How to fetch field from MySQL query result in bash

Question

I would like to get only the value of a MySQL query result in a bash script. For example the running the following command:

mysql -uroot -ppwd -e "SELECT id FROM nagios.host WHERE name='$host'" 

returns:

+----+ | id | +----+ | 0  | +----+ 

How can I fetch the value returned in my bash script?

Answer 1

Even More Compact:

id=$(mysql -uroot -ppwd -se "SELECT id FROM nagios.host WHERE name=$host"); echo $id; 

Answer 2

Use -s and -N:

> id=`mysql -uroot -ppwd -s -N -e "SELECT id FROM nagios.host WHERE name='$host'"` > echo $id 0 

From the manual:

--silent, -s

   Silent mode. Produce less output. This option can be given multiple    times to produce less and less output.     This option results in nontabular output format and escaping of    special characters. Escaping may be disabled by using raw mode; see    the description for the --raw option. 

--skip-column-names, -N

   Do not write column names in results. 

EDIT

Looks like -ss works as well and much easier to remember.