英文:
Why getting blank output when running "/bin/sh" with "-c" option
问题
你好，这是翻译的部分：
我正在以如下方式使用/bin/sh和-c选项运行，但却得到了空白输出：
[root@dockerhost dproj]# /bin/sh -c echo helloworld
[root@dockerhost dproj]#
为什么上述命令没有打印出helloworld？
我尝试阅读man页面，但无法理解任何内容。
从man sh中：
-c 从command_string操作数中读取命令。
从command_name操作数的值以及剩余的参数操作数中按顺序设置特殊参数0的值
（参见第2.5.2节，特殊参数）。
不应从标准输入读取任何命令。
英文:
I am running /bin/sh with -c option as below but getting a blank output:
[root@dockerhost dproj]# /bin/sh -c echo helloworld
[root@dockerhost dproj]#
Why the above command is not printing helloworld?
I have tried to read the man page but not able to understand anything.
From man sh:
-c Read commands from the command_string operand.
Set the value of special parameter 0
(see Section 2.5.2, Special Parameters) from the
value of the command_name operand and the positional
parameters ($1, $2, and so on) in sequence from the
remaining argument operands.
No commands shall be read from the standard input.
答案1
得分: 4
/bin/sh -c echo helloworld 运行 echo 命令，会打印一个空行。您的意思是输入：
/bin/sh -c "echo helloworld"
这将运行命令 echo helloworld。
-c 选项用于 sh，会导致第一个非选项参数被视为要在新 shell 中运行的命令字符串。其余的参数用于填充该 shell 的编号参数，从 $0 开始（即 shell 进程正在运行的“名称”）。这些参数不会自动解析为由该 shell 执行的任何实用程序。
因此，如果您使用 /bin/sh -c echo helloworld 调用 sh：
- 传递给 shell 解释器
sh的输入只是echo helloworld变成了sh会话中的$0
通常，shell 调用中的 $0 应该是 shell 的名称；这通常是默认设置的：
bash$ /bin/sh
$ echo $0
/bin/sh
$ exit
bash$ echo $0
bash
由 -c 给 shell 的命令会被解释为同样输入到 shell 本身那样，您可以在命令中使用 $n 来引用 shell 解释器的额外参数。如果您想这样做，需要记住对 -c 选项参数使用单引号，这样其内容就不会被外部 shell 解释。也许学习以下内容会有所帮助：
bash$ /bin/sh -c 'echo $0' fakename # 正确使用了单引号
fakename
bash$ /bin/sh -c "echo $0" fakename # 这里，$0 被外部 shell 展开了
bash
bash$ /bin/sh -c 'echo "$@"' sh arg1 arg2 arg3 # $0 不是 $@ 的一部分
arg1 arg2 arg3
英文:
/bin/sh -c echo helloworld runs the command echo, which prints a blank line. You meant to type:
/bin/sh -c "echo helloworld"
which runs the command echo helloworld.
The -c option to sh causes the first non-option argument to be treated as a string of commands to run in a new shell. The remaining arguments are used to fill in the that shell's numbered arguments, starting with $0 (the "name" under which the shell process is running.) These arguments are not automatically parsed to any utility executed by that shell.
So if you invoke sh with /bin/sh -c echo helloworld:
- the input passed to the shell interpreter
shis simplyecho helloworldbecomes$0in theshsession.
Normally, $0 in a shell invocation should be the name of the shell; that's what it will be set to by default:
bash$ /bin/sh
$ echo $0
/bin/sh
$ exit
bash$ echo $0
bash
Since the command given to the shell by -c is interpreted as though it were input to the shell itself, you can use $n in the command in order to refer to the extra arguments to the shell interpreter. If you want to do that, you need to remember to single-quote the -c option argument so that it's contents are not interpreted by the outer shell. Perhaps studying the following will help:
bash$ /bin/sh -c 'echo $0' fakename # Correctly single-quoted
fakename
bash$ /bin/sh -c "echo $0" fakename # Here, $0 is expanded by the outer shell
bash
bash$ /bin/sh -c 'echo "$@"' sh arg1 arg2 arg3 # $0 is not part of $@
arg1 arg2 arg3
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