put the bulk of my application logic into a file in lib, and then have a script in bin that looks something like this:
#!/usr/bin/env ruby
$: << File.join(File.dirname(__FILE__), "/../lib")
require 'app.rb'
App.new.run(ARGV)
Another advantage is that I find it easier to do unit testing if the loading the application logic doesn't automatically start executing it.
The above will work even when you're running the script from some other directory.
However, inside the same directory the shorter forms you refer to work as expected and at least for ruby 1.9 won't result in a double-require.
testa.rb
puts "start test A"
require 'testb'
require './testb'
puts "finish test A"
testb.rb
puts "start test B"
puts "finish test B"
running 'ruby testa.rb' will result in:
start test A
start test B
finish test B
finish test A
However, the longer form will work even from another directory (eg. ruby somedir/script.rb)
Put this in a standard library directory (somewhere that's already in your default loadpath $:):
# push-loadpath.rb
if caller.first
$: << File.expand_path(File.dirname(caller.first))
end
Then, this should work
% ls /path/to/
bin.rb lib1.rb lib2.rb #...
% cat /path/to/bin.rb
load 'push-loadpath.rb'
require 'lib1'
require 'lib2'
#...
caller gives you access to the current callstack, and tells you what file and where, so push-loadpath.rb uses that to add the file that load'd it to the loadpath.
Note that you should load the file, rather than require it, so the body can be invoked multiple times (once for each time you want to alter the loadpath).
Alternately, you could wrap the body in a method,
# push-loadpath.rb
def push_loadpath
$: << File.expand_path(File.dirname(caller.first))
end
This would allow you to require it, and use it this way:
% ls /path/to/
bin.rb lib1.rb lib2.rb #...
% cat /path/to/bin.rb
require 'push-loadpath'
push_loadpath
require 'lib1'
require 'lib2'
#...