From: Christopher Dicely <cmdicely@...>
Date: 2009-03-15T02:34:48+09:00
Subject: Re: Problem With eval and each_index

On Fri, Mar 13, 2009 at 7:32 PM, matt neuburg <matt@tidbits.com> wrote:
> Michael W. Ryder <_mwryder@worldnet.att.net> wrote:
>
>> I am trying to set a group of variables to values stored in a tab
>> delimited string.  I have no problem splitting the string into an array,
>> call it e, or in putting the variable names in a second array, call it
>> v.  The problem arises when I try to merge the two together.  If I enter:
>> irb(main):073:0> p v
>> ["name", "street", "city", "state", "zip", "*", "telephone"]
>> => nil
>> irb(main):074:0> p e
>> ["John Doe", "123 Main St", "Anywhere", "US", "01234-5678", "ab123",
>> "1234567890"]
>> => nil
>> irb(main):075:0> eval "#{v[0]} = e[0]"
>> => "John Doe"
>> irb(main):076:0> p name
>> "John Doe"
>> => nil
>>
>> it works as I want.  But when I try:
>> irb(main):077:0> v.each_index {|i| eval "#{v[i]} = e[0]"}
>> => ["name", "street", "city", "state", "zip", "*", "telephone"]
>>
>> and then enter: puts zip
>> it returns with a NameError saying that the variable was undefined.
>
> Inside a block, local variables are local to the block. So, you're
> creating all those local variables and then the block ends and they are
> all thrown away.
>
> Why do you need variables called "name" (etc.) anyway? Why not use a
> hash, so that h["name"] contains the name ("John Doe"), h["street"]
> contains the street, and so on? So:
>
> h = Hash.new
> v.zip(e).each {|k,v| h[k]=v}

or, better:

h = Hash[*v.zip(e).flatten]