Increment Variable for Ruby

Question

I'm new to ruby. Is there a way to shorten this code("increment var#")? I've created 26 entry for an alphabet.

require 'tk'
dis = { 'padx' =>5, 'pady' =>5}
root = TkRoot.new {title "Alphabet"}
elpar = {'height' => 25, 'width' => 25}
f1 = TkFrame.new{
    relief 'sunken'
    borderwidth 1
    background "black"
    height 800
    width 800
    dis
    pack
    }
f1.place('x' => 0, 'y' => 0)

el00 = TkEntry.new(f1)
el01 = TkEntry.new(f1)
el02 = TkEntry.new(f1)
          to             # Can I use loop here to shorten it?
el26 = TkEntry.new(f1)   


var1 = TkVariable.new
           to            #(1)
var2 = TkVariable.new


el00.textvariable = var1
var1.value = "A"
el00.place(elpar)
el00.place('x' => 0, 'y' => 0)
          to             #
el26.textvariable = var26
var26.value = "Z"
el26.place(elpar)
el26.place('x' => 728, 'y' => 0)

Tk.mainloop

(1) I've tried using loop here

  x = 1
  while x < 27
    "var#{x}"=TkVariable.new         # var1 is still undefined local variable if I use a loop.  
    "var" + x.to_s=TkVariable.new    # another variation that i've tried 
     x += 1
  end

Is there a way to shorten it? My codes are just an increment of 1?

Thanks, Jim

Possible duplicate of: stackoverflow.com/questions/16419767/ruby-dynamic-variable-name — Yevgeniy Anfilofyev
– Yevgeniy Anfilofyev, Commented May 23, 2015 at 7:56

spickermann · Accepted Answer · 2015-05-23 08:03:41Z

1

You could just store the entries in an array:

entries = Array.new(26) { TkEntry.new(f1) }

Then read your entries:

entries[0] #=> first entry
entries[1] #=> second entry

Use the character range from A to Z to set the arguments:

('A'..'Z').each_with_index do |character, index|
  variable = TkVariable.new
  variable.value = character
  entries[index].textvariable = variable 
  entries[index].place(elpar)
  entries[index].place('x' => 0, 'y' => 0)   # you did not explain how to calculate this
end

edited May 23, 2015 at 8:03

answered May 23, 2015 at 7:56

spickermann

108k9 gold badges115 silver badges147 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Jim Over a year ago

Hi spickermann, Thank you so much for your immediate response.. :D the codes you gave me, save me a lot of lines. The entries[index].place('x' => 0, 'y' => 0) it is an increment of 27 in its horizontal plane. By the way, when I tried changing ('A'..'Z') to a range of integer it didn't work? I'll explore more about this "each" iterator. Thanks again, Jim

Collectives™ on Stack Overflow

Increment Variable for Ruby

1 Answer 1

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related