Escape dollar sign in regexp for sed - regex

Question

I will introduce what my question is about before actually asking - feel free to skip this section!
Some background info about my setup
To update files manually in a software system, I am creating a bash script to remove all files that are not present in the new version, using diff:
for i in $(diff -r old new 2>/dev/null | grep "Only in old" | cut -d "/" -f 3- | sed "s/: /\//g"); do echo "rm -f $i" >> REMOVEOLDFILES.sh; done
This works fine. However, apparently my files often have a dollar sign ($) in the filename, this is due to some permutations of the GWT framework. Here is one example line from the above created bash script:
rm -f var/lib/tomcat7/webapps/ROOT/WEB-INF/classes/ExampleFile$3$1$1$1$2$1$1.class
Executing this script would not remove the wanted files, because bash reads these as argument variables. Hence I have to escape the dollar signs with "\$".
My actual question
I now want to add a sed-Command in the aforementioned pipeline, replacing this dollar sign. As a matter of fact, sed also reads the dollar sign as special character for regular expressions, so obviously I have to escape it as well.
But somehow this doesn't work and I could not find an explanation after googling a lot.
Here are some variations I have tried:
echo "Bla$bla" | sed "s/\$/2/g" # Output: Bla2
echo "Bla$bla" | sed 's/$$/2/g' # Output: Bla
echo "Bla$bla" | sed 's/\\$/2/g' # Output: Bla
echo "Bla$bla" | sed 's/#"\$"/2/g' # Output: Bla
echo "Bla$bla" | sed 's/\\\$/2/g' # Output: Bla
The desired output in this example should be "Bla2bla".
What am I missing?
I am using GNU sed 4.2.2
EDIT
I just realized, that the above example is wrong to begin with - the echo command already interprets the $ as a variable and the following sed doesn't get it anyway... Here a proper example:
Create a textfile test with the content bla$bla
cat test gives bla$bla
cat test | sed "s/$/2/g" gives bla$bla2
cat test | sed "s/\$/2/g" gives bla$bla2
cat test | sed "s/\\$/2/g" gives bla2bla
Hence, the last version is the answer. Remember: when testing, first make sure your test is correct, before you question the test object........

Answer 1

The correct way to escape a dollar sign in regular expressions for sed is double-backslash. Then, for creating the escaped version in the output, we need some additional slashes:
cat filenames.txt | sed "s/\\$/\\\\$/g" > escaped-filenames.txt
Yep, that's four backslashes in a row. This creates the required changes: a filename like bla$1$2.class would then change to bla\$1\$2.class.
This I can then insert into the full pipeline:
for i in $(diff -r old new 2>/dev/null | grep "Only in old" | cut -d "/" -f 3- | sed "s/: /\//g" | sed "s/\\$/\\\\$/g"; do echo "rm -f $i" >> REMOVEOLDFILES.sh; done
Alternative to solve the background problem
chepner posted an alternative to solve the backround problem by simply adding single-quotes around the filenames for the output. This way, the $-signs are not read as variables by bash when executing the script and the files are also properly removed:
for i in $(diff -r old new 2>/dev/null | grep "Only in old" | cut -d "/" -f 3- | sed "s/: /\//g"); do echo "rm -f '$i'" >> REMOVEOLDFILES.sh; done
(note the changed echo "rm -f '$i'" in that line)

Answer 2

There are other problems with your script, but file names containing $ are not a problem if you properly quote the argument to rm in the resulting script.
echo "rm -f '$i'" >> REMOVEOLDFILES.sh
or using printf, which makes quoting a little nicer and is more portable:
printf "rm -f '%s'" "$i" >> REMOVEOLDFILES.sh
(Note that I'm addressing the real problem, not necessarily the question you asked.)

Answer 3

There is already a nice answer directly in the edited question that helped me a lot - thank you!
I just want to add a bit of curious behavior that I stumbled across: matching against a dollar sign at the end of lines (e.g. when modifying PS1 in your .bashrc file).
As a workaround, I match for additional whitespace.
$ DOLLAR_TERMINATED="123456 $"
$ echo "${DOLLAR_TERMINATED}" | sed -e "s/ \\$/END/"
123456END
$ echo "${DOLLAR_TERMINATED}" | sed -e "s/ \\$$/END/"
sed: -e expression #1, char 13: Invalid back reference
$ echo "${DOLLAR_TERMINATED}" | sed -e "s/ \\$\s*$/END/"
123456END
Explanation to the above, line by line:
Defining DOLLAR_TERMINATED - I want to replace the dollar sign at the end of DOLLAR_TERMINATED with "END"
It works if I don't check for the line ending
It won't work if I match for the line ending as well (adding one more $ on the left side)
It works if I additionally match for (non-present) whitespace
(My sed version is 4.2.2 from February 2016, bash is version 4.3.48(1)-release (x86_64-pc-linux-gnu), in case that makes any difference)