如何从bash中获取MySQL查询结果中的字段

Question

我想在bash脚本中只获取MySQL查询结果的值.例如,运行以下命令:

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

收益:

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

如何获取bash脚本中返回的值？

Answer 1

使用-s和-N:

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

从手册:

--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.

编辑

看起来像-ss工作也好,更容易记住.

Answer 2

更紧凑:

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