Question

我试图给自己写一些方便的脚本，以便~~合法地放弃~~工作更有效率，这个问题突然出现了：

给定一个非常长的字符串$LONGEST_EVER_STRING和几个关键字字符串，如$A='foo bar'，$B='omg bbq'和$C='stack overflow'

如何在case语句中使用完全关键字匹配作为条件？

for word in $LONGEST_EVER_STRING; do
  case $word in
    any exact match in $A) do something ;;
    any exact match in $B) do something ;;
    any exact match in $C) do something ;;
    *)  do something else;;
  esac
done

我知道我可以这样写，但看起来真的很难看：

for word in $LONGEST_EVER_STRING; do
  if [[ -n $(echo $A | fgrep -w $word) ]]; then
    do something;
  elif [[ -n $(echo $B | fgrep -w $word) ]]; then
    do something;
  elif [[ -n $(echo $C | fgrep -w $word) ]]; then
    do something;
  else
    do something else;
  fi
done

有没有人有一个优雅的解决方案？非常感谢！

Answer 1

您可以使用函数在A，B，C变量中进行一点变换，然后：

shopt -s extglob
Ax="+(foo|bar)"
Bx="+(omg|bbq)"
Cx="+(stack|overflow)"
for word in $LONGEST_EVER_STRING; do
  case $word in
    $Ax) do something ;;
    $Bx) do something ;;
    $Cx) do something ;;
    *)  do something else;;
  esac
done

Answer 2

我只想为此定义一个函数。对于大型单词列表，它会比grep慢，但比多次启动grep要快。

##
# Success if the first arg is one of the later args.
has() {
  [[ $1 = $2 ]] || {
    [[ $3 ]] && has "$1" "${@:3}"
  }
}

$ has a b c && echo t || echo f
f
$ has a b c a d e f && echo t || echo f
t

Answer 3

/ etc / bashrc的“pathmunge”

的变体

for word in $LONGEST_EVER_STRING; do
    found_word=false
    for list in " $A " " $B " " $C "; do
        if [[ $list == *" $word "* ]]; then 
            found_word=true
            stuff with $list and $word
            break
        fi
    done
    $found_word || stuff when not found
done

我应该如何在case语句中使用完全关键字匹配作为条件？

3 个答案: