Using a bash variable to pass multiple -e clauses to sed [duplicate] - bash

Question

This question already has answers here:
Why does shell ignore quoting characters in arguments passed to it through variables? [duplicate]
(3 answers)
Closed 6 years ago.
I'm creating a variable from an array which build up multiple -e clauses for a sed command.
The resulting variable is something like:
sedArgs="-e 's/search1/replace1/g' -e 's/search2/replace2/g' -e 's/search3/replace3/g'"
But when I try to call sed with this as the argument I get the error sed: -e expression #1, char 1: unknown command: ''
I've tried to call sed the following ways:
cat $myFile | sed $sedArgs
cat $myFile | sed ${sedArgs}
cat $myFile | sed `echo $sedArgs`
cat $myFile | sed "$sedArgs"
cat $myFile | sed `echo "$sedArgs"`
and all give the same error.
UPDATE - Duplicate question
As has been identified, this is a 'quotes expansion' issue - I thought it was something sed specific, but the duplicate question that has been identified put me on the right track.
I managed to resolve the issue by creating the sedArgs string as:
sedArgs="-e s/search1/replace1/g -e s/search2/replace2/g -e s/search3/replace3/g"
and calling it with:
cat $myFile | sed $sedArgs
which works perfectly.
Then I took the advice of tripleee and kicked the useless cat out!
sed $sedArgs $myFile
also works perfectly.

Answer 1

Use BASH arrays instead of simple string:
# sed arguments in an array
sedArgs=(-e 's/search1/replace1/g' -e 's/search2/replace2/g' -e 's/search3/replace3/g')
# then use it as
sed "${sedArgs[#]}" file

Answer 2

Here is no sane way to do that, but you can pass the script as a single string.
sedArgs='s/search1/replace1/g
s/search2/replace2/g
s/search3/replace3/g'
: then
sed "$sedArgs" "$myFile"
The single-quoted string spans multiple lines; this is scary when you first see it, but perfectly normal shell script. Notice also how the cat is useless as ever, and how the file name needs to be quoted, too.