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JonMorehouse
JonMorehouse

Reputation: 1383

Bash Script Evaluate String Problems

I'm trying to create a script to automatically delete all of the tables from a database using shell.

The commented out variable $drop works fine, however when I try to substitute in the table

for table in $tables
do
    command="'drop table ${table}'"

    # drop=$(${login} -e 'drop table test') -- this works fine
    drop=$(${login} -e $command)
    echo $drop
    # echo -e "Removed table ${table}"
done

Upvotes: 1

Views: 213

Answers (1)

glenn jackman
glenn jackman

Reputation: 246764

(major edit)

The issue is with your use of quotes. In your code, since you do not quote $command it is subject to word splitting by the shell. The $login command receives these arguments: "-e", "'drop", "table", "table_name'" -- note the stray single quotes in the second and last elements.

Do this:

command="drop table $table"
drop=$($login -e "$command")

Upvotes: 1

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