Date: Sun, 5 Apr 2015 00:21:49 +0200 From: Polytropon <freebsd@edvax.de> To: Jon Radel <jon@radel.com> Cc: Nancy Belle <belle@antennex.com>, freebsd-questions <freebsd-questions@freebsd.org> Subject: Re: Chop and replace method?? Message-ID: <20150405002149.e11077b2.freebsd@edvax.de> In-Reply-To: <55202099.9090908@radel.com> References: <DM__150404113753_01984604235@mail.antennex.com> <55202099.9090908@radel.com>
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Allow me a little improvement for readability: On Sat, 04 Apr 2015 13:34:17 -0400, Jon Radel wrote: > sed -i .bak 's/href=\"..\/..\/..\/archival\/archive13\//href=\"/' *.html You're using \/ to distinguish the '/' characters from the / in the regex. But you don't have to - just replace the / of the sed command with something that is _not_ part of the search & replace expression, for example |, such as: % sed -i .bak 's|href="../../../archival/archive13/|href="|' *.html As you can see, escaping " inside '...' also isn't neccessary. Depending on the patterns in the input, you could add 'g' for global search & replace (here: like s|before|after|g). -- Polytropon Magdeburg, Germany Happy FreeBSD user since 4.0 Andra moi ennepe, Mousa, ...
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