Bash for loop: Backticks

Question

I have this code

$db=test-1
for T in `mysql -u$dbUser -p$dbPass -N -B -e 'show tables from '$db`;
do
count=$((count+1))
mysqldump --skip-comments --compact --skip-lock-tables -u$dbUser -p$dbPass $db $T > $GIT_MYSQL/$T.sql
done
done;

It gives me this error

ERROR 1064 (42000) at line 1: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '-1' at line 1

How can I resolve it?

1). The Bash backtick syntax is deprecated, prone to typos and mis-reading, and cannot be nested. (And it's painful to put \` in SO comments). Consider using the `$(...)` syntax instead. 2). You generally should use double-quotes around parameter expansions to prevent word-splitting, etc. See [BashGuide](http://mywiki.wooledge.org/BashGuide/Practices#Quoting) for details. You may also find [ShellCheck](http://www.shellcheck.net) helpful. — PM 2Ring, Jan 25 '16 at 08:18
and what do you mean by `$db=test-1` ? I would have expected `db=$(( test - 1))` or similar. Good luck. — shellter, Jan 25 '16 at 13:55
$db="test-1" If test-1 is the name of your database and for T in `mysql -u$dbUser -p$dbPass -N -B -e 'show tables from '$db'`. Like Jaco say, there a ' missing. — Joao Vitorino, Jan 25 '16 at 15:07

miken32 · Accepted Answer · 2016-01-26T00:16:59.063

You're not declaring your database name variable properly. Try doing this:

$db=test-1
echo "My database is called $db"

Compare the output with this:

db=test-1
echo "My database is called $db"

Taking into account comments above and proper (I hope) quoting, your script like this should work:

dbUser="user"
dbPass="pass"
db="test-1"
for T in $(mysql -u "$dbUser" -p"$dbPass" -N -B -e "show tables from $db")
do
    mysqldump --skip-comments --compact --skip-lock-tables -u "$dbUser" -p"$dbPass" "$db" "$T" > "$GIT_MYSQL/$T.sql"
done

Bash for loop: Backticks

1 Answers1