shell script question

[iranzo]

hello,
I work on HPUX
I want to replace on all
scripts in the machine
the expression :
" $EXPHOME/dir/ "
with this expression :
" $EXPDIR "
Thanks.

[Ross Zubritski]

Can you be a bit more specific as to what you are trying tom accomplish? Is your question how to replace the string $EXPHOME/dir/ in multiple scripts/files?

Thanks

RZ

[Pete Randall]

Something along the lines of:

for file in `cat listofscripts`
do
sed 's/$EXPHOME\/dir\//$EXPDIR/' $file > $file.out
mv $file.out $file
done


might do it (though I expect that's pretty crude - I didn't test anything).

Pete

[F. X. de Montgolfier · Accepted solution]

Hi,

you'll have to change INI_DIR to "/", of course...

the original files with extension $SHELL_EXT will be backed up to .bak before the change is made.

#cat test.sh
#!/usr/bin/ksh
INI_DIR=.
SHELL_EXT=".sh"
echo $0
for i in `find $INI_DIR -name "*$SHELL_EXT"|grep -v $0`
do
sed -e 's|$EXPHOME/dir/|$EXPDIR|g' $i >/tmp/tmp_file
j=`basename $i $SHELL_EXT`.bak
mv $i $j
mv /tmp/tmp_file $i
done

#./test.sh
./test.sh
find: cannot open ./.AgentSockets
find: cannot open ./DBI-1.18

#diff test2.bak test2.sh
1c1
< $EXPHOME/dir/
---
> $EXPDIR



Cheers,

FiX

[Robin Wakefield]

Hi,

Use perl to do an in-line replace:

perl -i -pe '{s+\$EXPHOME/dir/+\$EXPDIR+g}' filename

Put a loop around this to do it on all files.

rgds, Robin

[Steven E. Protter]

Here is a cool little perl script that will let you actually generate a file list by searching file contents for an expression and then go out and make the changes.

It actually uses sed, just like the guys in the posts above. It's just got a little front end.

You'll need to update the perl location, and its pretty well documented. Buzz back if you have any questions.

SEP

[Wilfred Chau_1]

# cd /
# ls -1 |while read i
do
cp $i $i.bak
sed -e "s/\$EXPHOME\/dir\//\$EXPDIR/" $i.bak > $i
done