perl one liner to replace part of file with other file - perl

Question

I'm having problems with this one liner:
perl -pe 's/FINDME/`cat rep.txt`/ge' in.txt
If i use it exactly like this, it works, but I also want to add some text before and after the replaced content:
perl -pe 's/FINDME/SOMETHING1`cat rep.txt`SOMETHING2/ge' in.txt
I get the error:
syntax error at -e line 1, near "SOMETHING1`cat rep.txt`"
Shouldn't the output of the command be treated like a string?
Adicionally, I'm also confused by the fact I can't replace to something with the character <
perl -pe 's/SOMETHING/<SOMETHINGELSE/ge' in.txt
Unterminated <> operator at -e line 1.
Escaping the < (\<) gives me the same error.

Answer 1

The problem is that you are using the e modifier to the regexp, which means to eval{} (or in other words, execute) the replacement string as a code snippet, but you are treating it like a shell replacement. The e modifier expects CODE, not TEXT.
So, a normal (global) replace would use:
s/FINDME/REPLACE/g
... and this is fine. However, when you use an e flag, the replacement is run as code. Thus:
s/FINDME/`cat foo.txt`/ge;
... is equivalent to ...
$replace = `cat foo.txt`;
s/FINDME/$replace/g;
So, you can see how this:
s/FINDME/SOMETHING`cat foo.txt`/ge;
... is equivalent to...
$replace = SOMETHING`cat foo.txt`;
s/FINDME/$replace/g;
... and this is clearly a syntax error. Try this way instead:
s/FINDME/"SOMETHING".`cat foo.txt`/ge;
and you will find that it works, because this is valid code:
$replace = "SOMETHING".`cat foo.txt`;
( You can of course put even more complex things in there; since what is going on behind the scenes is an eval{}, your code is actually doing this:
eval { "SOMETHING".`cat foo.txt`; }
s/FINDME/$_/g;
however I'm simplifying for ease of comprehension :-)