I'll show an example of what I'm trying to do.
child.rb
require 'parent'
class Child < Parent
...
end
parent.rb
class Parent
puts __FILE__
...
end
running this (e.g. in IRB):
require 'child'
child=Child.new
returns this:
/path/to/parent.rb
but i'm really trying to get the file child.rb instead.
How can i do this without moving the puts __FILE__ in to the child class directly?
There is no way in Ruby to get what you are asking for. Perhaps if you explained why you want this—what you are trying to accomplish—then I (or someone else) could help you further. As it stands you seem to be asking an XY problem.
Although it's not always the "file that required me", you can use $0 in this case instead of __FILE__ to get "the root file that is running" in the case when you have done ruby parent.rb. Note that this will not work with IRB, however. (Inside IRB $0 is always "irb" or something equally unhelpful.)
Redefine require to output caller_locations.first.absolute_path before calling the original require.
you can use some backtrace method like caller_locations(1,1)[0].path or caller
Related
Coming to Ruby from a PHP background, I'm used to being able to use require, require_once, include, or include_once which all have a similar effect, but the key being they continue to process code in the same scope where the include / require command was invoked.
Example:
sub.php
<?php
echo $foo;
main.php
<?php
$foo = 1234;
include('sub.php'); // outputs '1234'
When I first started using Ruby I tried to include / require / require_relative / load other .rb files, and after becoming a little frustrated with not having it work how I would expect it to I decided that there were better ways to go about breaking up large files and that Ruby didn't need to behave in the same way PHP did.
However, occasionally I feel that for testing purposes it would be nice to to load code from another .rb file in the way PHP does - in the same scope with access to all the same variables - without having to use class / instance variables or constants. Is this possible? Maybe somehow using a proc / binding / or eval command?
Again, I'm not advocating that this should be used during development - but I am curious if it is possible - and if so, how?
Yes, this is possible, although certainly not something I'd recommend doing. This works:
includer.rb:
puts var
include.rb:
var = "Hello!"
eval(File.read("include.rb"), binding)
Running this (Ruby 2.2.1, Ruby 1.9.3) will print Hello!. It works simply: eval takes an optional binding with which to evaluate the code it is passed, and Kernel#binding returns the current binding.
To have code run in same binding, you could simply eval the file contents as follows:
example.rb
class Example
def self.called_by_include
"value for bar"
end
def foo
puts "Called foo"
end
eval( File.read( 'included.rb' ) )
end
Example.new.bar
included.rb
BAR_CONSTANT = called_by_include
def bar
puts BAR_CONSTANT
end
Running ruby example.rb produces output
value for bar
The important thing is the eval( File.read( 'included.rb' ) ) code, which if you really wanted you could define as a class method on Object, to allow arbitrary source to be included with a convenience function*. The use of constants, class variables etc just shows influences working in both directions between the two pieces of source code.
It would be bad practice to use this in any production code. Ruby gives you much better tools for meta-programming, such as ability to use mix-ins, re-open classes, define methods from blocks etc.
* Something like this
class Object
def self.include_source filename
eval( File.read( filename ) )
end
end
And the line in example.rb would become just
include_source 'included.rb'
Again I have to repeat this is not such a great idea . . .
To import external .rb file in your code, I'm not sure but I think it have to be a gem.
Use require followed by the name of the gem you want to import.
Example
require 'foobar'
# do some stuff
Or you can use load to import entire rb file
load 'foobar.rb'
# do some stuff
Good luck and sorry for my english
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
I want to write a Ruby script something like this:
class Foo
# instance methods here
def self.run
foo = Foo.new
# do stuff here
end
end
# This code should only be executed when run as a script, but not when required into another file
unless required_in? # <-- not a real Kernel method
Foo.run
end
# ------------------------------------------------------------------------------------------
I want to be able to unit test it, which is why I don't want the code outside of the class to run unless I execute the script directly, i.e. ruby foo_it_up.rb.
I know I can simply put the Foo class in another file and require 'foo' in my script. In fact, that is probably a better way to do it, just in case Foo's functionality is needed somewhere else. So my question is more academic than anything, but I'd still be interested in knowing how to do this in Ruby.
This is usually done with
if __FILE__ == $0
Foo.run
end
but I prefer
if File.identical?(__FILE__, $0)
Foo.run
end
because programs like ruby-prof can make $0 not equal __FILE__ even when you use --replace-progname.
$0 refers to the name of the program ($PROGRAM_NAME), while __FILE__ is the name of the current source file.
I have taken this example exactly from the Ruby Cookbook. Unfortunately for me, like a whole lot of the examples in that book, this one does not work:
my file (Find.rb - saved both locally and to Ruby\bin):
require 'find'
module Find
def match(*paths)
matched=[]
find(*paths) { |path| matched << path if yield path }
return matched
end
module_function :match
end
I try to call it this way from IRB, according to the example the book provides:
irb(main):002:0> require 'Find'
=> false
irb(main):003:0> Find.match("./") { |p| ext = p[-4...p.size]; ext && ext.downcase == "mp3" }
It SHOULD return a list of mp3 files in my recursive directory. Instead, it does this:
NoMethodError: undefined method `match' for Find:Module
from (irb):3
from C:/Ruby192/bin/irb:12:in `<main>'
What gives? I'm new at this (although I MUST say that I'm farther along with Python, and much better at it!).
How can I get IRB to use my method?
I ran into this with irb on a Mac running Snow Leopard while using the default version of ruby (and irb of course) installed with OS X. I was able to get past it by including the module in IRB after loading the module or in the file after the module definition.
include module_name
I'm not sure if this is a defect or known behavior.
The only explanation is that the code you posted is not the code you are running, since both carefully reading it and simply cut&paste&running it shows absolutely no problems whatsoever.
What directory are you calling IRB from? Try calling it from the directory where your find.rb file is located. Also, I don't know if it makes any difference but convention is to name the file the lowercase version of the module / class. So the module would be Find and the file name would be find.rb. You shouldn't need the require call in the file itself.
So, start your command prompt window, cd into the directory that contains find.rb and run irb. In IRB you should be able to require "find" and it should return true. From there you should be able to call Find.match.
I know this question is already 3 years old, but since this is the first hit on google for the problem, and I had been banging my head against the wall all afternoon with the same problem doing the tutorial here: http://ruby.learncodethehardway.org/book/ex25.html, here goes: the function definition in the module should read
module Find
def Find.match(*paths)
...
end
end
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