在 bash 脚本中执行命令 [英] Execute a command in bash script

问题描述

当命令作为第一个参数传递给 shell 脚本时，我试图在 shell 脚本中执行命令(在我的情况下使用 sh 作为我的 shell).示例:

I'm trying to execute a command inside a shell script (in my case using sh as my shell) when the command is passed as the first argument to the shell script. Example:

sh command_execute.sh ls -l

假设我在命令行中输入了上述内容，理想情况下我希望脚本查看第一个参数(在本例中为 ls -l)，然后执行它.

Let's say I enter the above on the command line, ideally I would like the script to look at the first argument (which in this case would be ls -l), and execute it.

推荐答案

有两种方法可以做到这一点.一种方法是将您的 shell 脚本设置为

There are two ways you can do this. One way would be to set your shell script as

# File: command_execute.sh
$1

并像这样运行它:sh command_execute.sh 'ls -l'(注意ls -l 周围的单引号).这将做的是将完整的字符串 ls -l 传递给您的脚本，并且由于它是脚本的第一个参数，因此脚本中的 $1 将有效替换为 ls -l，然后该命令将被执行.

and run it like this: sh command_execute.sh 'ls -l' (notice the single quotes around ls -l). What this will do is the full string ls -l will be passed to your script, and since it is the first argument to the script, $1 in the script will be effectively replaced with ls -l, and then that command will be executed.

另一种方法是将其用作您的脚本:

The other way would be to use this as your script:

# File: command_execute.sh
"$@"

在这种情况下，"$@" 是传递给脚本的所有参数.如果你运行像 sh command_execute.sh ls -l 这样的脚本(注意在这种情况下 ls -l not 被引用)，那么 ls 作为参数 1 传递给脚本，-l 作为参数 2 传递给脚本.然后 "$@" 是有效地替换为ls -l，然后执行命令.

In this case, "$@" is all the arguments that were passed to the script. If you run the script like sh command_execute.sh ls -l (note in this case the ls -l is not quoted), then ls is passed to the script as argument 1, and -l is passed to the script as argument 2. Then "$@" is effectively replaced with ls -l, and then the command is executed.

哪个最好取决于您的要求.

Which of these is best depends on your requirements.

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