in reply to Re: printing the subscript separator
in thread printing the subscript separator
I want to say: $operator is something more than *just* a space or spaces. So saying as you suggest: "$operator does not match a space". Wouldn't do that.
You are right about the plus sign and I have removed it. Thanks. So now I have:
But I am still seeing the same problem with $;$operator !~ /[^\s]/
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Re^3: printing the subscript separator
by cog (Parson) on Feb 09, 2005 at 19:16 UTC |