Start of string?

ruby-talk
1

What't the most elegant way to find a substring match at the very start of a string?

Currently I'm using this approach:
a = "long string"
key = "long"
if a[0, key.length] == key then

...while this might look ok, it doesn't look so good with constant strings:
if a[0, 4] == "long" then
or
if a[0, "long".length] == "long" then

I guess what I'm looking for is something like:
if a.startswith("long") then

Is there any such solution?

Best regards,

Jari Williamsson

2

Hi --

···

On Tue, 6 Nov 2007, Jari Williamsson wrote:

What't the most elegant way to find a substring match at the very start of a string?

Currently I'm using this approach:
a = "long string"
key = "long"
if a[0, key.length] == key then

...while this might look ok, it doesn't look so good with constant strings:
if a[0, 4] == "long" then
or
if a[0, "long".length] == "long" then

I guess what I'm looking for is something like:
if a.startswith("long") then

Is there any such solution?

You could do:

   if a.index("long") == 0

David

--
Upcoming training by David A. Black/Ruby Power and Light, LLC:
   * Advancing With Rails, Edison, NJ, November 6-9
   * Advancing With Rails, Berlin, Germany, November 19-22
   * Intro to Rails, London, UK, December 3-6 (by Skills Matter)
See http://www.rubypal.com for details!

3

Alle martedì 6 novembre 2007, Jari Williamsson ha scritto:

What't the most elegant way to find a substring match at the very start
of a string?

Currently I'm using this approach:
a = "long string"
key = "long"
if a[0, key.length] == key then

...while this might look ok, it doesn't look so good with constant strings:
if a[0, 4] == "long" then
or
if a[0, "long".length] == "long" then

I guess what I'm looking for is something like:
if a.startswith("long") then

Is there any such solution?

Best regards,

Jari Williamsson

You can use a regexp:

if a.match /^long/ then
...

The ^ at the beginning of the regexp means that the regexp should be matched
at the beginning of the string (actually, of the line. If your string is
multiline, you need to use \A to match the beginning of the string).

If the string you want to look for is stored in a variable, you can use string
interpolation inside the regexp:

str = "long"
if a.match /^#{str}/ then

Of course, this can cause problems if str contains characters which are
special in a regexp. In this case, you need to quote it:

str = "long"
if a.match /^#{Regexp.quote(str)}/ then

I hope this helps

Stefano

4

Hi!

Take a look at Regexp class:

  http://www.ruby-doc.org/core/classes/Regexp.html

...after that your starts_with? method should be similar to this one:

class String
  def starts_with?(a_string)
    self =~ Regexp.new('^%s' % [ a_string ])
  end
end

a = "long string"
a.starts_with?("long") # => 0
a.starts_with?("not soo long") # => nil

!note: in ruby 0 (Fixnum) means true in a condition, so does ""
(String), but nil (NilClass) means false.

Cheers, Bence

Jari Williamsson wrote:

···

What't the most elegant way to find a substring match at the very start
of a string?

Currently I'm using this approach:
a = "long string"
key = "long"
if a[0, key.length] == key then

...while this might look ok, it doesn't look so good with constant strings:
if a[0, 4] == "long" then
or
if a[0, "long".length] == "long" then

I guess what I'm looking for is something like:
if a.startswith("long") then

Is there any such solution?

Best regards,

Jari Williamsson

5

Jari,

Is there any such solution?

Here is a regular expression one:

"longstring" =~ /^long/

HTH,
Peter

···

___
http://www.rubyrailways.com
http://scrubyt.org

6

How about;

if /^long/.match a

if a.match /^long/

if a.index('long').zero?

The last one probably has the best portability for your non-constant
keys.

Arlen

···

On Tue, 2007-11-06 at 21:30 +0900, Jari Williamsson wrote:

What't the most elegant way to find a substring match at the very start
of a string?

Currently I'm using this approach:
a = "long string"
key = "long"
if a[0, key.length] == key then

...while this might look ok, it doesn't look so good with constant strings:
if a[0, 4] == "long" then
or
if a[0, "long".length] == "long" then

I guess what I'm looking for is something like:
if a.startswith("long") then

Is there any such solution?

Best regards,

Jari Williamsson

7

Hi,

At Tue, 6 Nov 2007 22:02:27 +0900,
David A. Black wrote in [ruby-talk:277691]:

You could do:

   if a.index("long") == 0

rindex("long", 0) is faster for long but unmatching strings.

···

--
Nobu Nakada