Bash Mysql Command Substitution Login Failure

Question

Bash and MySQL versions:

GNU bash, version 4.2.25(1)-release (x86_64-pc-linux-gnu)
mysql  Ver 14.14 Distrib 5.5.37, for debian-linux-gnu (x86_64) using readline 6.2

I have a bash script that tries to capture the result of a MySQL query to a variable using command substitution with the MySQL password as a variable that contains a dollar sign and no matter which way I try it, it always either fails authentication with access denied message or the query returns the MySQL help command text. I've tried dozens of different combinations of both types of command substitution and escaping characters till the cows come home. Here is an example:

PASS='pass$word'; RESULT=`mysql -u user -p'${PASS}' -h RemoteHostName DBName -e "select count(*) from TableName;"`; echo "${RESULT}";

This will return "Access denied for user"...

PASS='pass$word'; RESULT=`mysql -u user -p\'${PASS}\' -h RemoteHostName DBName -e \"select count\(*\) from TableName\;\"`; echo "${RESULT}";

This will return the MySQL help text

Any help would be much appreciated, thanks in advance.

Answer 1

val0x00ff's answer was correct in the comment section above

RESULT=$(mysql -u user -p "${PASS}" -h RemoteHostName DBName -e "select count(*) from TableName;")

Answer 2

You need to escape the password value, not its usage:

PASS="pass\$word"; RESULT=`mysql -u user -p'${PASS}' -h RemoteHostName DBName -e "select count(*) from TableName;"`; echo "${RESULT}";

As everything after the $ is being interpreted as a reference to a variable named $word

[user@host ~]$ echo "pass$word"
pass
[user@host ~]$ echo "pass\$word"
pass$word