Problem with return variable in bash

Question

This mysql command line return me: count(*) 19. It\'s a big problem for me because I just want 19 to make a loop with this number. How can

Answer 1

You can use awk or cut to get second value.

awk

var = $(echo "select count(*) from shop_tab where catalog <> '';" | mysql -h abcdcef.com --port=3306 --user=root --password=hbbfe shop | awk '{ print $2 }')

cut

var = $(echo "select count(*) from shop_tab where catalog <> '';" | mysql -h abcdcef.com --port=3306 --user=root --password=hbbfe shop | cut -d' ' -f2)