I'm trying to do search/replace on multiple text files.
I'm using the following script:

#!/bin/sh
USAGE=`basename $0`
USAGE=$USAGE' <PATTERN> <SUBSTITUTE> <FILES PATH>'
if test $# -lt 3
then
echo "-I-: $USAGE"
exit 1
fi
PATTERN=$1
shift
SUBSTITUTION=$1
shift
for i;
do
mv $i $i.old;
sed -e 's/$PATTERN/$SUBSTITUTION/g' $i.old > $i;
done
\rm *.old

What happens is that if I echo the commands out to the shell and
execute them, the replacement will occur as expect. If I run to script
to execute the sed command, the resulting output will not have any
change.

Can someone please me understand this?

Thanks in advance,
Pedro Alves

On 24 Aug 2006 07:11:52 -0700, pedromalves@gmail.com
<pedromalves@gmail.com> wrote:
> I'm trying to do search/replace on multiple text files.
> I'm using the following script:
>
> #!/bin/sh
> USAGE=`basename $0`
> USAGE=$USAGE' <PATTERN> <SUBSTITUTE> <FILES PATH>'
> if test $# -lt 3
> then
> echo "-I-: $USAGE"
> exit 1
> fi
> PATTERN=$1
PATTERN="$1"
> shift
> SUBSTITUTION=$1
SUBSTITUTION="$1"
> shift
> for i;
> do
> mv $i $i.old;
mv "$i" "$i.old"
> sed -e 's/$PATTERN/$SUBSTITUTION/g' $i.old > $i;
> done
> \rm *.old
>
> What happens is that if I echo the commands out to the shell and
> execute them, the replacement will occur as expect. If I run to script
> to execute the sed command, the resulting output will not have any
> change.
>
> Can someone please me understand this?
>
Variables are not substituted inside single quotes!
sed -e "s/$PATTERN/$SUBSTITUTION/g" $i.old > $i;

pedromalves@gmail.com wrote:
> I'm trying to do search/replace on multiple text files.
> I'm using the following script:
>
> #!/bin/sh
> USAGE=`basename $0`
> USAGE=$USAGE' <PATTERN> <SUBSTITUTE> <FILES PATH>'
> if test $# -lt 3
> then
> echo "-I-: $USAGE"
> exit 1
> fi
> PATTERN=$1

If you are going to pick arbitrary PATTERN/SUBSTITION, better escape
the delimiter which will be used in the later sed command, i.e. the
slash in the following example:

PATTERN=$(printf "%s" $1 | sed 's#/#\\/#g')

similar to the other variable.

> shift
> SUBSTITUTION=$1
> shift
> for i;
> do

This 'for' loop looks problematic, i.e. where did you increment the
value of i? how about change this structure to be like:

while [ $# -ne 0 ]
do
#do stuff on $1
shift
done

> mv $i $i.old;
> sed -e 's/$PATTERN/$SUBSTITUTION/g' $i.old > $i;

change your sed line to be enclosed by double-quotes:

sed -e "s/$PATTERN/$SUBSTITUTION/g" $i.old > $i;

Xicheng

On 24 Aug 2006 08:16:38 -0700, Xicheng Jia
<xicheng@gmail.com> wrote:
>> for i;
>> do
>
> This 'for' loop looks problematic, i.e. where did you increment the
> value of i?
>
'for i' is equivalent to 'for i in "$@"'


--
"Nominal fee". What an ugly sentence. It's one of those things that
implies that if you have to ask, you can't afford it.
-- Linus Torvalds

On 2006-08-24, Bill Marcum wrote:
> On 24 Aug 2006 07:11:52 -0700, pedromalves@gmail.com
> <pedromalves@gmail.com> wrote:
>> I'm trying to do search/replace on multiple text files.
>> I'm using the following script:
>>
>> #!/bin/sh
>> USAGE=`basename $0`
>> USAGE=$USAGE' <PATTERN> <SUBSTITUTE> <FILES PATH>'
>> if test $# -lt 3
>> then
>> echo "-I-: $USAGE"
>> exit 1
>> fi
>> PATTERN=$1
> PATTERN="$1"

That doesn't make any difference.

>> shift
>> SUBSTITUTION=$1
> SUBSTITUTION="$1"

Ditto.

>> shift
>> for i;
>> do
>> mv $i $i.old;
> mv "$i" "$i.old"
>> sed -e 's/$PATTERN/$SUBSTITUTION/g' $i.old > $i;
>> done
>> \rm *.old
>>
>> What happens is that if I echo the commands out to the shell and
>> execute them, the replacement will occur as expect. If I run to script
>> to execute the sed command, the resulting output will not have any
>> change.
>>
>> Can someone please me understand this?
>>
> Variables are not substituted inside single quotes!
> sed -e "s/$PATTERN/$SUBSTITUTION/g" $i.old > $i;

That does.

--
Chris F.A. Johnson, author <http://cfaj.freeshell.org>
Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
===== My code in this post, if any, assumes the POSIX locale
===== and is released under the GNU General Public Licence