几个表达式中的 Perl 和 Sed 字符串替换

Question

我想以非贪婪匹配方式进行字符串替换

• 删除所有前导和尾随破折号、撇号（当这些符号出现在单词中间时，必须保留）

• 将多个空格转化为1个空格

例子：

--ONE   Tw'o--   -333-   -'FO-UR'

必须成为

ONE Tw'o 333 FO-UR

我无法得到确切的结果。你能帮我更正下面的 perl 和 sed 语法吗？

$ echo "--ONE   Tw'o--   -333-   -'FO-UR'" \
  | perl -pe "s/[-']+(.+?)/\1/g"           \
  | perl -pe "s/(.+?)[-']+/\1/g"           \
  | perl -pe "s/\s+/ /g"

Result (perl): "ONE Two 333 FOUR"

$ echo "--ONE   Tw'o--   -333-   -'FO-UR'" \
  | sed -r -e "s/[-']+(.+?)/\1/g"          \
    -e "s/(.+)[-']+/\1/g"                  \
    -e "s/\s+/ /g"

Result (sed): "ONE Tw'o-- -333- -'FO-UR"

Answer 1

这里是 perl 版本：

echo "--ONE   Tw'o--   -333-   -'FO-UR'" | perl -ne "s|-'||g; s|'-||g; s|^'||; s|'$||; s|^-+||; s|-+$||; s|-+\s+| |g; s|\s+-+| |g; s|\s+| |g; s|\s+$||; print;"

ONE Tw'o 333 FO-UR

sed版本基本一致：

echo "--ONE   Tw'o--   -333-   -'FO-UR'" | sed -r -e "s|-'||g; s|'-||g; s|^'||; s|'$||; s|^-+||; s|-+$||; s|-+\s+| |g; s|\s+-+| |g; s|\s+| |g; s|\s+$||;"

ONE Tw'o 333 FO-UR

所用正则表达式的注释：

s|-'||g;     # Remove dash followed by quote everywhere
s|'-||g;     # Remove quote followed by dash everywhere
s|^'||;      # Remove leading quote
s|'$||;      # Remove trailing quote
s|^-+||;     # Remove leading dash characters
s|-+$||;     # Remove trailing dash characters
s|-+\s+| |g; # Replace dash characters followed by whitespace with 1 space everywhere
s|\s+-+| |g; # Replace whitespace followed by dash characters with 1 space everywhere
s|\s+| |g;   # Replace multiple spaces with 1 space
s|\s+$||;    # Remove trailing spaces

Answer 2

在perl 中使用环视很容易：

s='"asd,f",,,"as,df","asdf"asdf"'
perl -pe 's/(?<!\w)-|-(?!\w)//g' <<< "$s"
ONE Tw'o 333 'FO-UR'

(?<!\w)- # Lookbehind meaning match - if not preceded by a word character
|        # regex alternation
(?!\w)-  # Lookahead meaning match - if not followed by a word character