sed 's|(http:\/\/[^\/]+\/).*|\1|'

在sed中模拟惰性(非贪婪)量词

以及所有其他正则表达式口味!

Finding first occurrence of an expression: POSIX ERE (using -r option) Regex: (EXPRESSION).*|. Sed: sed -r ‍'s/(EXPRESSION).*|./\1/g' # Global `g` modifier should be on Example (finding first sequence of digits) Live demo: $ sed -r 's/([0-9]+).*|./\1/g' <<< 'foo 12 bar 34' 12 How does it work? This regex benefits from an alternation |. At each position engine tries to pick the longest match (this is a POSIX standard which is followed by couple of other engines as well) which means it goes with . until a match is found for ([0-9]+).*. But order is important too. Since global flag is set, engine tries to continue matching character by character up to the end of input string or our target. As soon as the first and only capturing group of left side of alternation is matched (EXPRESSION) rest of line is consumed immediately as well .*. We now hold our value in the first capturing group. POSIX BRE Regex: \(\(\(EXPRESSION\).*\)*.\)* Sed: sed 's/\(\(\(EXPRESSION\).*\)*.\)*/\3/' Example (finding first sequence of digits): $ sed 's/\(\(\([0-9]\{1,\}\).*\)*.\)*/\3/' <<< 'foo 12 bar 34' 12 This one is like ERE version but with no alternation involved. That's all. At each single position engine tries to match a digit. If it is found, other following digits are consumed and captured and the rest of line is matched immediately otherwise since * means more or zero it skips over second capturing group \(\([0-9]\{1,\}\).*\)* and arrives at a dot . to match a single character and this process continues. Finding first occurrence of a delimited expression: This approach will match the very first occurrence of a string that is delimited. We can call it a block of string. sed 's/\(END-DELIMITER-EXPRESSION\).*/\1/; \ s/\(\(START-DELIMITER-EXPRESSION.*\)*.\)*/\1/g' Input string: foobar start block #1 end barfoo start block #2 end -EDE: end -SDE: start $ sed 's/\(end\).*/\1/; s/\(\(start.*\)*.\)*/\1/g' Output: start block #1 end First regex \(end\).* matches and captures first end delimiter end and substitues all match with recent captured characters which is the end delimiter. At this stage our output is: foobar start block #1 end. Then the result is passed to second regex \(\(start.*\)*.\)* that is same as POSIX BRE version above. It matches a single character if start delimiter start is not matched otherwise it matches and captures the start delimiter and matches the rest of characters.

直接回答你的问题

使用方法#2(带分隔符的表达式)，你应该选择两个合适的表达式:

艾德:[^]\ / SDE: http:

用法:

$ sed 's/\([^:/]\/\).*/\1/g; s/\(\(http:.*\)*.\)*/\1/' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'

输出:

http://www.suepearson.co.uk/

注意:对于相同的分隔符，这将不起作用。

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="/"

您还应该考虑没有匹配界限的情况。你是否想输出这一行。如果不匹配，我这里的示例不会输出任何内容。

你需要前缀到第三个/，所以选择两次字符串的任何长度不包含/和后面的/，然后字符串的任何长度不包含/，然后匹配/后面的任何字符串，然后打印选择。这个想法适用于任何单个的char delims。

echo http://www.suepearson.co.uk/product/174/71/3816/ | \
  sed -nr 's,(([^/]*/){2}[^/]*)/.*,\1,p'

使用sed命令，您可以快速删除前缀或delim选择，如:

echo 'aaa @cee: { "foo":" @cee: " }' | \
  sed -r 't x;s/ @cee: /\n/;D;:x'

这比一次吃焦肉快多了。

如果之前匹配成功，跳转到标签。在第一道线/之前加\n。移除到第一个\n。如果添加了\n，则跳转到结束并打印。

如果有开始和结束delim，很容易删除结束delim，直到你到达你想要的第n -2个元素，然后做D技巧，在结束delim后删除，如果不匹配跳转到删除，在开始delim和打印之前删除。这仅在开始/结束分隔成对出现时有效。

echo 'foobar start block #1 end barfoo start block #2 end bazfoo start block #3 end goo start block #4 end faa' | \
  sed -r 't x;s/end//;s/end/\n/;D;:x;s/(end).*/\1/;T y;s/.*(start)/\1/;p;:y;d'

sed 's|(http:\/\/[^\/]+\/).*|\1|'

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|'