Referencing mysql return in shellscript

Question

I need to reference the output of this mysql code

mysql -h xx.xx.x.xx -u xxxxxx -pxxxxxx xxxx_xxxx < cmd.sql

that does a count(*) to check if that db table has been filled or not. Trying to grab the return for an if statement later in the script.

I've tried both

count=$(mysql -h xx.xx.x.xx -u xxxxxx -pxxxxxx xxxx_xxxx < cmd.sql)
if [ $count > 0 ]; then
    echo "record found\n"
fi

as well as

count=$(mysql -h xx.xx.x.xx -u xxxxxx -pxxxxxx xxxx_xxxx "select count(*) from xxxx_x")
if [ $count > 0 ]; then
    echo "record found\n"
fi

but they both got a "[: count(*): unary operator expected" error.

I even tried to "select count(specific_column) from xxx_xx" with the same error.

score 1 · Accepted Answer · edited May 23 '17 at 10:25

1

You want

if [ $count -gt 0 ]; then

The > is the output redirection operator (analogous to <, which you're using above)

To capture the last line, just pipe your results through tail -1 (see here)

edited May 23 '17 at 10:25

Community

answered Jul 19 '13 at 15:47

Brian Agnew

Thanks. I knew that was a problem but since the error wasn't on that line I didn't figure that it would be the issue. It fixed that error but now I get a "too many arguments" error. I suspect because the mysql command outputs 2 lines "count(*) | 0" How would I edit my code to only read the second line? – N Schleder Jul 19 '13 at 15:52

1 Answers1