Start of string?

Ruby
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 W.

2

Hi –

On Tue, 6 Nov 2007, Jari W. wrote:

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

3

Alle martedì 6 novembre 2007, Jari W. ha scritto:

or
Jari W.
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:

class Regexp - RDoc Documentation

…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

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

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.

7

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

or
Jari W.
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