正则表达式使用字符拆分并用不同的字符串替换第一部分
Regex split using a character and replace the first part with a different string
我有如下字符串列表:
Austria: ac.at
Bangladesh: ac.bd
Belgium: ac.be
China: ac.cn
Cyprus: ac.cy
India: ac.in
我需要得到如下列表:
$Resolved URL$ LIKE "*ac.at/*"
$Resolved URL$ LIKE "*ac.bd/*"
$Resolved URL$ LIKE "*ac.be/*"
$Resolved URL$ LIKE "*ac.cn/*"
$Resolved URL$ LIKE "*ac.cy/*"
$Resolved URL$ LIKE "*ac.id/*"
使用正则表达式替换执行此操作的最佳方法是什么?
您可以在正则表达式模式下尝试以下查找和替换:
Find: ^.*?(\S+)$
Replace: $Resolved URL$ LIKE "*/*"
下面是对正则表达式的解释:
^ from the start of the string
.*? consume (but do not capture) all content
(\S+) up until the final collection of non word characters and capture them
$ end of the input
只需匹配您希望保留的五个字符并围绕它构建替换字符串。您可以使用以下正则表达式。
[a-z]{2}\.[a-z]{2}
替换字符串是
$Resolved URL$ LIKE "*[=11=]/*"
其中 [=12=]
是匹配的字符串(例如,"ac.at"
)。
我有如下字符串列表:
Austria: ac.at
Bangladesh: ac.bd
Belgium: ac.be
China: ac.cn
Cyprus: ac.cy
India: ac.in
我需要得到如下列表:
$Resolved URL$ LIKE "*ac.at/*"
$Resolved URL$ LIKE "*ac.bd/*"
$Resolved URL$ LIKE "*ac.be/*"
$Resolved URL$ LIKE "*ac.cn/*"
$Resolved URL$ LIKE "*ac.cy/*"
$Resolved URL$ LIKE "*ac.id/*"
使用正则表达式替换执行此操作的最佳方法是什么?
您可以在正则表达式模式下尝试以下查找和替换:
Find: ^.*?(\S+)$
Replace: $Resolved URL$ LIKE "*/*"
下面是对正则表达式的解释:
^ from the start of the string
.*? consume (but do not capture) all content
(\S+) up until the final collection of non word characters and capture them
$ end of the input
只需匹配您希望保留的五个字符并围绕它构建替换字符串。您可以使用以下正则表达式。
[a-z]{2}\.[a-z]{2}
替换字符串是
$Resolved URL$ LIKE "*[=11=]/*"
其中 [=12=]
是匹配的字符串(例如,"ac.at"
)。