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.

Community · Accepted Answer · 2017-05-23 10:25:34Z

0

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, 2017 at 10:25

CommunityBot

11 silver badge

answered Jul 19, 2013 at 15:47

Brian Agnew

273k38 gold badges342 silver badges443 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

user2507193 Over a year ago

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?

Collectives™ on Stack Overflow

Referencing mysql return in shellscript

1 Answer 1

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related