Re: determine NS release/level using regexp

* Esbrook, Scott meinte:

> RE nsRegExp = new RE("(netscape*)/");

this matches:
'n' followed by 'e' fb 't' fb 's' ... fb 'p' fb 'e' fb 'null or more "e"'
fb '/'
 
> boolean matched = nsRegExp.match("mozilla/5.0 (windows; u; windows nt 5.0;
> en-us; rv:0.9.2) gecko/20010726 netscape6/6.1");

return false, cause 'e' is followed by '6' not 'null or more "e"'

.sw
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