bash 从bash中的变量在for循环中执行命令
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Execute command inside for loop from variable in bash
提问by tom
Trying to run commands defined in variables inside a for loop:
尝试运行在 for 循环内的变量中定义的命令：
somevar="Bit of text"
cmd1="command \"search '$somevar' here\""
cmd2="command \"search '$somevar' there\""
for cmd in cmd1 cmd2 ; do
eval $$cmd
ssh server1 eval $$cmd
done
I've put in the variations I have to consider such as the ssh inside the loop etc as these are needed in my script. I think the eval is the right direction, but the way that the quotes inside the command get interpreted comes through wrong.
我已经加入了我必须考虑的变化，例如循环内的 ssh 等，因为我的脚本需要这些。我认为 eval 是正确的方向，但是解释命令中的引号的方式是错误的。
回答by tom
Consider this broken example:
考虑这个破碎的例子：
$ cmd1="touch \"file with spaces\""
$ $cmd1
Quoting is handled before $cmd1is expanded, so instead of one file this will create three files called "file, with, and spaces". One can use eval $cmdto force quote removal after the expansion.
引用在$cmd1展开之前处理，因此这将创建三个文件"file，而不是一个文件，名为with、 和spaces"。可以使用eval $cmd在扩展后强制删除引用。
Even though it uses eval, the line eval \$$cmdhas that same quoting problem since \$$cmdexpands to $cmd1, which is then evaluated by evalwith the same behaviour as the broken example.
即使它使用eval，该行eval \$$cmd也有相同的引用问题，因为\$$cmd扩展到$cmd1，然后用eval与损坏的示例相同的行为对其进行评估。
The argument to evalmust be the actual command, not the expression $cmd1. This can be done using variable indirection: eval "${!cmd}".
的参数eval必须是实际的命令，而不是表达式$cmd1。这可以使用变量间接完成：eval "${!cmd}"。
When running this through SSH there is no need for the evalbecause the remote shell also performs quote removal.
当通过 SSH 运行时，不需要 ，eval因为远程 shell 也会执行引用删除。
So here is the fixed loop:
所以这是固定循环：
for cmd in cmd1 cmd2 ; do
eval "${!cmd}"
ssh server1 "${!cmd}"
done
An alternative to indirection is to iterate over the values of cmd1and cmd2instead of their names:
间接的替代方法是迭代 和 的值cmd1而cmd2不是它们的名称：
for cmd in "$cmd1" "$cmd2" ; do
eval "$cmd"
ssh server1 "$cmd"
done
回答by Maxime Chéramy
I see two solutions, either you change your loop to:
我看到两种解决方案，要么将循环更改为：
for cmd in "$cmd1" "$cmd2" ; do
ssh server1 $cmd
done
or to:
或者：
for cmd in cmd1 cmd2 ; do
ssh server1 ${!cmd}
done
回答by anubhava
Instead of eval \$$cmdyou need to use:
而不是eval \$$cmd你需要使用：
res=$(eval "$cmd")
ssh server1 "$res"
