regex - 连字符和下划线在 sed 中不兼容

Question

我很难让 sed 识别其模式字符串中的连字符和下划线。

有谁知道为什么

[a-z|A-Z|0-9|\-|_]

在以下示例中的工作方式如下

[a-z|A-Z|0-9|_]

?

$  cat /tmp/sed_undescore_hypen
lkjdaslf lkjlsadjfl dfpasdiuy service-type = service-1; jaldkfjlasdjflk address = address1; kldjfladsf
lkjdaslf lkjlsadjfl dfasdf  service-type = service_1; jaldkfjlasdjflk address = address1; kldjfladsf

$  sed 's/.*\(service-type = [a-z|A-Z|0-9|\-|_]*\);.*\(address = .*\);.*/\1    \2/g' /tmp/sed_undescore_hypen
lkjdaslf lkjlsadjfl dfpasdiuy service-type = service-1; jaldkfjlasdjflk address = address1; kldjfladsf
service-type = service_1    address = address1

$  sed 's/.*\(service-type = [a-z|A-Z|0-9|\-]*\);.*\(address = .*\);.*/\1    \2/g' /tmp/sed_undescore_hypen
service-type = service-1    address = address1
lkjdaslf lkjlsadjfl dfasdf  service-type = service_1; jaldkfjlasdjflk address = address1; kldjfladsf

$  sed 's/.*\(service-type = [a-z|A-Z|0-9|_]*\);.*\(address = .*\);.*/\1    \2/g' /tmp/sed_undescore_hypen
lkjdaslf lkjlsadjfl dfpasdiuy service-type = service-1; jaldkfjlasdjflk address = address1; kldjfladsf
service-type = service_1    address = address1

Answer 1

如前所述，您不需要任何东西来分隔括号表达式中的范围。所要做的就是添加 |到表达式匹配的字符。
然后，要添加连字符，您可以将其作为表达式中的第一个或最后一个字符:

[a-zA-Z0-9_-]

最后，范围如 a-z不一定意味着 abcd...xyz ，取决于您的语言环境。您可以改用 POSIX 字符类:

[[:alnum:]_-]

哪里 [:alnum:]对应于您的语言环境的所有字母数字字符。在 C locale，对应于 0-9A-Za-z .