SED:在同一行上匹配2种模式 [英] SED: Matching on 2 patterns on the same line

问题描述

我想使用sed删除一行，如果它与同一行中的2个正则表达式匹配. EG该行以/*开头，以*/结束(注释).以下脚本将完成大部分操作. sed -e'/^/*/d'-e'/*/$/d'文件名 该脚本将删除所有以*开头和以*/结尾的行.我希望它仅在满足两个条件而不是一个条件时才删除该行.

Hi I want to delete a line using sed if it matches 2 regular expressions in the same line. EG the line starts with /* and end with */ (comment). The following script will do most of that. sed -e '/^/*/ d' -e '/*/$/ d' filename This script will remove all lines that start with * and end with */. I want it to remove the line only if is meets both criteria not one.

推荐答案

尝试

sed '/^\/\*.*\*\/$/ d' filename

这里的关键是，您可以将两个正则表达式模式组合为一个，只需将它们与.*连接即可，该模式匹配任意数量的任何字符".当然，这将强制在两者之间进行排序.为了使此特定模式匹配，第一个模式^\/\*必须出现在第二个模式\*\/$之前.

The key here is that you can sorta combine two regex patterns into one, simply by connecting them with .*, which matches "any number of any character". Of course, this enforces an ordering between the two. The first pattern ^\/\* must occur before the second one \*\/$ for this particular pattern to match.

此外，由于*在正则表达式中具有特殊含义，因此请务必转义其斜线，就像必须转义斜线一样.

Also, since * has special meaning in regex, be sure to escape your astrices, just as you have to escape your slashes.

这篇关于SED:在同一行上匹配2种模式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！