linux shell 截取内容，赋值给变量

Posted 2023-04-09

询问个linux系统shell脚本问题，
通过grep已经截取到一行数据
a 123
怎么把这2个数据，赋值给变量 var1 var2？

参考技术A 说具体点吧。追问

具体就是：
myvar=8;#初始值，获取到的需要保存到这个变量
#从数据库获取指定配置，
mysql -h$config_host -u$user \
-p$pwd $dbname -e "select id,config from $tablename where id=1" |
grep -e "^[^i]"
#grep 去掉列名行，只留下数据行，例如"1 8096"
需要在myvar中 保存8096这个数值，，，，

参考技术B var1=`grep xxx`本回答被提问者采纳 参考技术C 关注中！

Linux Shell中的命令替换

Shell脚本中最有用的特性之1就是可以从命令输出中提取信息，并将其赋给变量。有两种方法可以将命令输出赋值给变量。

• 反引号字符(`)

反引号字符位于键盘上ESC键的下面。赋值给变量的情势以下：

test=`date`

# demo
$ test=`date`
$ echo $test
2017年 1月11日 星期3 21时26分04秒 CST

两个反引号字符之间是shell命令，shell命令的履行结果赋值给test，注意=两边不能有空格。

• $()情势

赋值给变量的情势以下：

test=$(date)

# demo
$ test=$(date)
$ echo $test
2017年 1月11日 星期3 21时26分35秒 CST

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