通过 bash 脚本将参数传递给 /bin/bash
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Passing arguments to /bin/bash via a bash script
提问by JCOidl
I am writing a bash script that takes a number of command line arguments (possibly including spaces) and passes all of them to a program (/bin/some_program) via a login shell. The login shell that is called from the bash script will depend on the user's login shell. Let's suppose the user uses /bin/bash as their login shell in this example... but it might be /bin/tcsh or anything else.
我正在编写一个 bash 脚本，它接受许多命令行参数（可能包括空格）并通过登录 shell 将所有这些参数传递给一个程序 (/bin/some_program)。从 bash 脚本调用的登录 shell 将取决于用户的登录 shell。让我们假设用户在这个例子中使用 /bin/bash 作为他们的登录 shell……但它可能是 /bin/tcsh 或其他任何东西。
If I know how many arguments will be passed to some_program, I can put the following lines in my bash script:
如果我知道将有多少参数传递给 some_program，我可以在我的 bash 脚本中加入以下几行：
#!/bin/bash
# ... (some lines where we determine that the user's login shell is bash) ...
/bin/bash --login -c "/bin/some_program \"\" \"\""
and then call the above script as follows:
然后调用上面的脚本如下：
my_script "this is too" cool
With the above example I can confirm that some_program receives two arguments, "this is too" and "cool".
通过上面的例子，我可以确认 some_program 收到两个参数，“这太”和“酷”。
My question is... what if I don't know how many arguments will be passed? I'd like to pass all the arguments that were sent to my_script along to some_program. The problem is I can't figure out how to do this. Here are some things that don't work:
我的问题是...如果我不知道将传递多少个参数怎么办？我想将发送到 my_script 的所有参数传递给 some_program。问题是我不知道如何做到这一点。以下是一些不起作用的事情：
/bin/bash --login -c "/bin/some_program $@" # --> 3 arguments: "this","is","too"
/bin/bash --login -c /bin/some_program "$@" # --> passes no arguments
回答by themel
Quoting the bash manual for -c:
引用 bash 手册-c：
If the -c option is present, then commands are read from string. If there are arguments after the string, they are assigned to the positional parameters, starting with $0.
如果存在 -c 选项，则从字符串中读取命令。如果字符串后面有参数，则将它们分配给位置参数，从 $0 开始。
Works for me:
对我有用：
$ cat x.sh
#!/bin/bash
/bin/bash --login -c 'echo 1: 2: 3:' echo "$@"
$ ./x.sh "foo bar" "baz" "argh blargh quargh"
1:foo bar 2:baz 3:argh blargh quargh
I don't know how you arrived at the "passes no arguments" conclusion, maybe you missed the $0bit?
我不知道你是如何得出“不通过论证”的结论的，也许你错过了这$0一点？
回答by geirha
Avoid embedding variables into other scripts, pass them on as arguments instead. In this case:
避免将变量嵌入到其他脚本中，而是将它们作为参数传递。在这种情况下：
bash --login -c 'some_program "$@"' some_program "$@"
The first argument after -c '...' is taken as $0, so I just put in some_program there.
-c '...' 之后的第一个参数被视为 $0，所以我只是把 some_program 放在那里。
On a side note, it's an odd requirement to require a login shell. Doesn't the user log in?
附带说明一下，需要登录 shell 是一个奇怪的要求。用户没有登录吗？
