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)

Answer 2

Pipe the result through tail to cut off the header:

... shop | tail -1)

Answer 3

An easier way is :

var=$(mysql -h abcdcef.com --port=3306 --user=root --password=hbbfe --batch --skip-column-names -Dshop -e "select count(*) from shop_tab where catalog <> ''")

Moreover, I'll preconize the use of function in order to easily add options to the MySQL command without having to modifying all your script.

function MysqlQuery() {
    mysql -h abcdcef.com --port=3306 --user=root --password=hbbfe --batch --skip-column-names -D "$1" -e "$2";
}

va=$(MysqlQuery Shop "SELECT COUNT(*) FROM shop_tab WHERE catalog <> ''")
vaABC=$(MysqlQuery Shop "SELECT COUNT(*) FROM shop_tab WHERE catalog <> 'abc'")
vadef=$(MysqlQuery Shop "SELECT COUNT(*) FROM shop_tab WHERE catalog <> 'def'")
# ...

I find this more readable too...