How does bash expand escaped characters when dereferencing variables

Question

If I quit using variables, and just write the regexes directly into to the last sed command, everything works. But as it is here, no substitutions are done?

#!/bin/bash
#html substitutions
ampP="\&"
ampR="&"

ltP="\<"
ltR="<"

gtP="\&gt;"
gtR=">"

quotP="\&quot;"
quotP2='\&#8220;'
quotP3="\&#8221;"
quotR="\""

tripDotP="\&#8230"
tripDotR="..."

tickP="\&#8217;"
tickR="\´"

#get a random page, and filter out the quotes
#pick a random quote
#translate wierd html symbols
curl "www.yodaquotes.net/page/$((RANDOM % 9 +1))/" -Ls | sed -nr 's/.*data-text=\"([^\"]+)\".*/\1/p' \
| sort -R | head -n1 \
| sed 's/"$ampP"/"$ampR"/g; s/$ltP/$ltR/g; s/$gtP/$gtR/g; s/$quotP/$quotR/g; s/"$quotP2"/"$quotR"/g; s/$quotP3/$quotR/g; s/$tripDotP/$tripDotR/g; s/$stickP/$stickR/g'

Answer 1

This sed isn't going to work:

sed 's/"$ampP"/"$ampR"/g'

because of wrong shell quoting. Your shell variables won't be expanded at all in single quotes. Try using this form:

sed "s~$ampP~$ampR~g"

Answer 2

Debugging 101: Let's just echo what sed receives:

echo 's/"$ampP"/"$ampR"/g; s/$ltP/$ltR/g; s/$gtP/$gtR/g; s/$quotP/$quotR/g; s/"$quotP2"/"$quotR"/g; s/$quotP3/$quotR/g; s/$tripDotP/$tripDotR/g; s/$stickP/$stickR/g'

s/"$ampP"/"$ampR"/g; s/$ltP/$ltR/g; s/$gtP/$gtR/g; s/$quotP/$quotR/g; s/"$quotP2"/"$quotR"/g; s/$quotP3/$quotR/g; s/$tripDotP/$tripDotR/g; s/$stickP/$stickR/g

That doesn't look right now, does it?

There's no variable substitution within single quotes in bash. That's why we have two different quotes, so you can decide which one is more appropriate for the task.

For readability I would suggest putting each sed command within double quotes.

Like this: "s/$ampP/$ampR/g"