sed：仅删除引号内的所有非字母数字字符

Question

假设我有这样一个字符串：

Output:   
I have some-non-alphanumeric % characters remain here, I "also, have_+ some & .here"

我想仅删除引号内的非字母数字字符逗号、句点或空格除外:

Desired Output:    
I have some-non-alphanumeric % characters remain here, I "also, have some  .here"

我尝试了以下 sed 命令匹配字符串并删除引号内的内容，但它会删除引号内的所有内容，包括引号：

sed '/characters/ s/\("[^"]*\)\([^a-zA-Z0-9\,\. ]\)\([^"]*"\)//g'

感谢任何帮助，最好使用 sed，以获得所需的输出。提前致谢！

Answer 1

Sed 不是执行此操作的正确工具。这是通过 Perl 实现的。

perl -pe 's/[^a-zA-Z0-9,.\s"](?!(?:"[^"]*"|[^"])*$)//g' file

示例：

$ echo 'I have some-non-alphanumeric % characters remain here, I "also, have_+ some & .here"' | perl -pe 's/[^a-zA-Z0-9,.\s"](?!(?:"[^"]*"|[^"])*$)//g'
I have some-non-alphanumeric % characters remain here, I "also, have some  .here"

Regex Demo

Answer 2

您需要多次重复替换以删除所有非字母数字字符。在 sed 中执行这样的循环需要标签并使用 b 和 t 命令：

sed '
# If the line contains /characters/, just to label repremove
/characters/ b repremove
# else, jump to end of script
b
# labels are introduced with colons
:repremove
# This s command says: find a quote mark and some stuff we do not want
# to remove, then some stuff we do want to remove, then the rest until
# a quote mark again. Replace it with the two things we did not want to
# remove
s/\("[a-zA-Z0-9,. ]*\)[^"a-zA-Z0-9,. ][^"a-zA-Z0-9,. ]*\([^"]*"\)//
# The t command repeats the loop until we have gotten everything
t repremove
'

（即使没有 [^"a-zA-Z0-9,. ]* 也能正常工作，但在连续包含许多非字母数字字符的行上会变慢）

虽然另一个答案是正确的，因为在 perl 中这样做要容易得多。

sed：仅删除引号内的所有非字母数字字符

sed: remove all non-alphanumeric characters inside quotations only

regex

bash

sed

alphanumeric

non-alphanumeric