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Re: Replacing sed for zsh portability
- X-seq: zsh-users 9317
- From: Peter Stephenson <pws@xxxxxxx>
- To: Zsh Users <zsh-users@xxxxxxxxxx>
- Subject: Re: Replacing sed for zsh portability
- Date: Mon, 15 Aug 2005 12:22:44 +0100
- In-reply-to: <20050815103557.GA1411@DervishD>
- Mailing-list: contact zsh-users-help@xxxxxxxxxx; run by ezmlm
- References: <20050815103557.GA1411@DervishD>
DervishD wrote:
> #<=
> # This is an example of documentation
> # Here are more lines
> #
> # Yet even more lines here
> ...
> #=>
>
> Since I cannot use sed for the entire work unless I complicate
> the script (doing multiple substitutions and the like), I want to get
> rid of the current script and doing all the job in ZSH. It doesn't
> matter if it is very slow, speed here is not an issue.
>
> Can this be done in zsh easily or the only way is to use a while
> loop to ignore lines before the delimiter and another to process the
> documentation itself?
A while loop is the obvious way. You can read in the entire file
(file="$(<file)") and substitute on that. My example succeeded with this:
print -r ${(S)file//$'\n#<='*$'\n#=>'}
although the following is better since it takes account of the fact
that a comment may be at the start:
print -r ${(S)file//($'\n'|(#s))\#\<\=*$'\n#=>'}
This requires extended_glob.
> Another question, related to this. It seems that the X|Y glob
> operator doesn't work left-to-right, but shortest-first. If I do
> this:
>
> print ${line#\#(<=|=>| |)}
>
> then the 'nothing' at the end is used as the match, and the
> delimiters are not matched.
You're missing the fact that your substitution is of the form ${line#...}.
This is explicitly documented to remove the shortest matching chunk at
the head of the line. Try adding another "#":
print ${line##\#(<=|=>| |)}
--
Peter Stephenson <pws@xxxxxxx> Software Engineer
CSR PLC, Churchill House, Cambridge Business Park, Cowley Road
Cambridge, CB4 0WZ, UK Tel: +44 (0)1223 692070
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