$char

By opticlens on Dec 23, 2007

See comments below

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sunburst   -  Dec 25, 2007

Agh, and replace /(d+)/g with /([backslash]d+)/g :X

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sunburst   -  Dec 25, 2007

Just replace $asc(1) and $chr(1) with $chr(backslash1) and $asc(backslash1)

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sunburst   -  Dec 25, 2007

I think using $regsubex would be better in this case. Here is an updated version of the code:

Same syntax and all, just uses $regsubex.

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Rebellious   -  Dec 23, 2007

Very nice job on this.

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opticlens   -  Dec 23, 2007

comments, rants, criticism?

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