Put results of mysql command line query into bash script variable

Question

I am having a really difficult time figuring out how to get the results of a mysql query into a bash shell script variable.

testDBName="some_database_name"
mysqlUser="root"
mysqlPassword="password"
mysqlCmd="/Applications/MAMP/Library/bin/mysql -u $mysqlUser -p $mysqlPassword -B -N"
cmdRes=$($mysqlCmd -e 'SHOW DATABASES LIKE "$testDBName"') #THIS IS THE TROUBLESOME LINE
if [ "$cmdRes" = "" ]; then
    echo "Table does not exist"
else
    echo "Table already exists"
fi

The line that is giving me the trouble is "cmdRes=...". If I just hard-code a table name into that it works fine, like this:

cmdRes=$($mysqlCmd -e 'SHOW DATABASES LIKE "some_database_name"')

But I cannot wrap my head around how/why/what is going on when I have a $variable inside that thing. Based on answers to a lot of other similar-but-different questions, I've tried putting different portions of the string into different variables to cut down on quoting, I've tried single quotes, double quotes, backslashes, double-single-double quotes, curly braces, running eval, etc. -- but nothing is working.

Thanks for any help you can provide.

paulsm4 · Accepted Answer · 2011-09-04 21:10:12Z

5

The problem is that the shell won't expand anything inside the single quotes (').

One possible solution:

cmdRes=$($mysqlCmd -e "SHOW DATABASES LIKE '$testDBName'")

edited Sep 4, 2011 at 21:10

answered Sep 4, 2011 at 20:06

paulsm4

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1 Comment

Jordan Lev Over a year ago

Wait, I just tried that without the backslashes and it works! If you edit your answer I'll mark it correct (I don't have edit priveleges). Thanks!

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