bash bash脚本更改目录并使用参数执行命令
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bash script to change directory and execute command with arguments
提问by frodo
I am trying to do the following task:
write a shell script called changedirwhich
takes a directory name, a command name and (optionally) some additional arguments.
The script will then change into the directory indicated, and
executes the command indicated with the arguments provided.
我正在尝试执行以下任务：编写一个名为的 shell 脚本changedir，它接受一个目录名、一个命令名和（可选）一些附加参数。然后脚本将切换到指定的目录，并执行使用提供的参数指定的命令。
Here an example:
这里有一个例子：
$ sh changedir /etc ls -al
This should change into the /etcdirectory and run the command ls -al.
这应该更改为/etc目录并运行命令ls -al。
So far I have:
到目前为止，我有：
#!/bin/sh
directory=; shift
command=; shift
args=; shift
cd $directory
$command
If I run the above like sh changedir /etc lsit changes and lists the directory. But if I add arguments to the lsit does not work. What do I need to do to correct it?
如果我像上面一样运行sh changedir /etc ls它会更改并列出目录。但是如果我向ls它添加参数，它就不起作用。我需要做什么来纠正它？
回答by CB Bailey
You seemed to be ignoring the remainder of the arguments to your command.
您似乎忽略了命令的其余参数。
If I understand correctly you need to do something like this:
如果我理解正确，你需要做这样的事情：
#!/bin/sh
cd "" # change to directory specified by arg 1
shift # drop arg 1
cmd="" # grab command from next argument
shift # drop next argument
"$cmd" "$@" # expand remaining arguments, retaining original word separations
A simpler and safer variant would be:
一个更简单、更安全的变体是：
#!/bin/sh
cd "" && shift && "$@"
回答by dgasper
Since there can probably be more than a single argument to a command, i would recommend using quotation marks. Something like this:
由于命令的参数可能不止一个，我建议使用引号。像这样的东西：
sh changedir.sh /etc "ls -lsah"
Your code would be much more readable if you ommited the 'shift':
如果您省略“shift”，您的代码将更具可读性：
directory=;
command=;
cd $directory
$command
or simply
或者干脆
cd DIRECTORY_HERE; COMMAND_WITH_ARGS_HERE
