Bash-字符串，命令和转义(哦，我的！) [英] Bash - Strings, Commands and Escaping (oh, my!)

问题描述

我现在正在浪费太多时间，试图弄清这么简单的东西....

I'm wasting so much time right now trying to figure out something so simple....

伪代码(几种语法的混合，抱歉):

pseudo code (mixture of several syntax's, sorry):

cmd1 = "find /my/starting/path -type f | grep -v -f /my/exclude/files"
cmd2 = " nl -ba -s'  ' "
cmd3 = " | xargs mv -t /move/here/dir " 

echo run_command_and_return_output($cmd1$cmd2)

$cmd1$cmd3  # just this now... 

# i don't actually want a function... but the name explains what i want to do
function run_command_and_return_output(){ /* magic */ }

---

这有效....

---

this works....

FIND=$(find $LOG_DIR -type f | grep -v -f $EXCLUDE | nl -ba -s'   ')

printf "%s\n" "$FIND"

这不是...

NL="nl -ba -s'  '"
FIND=$(find $LOG_DIR -type f -mtime +$ARCH_AGE | grep -v -f $EXCLUDE | $NL)

printf "%s\n" "$FIND"

这也不是...

NL='nl -ba -s'\''   '\'' '

这绝对可以，但是:

find /my/starting/path -type f | grep -v -f /my/exclude/files |  nl -ba -s'  ' 

或

FIND=$(find $LOG_DIR -type f -mtime +$ARCH_AGE | grep -v -f $EXCLUDE | nl -ba -s'  ' )

推荐答案

简写形式:扩展$foo不带引号的内容通过字符串拆分和glob扩展运行，但不是语法分析.这意味着在不同上下文中使用引号和转义符的字符不会被视为语法，而只会被视为数据.

Short form: Expanding $foo unquoted runs the content through string-splitting and glob expansion, but not syntactical parsing. This means that characters which would do quoting and escaping in a different context aren't honored as syntax, but are only treated as data.

如果要通过语法分析来运行字符串，请使用eval，但请注意警告，警告很大，而且影响安全性.

If you want to run a string through syntactical parsing, use eval -- but mind the caveats, which are large and security-impacting.

更好的方法是使用正确的工具完成工作-在shell数组中构建单个简单命令(而不是管道！)，并使用函数作为构造复杂命令的可组合单元. BashFAQ#50 描述了这些工具-并深入讨论了哪种工具合适.什么时候.

Much better is to use the right tools for the job -- building individual simple commands (not pipelines!) in shell arrays, and using functions as the composable unit for constructing complex commands. BashFAQ #50 describes these tools -- and goes into in-depth discussion on which of them is appropriate when.

更具体一点:

nl=( nl -ba -s'  ' )
find_output=$(find "$log_dir" -type f -mtime "+$arch_age" | grep -v -f "$exclude" | "${nl[@]}")
printf "%s\n" "$find_output"

...将是正确的，因为它以数组的形式跟踪简单命令nl.

...would be correct, since it tracks the simple command nl as an array.

这篇关于Bash-字符串，命令和转义(哦，我的！)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！