在命令行中在unix中传递参数

How can I pass arguments to a command in unix? For example, if I have to open a file:

如何在unix中将参数传递给命令？例如，如果我必须打开一个文件：

R> vi john/pic/mars/NASA/rover.txt

In the above vi command, I want to replace "mars" with a variable, and pass the variable value in the same line, as in:

在上面的vi命令中，我想用变量替换“mars”，并在同一行中传递变量值，如：

R> vi john/pic/$variable/NASA/rover.txt | $varaiable=pluto

Of course this doesn't work. But I hope my question is clear. Can anyone help me with this?

当然这不起作用。但我希望我的问题很明确。谁能帮我这个？

1 个解决方案

Simply move the variable definition to the beginning of the command line, as in:

只需将变量定义移动到命令行的开头，如下所示：

variable=pluto; vi john/pic/$variable/NASA/rover.txt

or even:

甚至：

variable=pluto && vi john/pic/$variable/NASA/rover.txt

OBS:

OBS：

notice you can't use $ while defining variables, only while using their values;
注意，只有在使用它们的值时，才能在定义变量时使用$;
piping your vi command to a variable assignment does not make much sense, although you could achieve some clearer parameterization from:

尽管你可以从以下方面获得更清晰的参数化，但是将vi命令输入到变量赋值没有多大意义。
```
function opener() {
    vi john/pic/$1/NASA/rover.txt
}
$ opener "pluto"
```

Simply move the variable definition to the beginning of the command line, as in:

只需将变量定义移动到命令行的开头，如下所示：

variable=pluto; vi john/pic/$variable/NASA/rover.txt

or even:

甚至：

variable=pluto && vi john/pic/$variable/NASA/rover.txt

OBS:

OBS：

notice you can't use $ while defining variables, only while using their values;
注意，只有在使用它们的值时，才能在定义变量时使用$;
piping your vi command to a variable assignment does not make much sense, although you could achieve some clearer parameterization from:

尽管你可以从以下方面获得更清晰的参数化，但是将vi命令输入到变量赋值没有多大意义。
```
function opener() {
    vi john/pic/$1/NASA/rover.txt
}
$ opener "pluto"
```