使用sed，我通常通过搜索除分隔符以外的任何东西来实现非贪婪搜索，直到分隔符:

echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*\)/.*;\1;p'

输出:

http://www.suon.co.uk

这是:

don't output -n search, match pattern, replace and print s/<pattern>/<replace>/p use ; search command separator instead of / to make it easier to type so s;<pattern>;<replace>;p remember match between brackets \( ... \), later accessible with \1,\2... match http:// followed by anything in brackets [], [ab/] would mean either a or b or / first ^ in [] means not, so followed by anything but the thing in the [] so [^/] means anything except / character * is to repeat previous group so [^/]* means characters except /. so far sed -n 's;\(http://[^/]*\) means search and remember http://followed by any characters except / and remember what you've found we want to search untill the end of domain so stop on the next / so add another / at the end: sed -n 's;\(http://[^/]*\)/' but we want to match the rest of the line after the domain so add .* now the match remembered in group 1 (\1) is the domain so replace matched line with stuff saved in group \1 and print: sed -n 's;\(http://[^/]*\)/.*;\1;p'

如果你想在域名后面加上反斜杠，那么在组中再加一个反斜杠来记住:

echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*/\).*;\1;p'

输出:

http://www.suon.co.uk/

因为您特别指出您正在尝试使用sed(而不是perl、cut等)，所以请尝试分组。这就避免了非贪婪标识符可能无法被识别。第一组是协议(即协议)。'http://'， 'https://'， 'tcp://'，等等)。第二组是域:

echo "http://www.suon.co.uk/product/1/7/3/" | sed "s|^\(.*//\)\([^/]*\).*$|\1\2|"

如果您不熟悉分组，请从这里开始。

另一种方法，不使用正则表达式，是使用字段/分隔符方法，如

string="http://www.suepearson.co.uk/product/174/71/3816/"
echo $string | awk -F"/" '{print $1,$2,$3}' OFS="/"

另一个sed版本:

sed 's|/[:alnum:].*||' file.txt

它匹配/后面跟着一个字母数字字符(所以不是另一个正斜杠)以及其余字符，直到行尾。之后，它将什么也没有替换掉。删除它。)

以下是你可以用两步方法和awk完成的事情:

A=http://www.suepearson.co.uk/product/174/71/3816/  
echo $A|awk '  
{  
  var=gensub(///,"||",3,$0) ;  
  sub(/\|\|.*/,"",var);  
  print var  
}'  

输出: http://www.suepearson.co.uk

希望有帮助!

基本的和扩展的Posix/GNU regex都不能识别非贪婪的量词;你需要稍后的正则表达式。幸运的是，这个上下文的Perl regex非常容易获得:

perl -pe 's|(http://.*?/).*|\1|'