Untested code. $str = mississippi; $str =~ m/m(.*i)/; print $1; # Should output mississippi
$str = mississippi; $str =~ m/m(.*i)(.*pi)/; This requires $2 to have .*pi in it. Since this is a greedy RegEx, $1 will grab mississi (must end by an i) and leave as little as possible for $2 - ie ppi. Um. Hope this post helped! On 5/31/07, Sharan Basappa <[EMAIL PROTECTED]> wrote:
Thanks a lot Paul .. For this rule : $str = mississippi; $str =~ m/m(.*i)(.*pi)/; My initial understanding was that .*i would match all the way till last char i. This would indeed be true if .*i was not followed by .*pi. Do you agree ?
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