Start of string?

[Jari Williamsson]

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

 

[David A. Black]

Hi --

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

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* Intro to Rails, London, UK, December 3-6 (by Skills Matter)
See http://www.rubypal.com for details!

 

[Stefano Crocco]

Alle marted=EC 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 =3D "long string"
key =3D "long"
if a[0, key.length] =3D=3D key then

...while this might look ok, it doesn't look so good with constant string= s:
if a[0, 4] =3D=3D "long" then
or
if a[0, "long".length] =3D=3D "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
=2E..

The ^ at the beginning of the regexp means that the regexp should be matche=
d=20
at the beginning of the string (actually, of the line. If your string is=20
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 str=
ing=20
interpolation inside the regexp:

str =3D "long"
if a.match /^#{str}/ then

Of course, this can cause problems if str contains characters which are=20
special in a regexp. In this case, you need to quote it:

str =3D "long"
if a.match /^#{Regexp.quote(str)}/ then


I hope this helps

Stefano

 

[Golda Bence]

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

 

[Peter Szinek]

Jari,

Is there any such solution?

Here is a regular expression one:

"longstring" =~ /^long/

HTH,
Peter
___
http://www.rubyrailways.com
http://scrubyt.org

 

[Nobuyoshi Nakada]

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.