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Re: Calling Bash with /S
From: |
Roman Rakus |
Subject: |
Re: Calling Bash with /S |
Date: |
Thu, 23 Jun 2011 11:57:27 +0200 |
User-agent: |
Mozilla/5.0 (X11; U; Linux x86_64; en-US; rv:1.9.2.15) Gecko/20110307 Fedora/3.1.9-0.39.b3pre.fc14 Lightning/1.0b3pre Thunderbird/3.1.9 |
On 06/22/2011 06:59 PM, sweinberger wrote:
Hi All,
I am calling bash from dmake. The make file initiates the call to bash as
follows:
/bin/bash /S /c ""
/bin/bash -S -c ?
Anyway, -S is not valid option in Fedora's bash version 4.1.7.
The command to execute is between the double quotes. For instance
In Make File:
@ls -l ./SomeDir/*
Bash Command
/bin/bash /S /c "ls -l ./SomeDir/*"
The /S from what I see enables regular expression processing in the bash
command line, but maybe I might be wrong. In any event, the problem that I
have is the asterisk. Bash does not like it. I tried escaping it with a
backslash up to nine levels but that did not work; should only require one.
Pathname expansion is on by default. Can be turned off by `set -f' or
invoking bash with `-f' option.
Yes, the actual command that I use is not ls. In the real example, I am
using "rm -f", but ls is a good example too and safer for experimentation.
How do I write the command line, so that it works? "ls -l ./SomeDir/\*"
does not work, nor anything else that I tried so far.
Explain what means `does not work'. What it does?
Thanks in advance,
Sarah
RR