Re: [code] Dealing with % before a snippet-placeholder

From: docmorchel . <docmorchel.att.gmail.com>
Date: Thu, 12 Nov 2015 19:12:35 +0100

Resolved with 8.4 :)

re1 = '%%%1(digits)', works

Regards,
Martin

> From: docmorchel . <docmorchel.att.gmail.com>
> Date: Wed, 11 Nov 2015 23:22:01 +0100
> Hi all,
> How would I get a percent sign in front of the substituted placeholder?
> The snippet result (actually part of it) should consist of a percent sign
> followed by some digits, eg. '%12'. My two attempts, which failed were:
> snippets.pm = {
> re1 = '%%%1(digits)',
> re2 = '%%1(digits)',
> }

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Received on Thu 12 Nov 2015 - 13:12:35 EST

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