How to fetch field from MySQL query result in bash

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?

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.

Even More Compact:

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

Try:

mysql -B --column-names=0 -uroot -ppwd -e "SELECT id FROM nagios.host WHERE name='$host'"

-B will print results using tab as the column separator and

--column-names=0 will disable the headers.

I tried the solutions but always received empty response.

In my case the solution was:

#!/bin/sh

FIELDVALUE=$(mysql -ss -N -e "SELECT field FROM db.table where fieldwhere = '$2'")

echo $FIELDVALUE