Reputation: 1234
I want to replace (whole string)
$(TOPDIR)/$(OSSCHEMASDIRNAME)
with
/udir/makesh/$(OSSCHEMASDIRNAME)
in a makefile
I tried with
perl -pi.bak -e "s/\$\(TOPDIR\)\/\$\(OSSCHEMASDIRNAME\)/\/udir\/makesh\/\$\(OSSCHEMASDIRNAME\)/g " makefile
but i am getting unmatched parentheses error
Upvotes: 1
Views: 368
Reputation: 67900
When in doubt about escaping, you can simply use quotemeta
or \Q ... \E
.
perl -pe 's#\Q$(TOPDIR)\E(?=/\Q$(OSSCHEMASDIRNAME)\E)#/udir/makesh#;'
Note the use of a look-ahead assertion to save us the trouble of repeating the trailing part in the substitution.
A quotemeta
solution would be something like:
perl -pe 'BEGIN { $dir = quotemeta(q#$(TOPDIR)/$(OSSCHEMASDIRNAME)#); }
s#$dir#/udir/makesh/$(OSSCHEMASDIRNAME)#;'
Of course, you don't need to use an actual one-liner. When the shell quoting is causing troubles, the simplest option of them all is to write a small source file for your script:
s#\Q$(TOPDIR)\E(?=/\Q$(OSSCHEMASDIRNAME)\E)#/udir/makesh#;
And run with:
perl -p source.pl inputfile
Upvotes: 1
Reputation: 17131
First off, you don't need to use /
for regular expressions. They're just canonical. You can use pretty much anything. Thus your code can become (simplify away some \
):
perl -pi.bak -e "s|\$\(TOPDIR\)/\$\(OSSCHEMASDIRNAME\)|/udir/makesh/\$\(OSSCHEMASDIRNAME\)|g " makefile
Now to actually address your issue, because you're using "
instead of '
, the shell attempts to figure out what $\
means which is then replaced with (presumably) nothing. So what you really want is:
perl -p -i.bak -e 's|\$\(TOPDIR\)/\$\(OSSCHEMASDIRNAME\)|/udir/makesh/\$\(OSSCHEMASDIRNAME\)|g' makefile
Upvotes: 2
Reputation: 103
You have to "double" escape the dollar sign. Like this:
echo "\$(TOPDIR)/\$(OSSCHEMASDIRNAME)" | perl -p -e "s/\\$\(TOPDIR\)\/\\$\(OSSCHEMASDIRNAME\)/\/udir\/makesh\/\\$\(OSSCHEMASDIRNAME\)/g"
Upvotes: 2