grep for converting decimal to ascii characters?

[Ken G. Brown]

I'm trying to figure out a way of converting a list of decimal representations of ascii characters, to their ascii character so I do not need to do it manually.

(\d+) finds the decimal numbers and \1 in the replace field will put them back in, but how do I get them put back in as ascii characters?

I'd like this:

35
123
86
58

to end up something like this:

35 #
123 {
86 V
58 :

or even better have the multiple lines end up like this on one line but I think I can eventually figure that out if I really want.

#{V:

Thx for any tips.

Ken G. Brown

[Jan Leppink]

Hi Ken,
I think you must realise that ascii encoding is beyond the scope of grep. Especially for codes
above 128 there are a lot of encodings that grep knows nothing about (and shouldn't).

But I think there is a workaround by creating pattern and replacement arrays to get what you
want. You can use Perl to create these arrays, But maybe it is easier to write the whole
replacement in Perl and call that script from BBEdit.

Hope this helps you,
Jan Leppink.

Post by Ken G. Brown
I'm trying to figure out a way of converting a list of decimal representations of ascii

characters, to their

Post by Ken G. Brown
ascii character so I do not need to do it manually.
(\d+) finds the decimal numbers and \1 in the replace field will put them back in, but how do

I get them

Post by Ken G. Brown
put back in as ascii characters?
35
123
86
58
35 #
123 {
86 V
or even better have the multiple lines end up like this on one line but I think I can eventually

figure that

Post by Ken G. Brown
out if I really want.
Thx for any tips.
Ken G. Brown

[Ken G. Brown]

Got a Perl filter that does what I need from Ronald on the BBedit Scripting list.
Don't need the ones over 128 right now.
Thx,
Ken

Date: Mon, 17 Mar 2008 10:48:26 -0400
Subject: Re: regex for converting decimal to ascii?
Content-Type: text/plain; charset=us-ascii

Post by Ken G. Brown
I'm trying to figure out a way of converting a list of decimal representations of ascii characters, to their ascii character so I do not need to do it manually.
(\d+) finds the decimal numbers and \1 in the replace field will put them back in, but how do i get them put back in as ascii characters?
Thx for any tips.

Here's a Perl solution, which you can run as a filter on the selected text.
#!perl -p
s/(\d+)/chr($1)/ge;

Ronald
Hi Ken,
I think you must realise that ascii encoding is beyond the scope of grep. Especially for codes
above 128 there are a lot of encodings that grep knows nothing about (and shouldn't).
But I think there is a workaround by creating pattern and replacement arrays to get what you
want. You can use Perl to create these arrays, But maybe it is easier to write the whole
replacement in Perl and call that script from BBEdit.
Hope this helps you,
Jan Leppink.

I'm trying to figure out a way of converting a list of decimal representations of ascii

characters, to their

ascii character so I do not need to do it manually.
(\d+) finds the decimal numbers and \1 in the replace field will put them back in, but how do

I get them

put back in as ascii characters?
35
123
86
58
35 #
123 {
86 V
or even better have the multiple lines end up like this on one line but I think I can eventually

figure that

out if I really want.
Thx for any tips.
Ken G. Brown

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