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 I do?

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

Try "select count(*) AS res from shop_tab where catalog <> '';" — Yasen Zhelev, Jul 20 '11 at 14:44
Duplicate with : http://stackoverflow.com/questions/6759397/declare-variable-in-bash/6759568#6759568 — Arnaud F., Jul 20 '11 at 14:48

score 6 · Answer 1 · answered Jul 20 '11 at 14:48

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...

score 2 · Answer 2 · answered Jul 20 '11 at 14:47

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)

score 2 · Accepted Answer · answered Jul 20 '11 at 14:47

2

Pipe the result through tail to cut off the header:

... shop | tail -1)

answered Jul 20 '11 at 14:47

Aaron Digulla

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Problem with return variable in bash

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