On Sat, Apr 17, 2010 at 12:02:40PM -0400, Daniel D Jones wrote:
On Saturday 17 April 2010 00:09:28 Michael Elkins wrote:
On Fri, Apr 16, 2010 at 08:15:38PM -0400, Daniel D Jones wrote:
What I'm trying to do is pretty simple. Getting it to work is turning out
not to be. What I want to do
On Saturday 17 April 2010 00:09:28 Michael Elkins wrote:
On Fri, Apr 16, 2010 at 08:15:38PM -0400, Daniel D Jones wrote:
What I'm trying to do is pretty simple. Getting it to work is turning out
not to be. What I want to do is call a bash script with a couple of
arguments, and, within the
On Sat, Apr 17, 2010 at 12:02:40PM -0400, Daniel D Jones wrote:
That was the first thing I tried and sed gave me an error:
sed: -e expression #1, char 18: unknown option to `s'
I just went back and tried it again and it worked, so I have no idea what I
did the first time that made it not work.
Daniel D Jones ddjo...@riddlemaster.org wrote:
What I want to do is call a bash script with a couple of arguments, and,
within the script, call sed to use those args to replace two
placeholders in a file:
bashscript SUB1 SUB2
This line inside bashscript doesn't work:
sed -e
* 2010-04-17 09:34 (-0700), Michael Elkins wrote:
You can run into that sort of problem if your pattern to replace
contains any forward slashes (/) in it. If you need to such an
expansion, you probably want to do it in two passes, first doing a / to
\/ substitution on your replacement
What I'm trying to do is pretty simple. Getting it to work is turning out not
to be. What I want to do is call a bash script with a couple of arguments,
and, within the script, call sed to use those args to replace two placeholders
in a file:
bashscript SUB1 SUB2
This line inside bashscript
On Fri, Apr 16, 2010 at 08:15:38PM -0400, Daniel D Jones wrote:
What I'm trying to do is pretty simple. Getting it to work is turning out not
to be. What I want to do is call a bash script with a couple of arguments,
and, within the script, call sed to use those args to replace two
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