Unix shell and the sed command...

Dear all,

I have a little shell script to perform a sed on a list of files.
I'm using bash on Linux.  The list is correct.
The only problem is that the sed seems not to allow variable arguments.
If instead of the line
  cat $i | sed 's/$pattern/$replacement/g' > $j
I had the line
  cat $i | sed 's/mypattern/myreplacement/g' > $j
it would work fine.

So the question is: How to pass variables to a sed command?


Regards
Stephane

#!/bin/sh

# The parameters of the script
pattern=$1
replacement=$2

# A temporary file
tmpfile=/tmp/dump.txt

# Store the search result in the file
# The find prints only the regular files (not the directories)
# The grep prints only the first file occurence without the found line
grep --files-with-matches $pattern `find . -name "*" -type f` > $tmpfile

# The following lines redefine the environment variable
# that contains the characters for word separators
SIFS=$IFS
IFS='
'

echo "Replace $pattern by $replacement"          
echo

for line in `cat $tmpfile`
do
  # Get the first field of the list
  i=`echo $line`
  # Replace the string using a temporary file
  j=`echo $i'.new'`
#  cat $i | sed 's/$pattern/$replacement/g' > $j
  cat $i | sed 's/mypattern/myreplacement/g' > $j
  echo "File is : $i"          
  mv $j "$i"
  rm -fr $j
done

# Reset the environment variable
IFS=$SIFS

exit $?