我似乎无法在“-c”选项字符串之后使用带有参数的 Bash“-c”选项
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I can't seem to use the Bash "-c" option with arguments after the "-c" option string
提问by Chris Markle
The man page for Bash says, regarding the -coption:
Bash 的手册页说，关于-c选项：
-c stringIf the-coption is present, then commands are read fromstring. If there are arguments after the string, they are assigned to the positional parameters, starting with$0.
-c string如果该-c选项存在，则从 中读取命令string。如果字符串后面有参数，则将它们分配给位置参数，以$0.
So given that description, I would think something like this ought to work:
因此，鉴于该描述，我认为这样的事情应该可行：
bash -c "echo arg 0: arg 0: -bash, arg 1:
, arg 1: " arg1
but the output just shows the following, so it looks like the arguments after the -cstring are not being assigned to the positional parameters.
但输出仅显示以下内容，因此看起来-c字符串之后的参数没有分配给位置参数。
$ bash -c 'echo arg 0: bash -c "echo arg 0: $0, arg 1: $1" arg0 arg1
, arg 1: ' arg1 arg2
arg 0: arg1, arg 1: arg2
I am running a fairly ancient Bash (on Fedora 4):
我正在运行一个相当古老的 Bash（在 Fedora 4 上）：
[root@dd42 trunk]# bash --version GNU bash, version 3.00.16(1)-release (i386-redhat-linux-gnu) Copyright (C) 2004 Free Software Foundation, Inc.
[root@dd42 trunk]# bash --version GNU bash, version 3.00.16(1)-release (i386-redhat-linux-gnu) 版权所有 (C) 2004 Free Software Foundation, Inc.
I am really trying to execute a bit of a shell script with arguments. I thought -clooked very promising, hence the issue above. I wondered about using eval, but I don't think I can pass arguments to the stuff that follows eval. I'm open to other suggestions as well.
我真的想用参数执行一些shell脚本。我认为-c看起来很有希望，因此出现了上述问题。我想知道如何使用 eval，但我不认为我可以将参数传递给 eval 之后的内容。我也愿意接受其他建议。
回答by martin clayton
You need to use single quotes to prevent interpolation happening in your calling shell.
您需要使用单引号来防止在调用 shell 中发生插值。
arg 0: arg0, arg 1: arg1Or escape the variables in your double-quoted string. Which to use might depend on exactly what you want to put in your snippet of code.
或者转义双引号字符串中的变量。使用哪个可能取决于您想要在代码片段中放入的内容。
回答by NawaMan
Because '$0' and '$1' in your string is replaced with a variable #0 and #1 respectively.
因为字符串中的' $0' 和 ' $1' 分别替换为变量 #0 和 #1。
Try :
尝试 ：
#!/bin/bash
echo 0:$ bash -c './foo.sh ${1+"$@"}' foo "bar baz"
0:./foo.sh
1:bar baz
2:
echo 1:
echo 2:
In this code $of both are escape so base see it as a string $and not get replaced.
在这段代码中$，两者都是转义的，因此基本将其视为字符串$而不是被替换。
The result of this command is:
这个命令的结果是：
$ bash -c ./foo.sh foo "bar baz"
0:./foo.sh
1:
2:
Hope this helps.
希望这可以帮助。
回答by z0r
martin is right about the interpolation: you need to use single quotes. But note that if you're trying to pass arguments to a command that is being executed withinthe string, you need to forward them on explicitly. For example, if you have a script foo.shlike:
马丁关于插值是正确的：你需要使用单引号。但请注意，如果你想传递参数给正在执行的命令中的字符串，你需要明确它们转发上。例如，如果您有一个脚本foo.sh，例如：
$ bash -c './foo.sh $@' foo "bar baz"
0:./foo.sh
1:bar
2:baz
Then you should call it like this:
那么你应该这样称呼它：
##代码##Or more generally bash -c '${0} ${1+"$@"}' <command> [argument]...
或更一般地 bash -c '${0} ${1+"$@"}' <command> [argument]...
This means you can pass in arguments to sub-processes without embedding them in the command string, and without worrying about escaping them.
这意味着您可以将参数传递给子进程，而无需将它们嵌入命令字符串中，也无需担心转义它们。
回答by tyranid
Add a backslash to the $0(i.e., \$0), otherwise your current shell escapes $0to the name of the shell before it even gets to the subshell.
给$0(ie, \$0)添加一个反斜杠，否则你当前$0的 shell 会在它到达子 shell 之前转义到shell的名称。
