bash bash脚本如何从变量执行命令
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bash script how to execute a command from a variable
提问by ooXei1sh
I am trying to alter the bash function below to execute each command argument. But when I run this script the first echo works as intended but the second echo that attempts to append to the scratch.txt file does not actually execute. It just gets echo'd into the prompt.
我正在尝试更改下面的 bash 函数以执行每个命令参数。但是当我运行这个脚本时，第一个回声按预期工作，但试图附加到scratch.txt 文件的第二个回声实际上没有执行。它只是被回显到提示中。
#!/bin/sh
clear
function each(){
while read line; do
for cmd in "$@"; do
cmd=${cmd//%/$line}
printf "%s\n" "$cmd"
$cmd
done
done
}
# pipe in the text file and run both commands
# on each line of the file
cat scratch.txt | each 'echo %' 'echo -e "%" >> "scratch.txt"'
exit 0
how do I get the $cmd variable to execute as a command?
如何让 $cmd 变量作为命令执行？
I found the original code from answer 2 here: xargs with multiple commands as argument
我在这里找到了答案 2 的原始代码： xargs with multiple commands as argument
回答by ghoti
You want eval. It's evil. Or at least, dangerous. Read all about it at BashFAQ #48.
你要eval。这是邪恶的。或者至少，危险。在BashFAQ #48阅读所有相关信息。
