You should generally avoid ls in scripts.
Also, you should generally prefer the modern POSIX $(command substitution) syntax like you already do in several other places in your script; the obsolescent backtick `command substitution` syntax is clunky and somewhat more error-prone.
If this works in the current directory but fails in others, it means that you have a file matching the regex in the current directory, but not in the other directory.
Anyway, the idiomatic way to do what you appear to be attempting is simply
cat *email?csv* >e.csv
If you meant to match a literal dot, that's \. in a regular expression. The ? is a literal interpretation of what your grep actually did; but in the following, I will assume you actually meant to match *email.csv* (or in fact probably even *email.csv without a trailing wildcard).
If you want to check if there are any files, and avoid creating e.csv if not, that's slightly tricky; maybe try
for file in *email.csv*; do
test -e "$file" || break
cat *email.csv* >e.csv
break
done
Alternatively, look into the nullglob feature of Bash. See also Check if a file exists with wildcard in shell script.
On the other hand, if you just want to check whether email.csv exists, without a wildcard, that's easy:
if [ -e email.csv ]; then
cat email.csv >e.csv
fi
In fact, that can even be abbreviated down to
test -e email.csv && cat email.csv >e.csv
As an aside, read can perfectly well split a line into tokens.
#!/bin/bash
yesterday=$(date --date "$c days ago" +%F)
while IFS=, read -r dir country _
do
cd "path/$dir" # notice proper quoting, too
cat *email.csv* > e.csv
# probably don't forget to cd back
cd ../..
done < "s.csv"
If this is in fact all your script does, probably do away with the silly and slightly error-prone cd;
while IFS=, read -r dir country _
do
cat "path/$dir/"*email.csv* > "path/$dir/e.csv"
done < "s.csv"
See also When to wrap quotes around a shell variable.
