不幸的是，如前所述，sed不支持这一点。 为了克服这个问题，我建议使用次优方法(实际上甚至更好)，使用类似vim sed的功能。

在.bash-profile中定义

vimdo() { vim $2 --not-a-term -c "$1"  -es +"w >> /dev/stdout" -cq!  ; }

这将创建无头vim来执行命令。

现在你可以这样做:

回声路径美元| vimdo“% s_ \ c: [a-zA-Z0-9 \ \ /] python (a-zA-Z0-9 \ {-} \\/]\{-}:__ g”,

过滤掉$PATH中的python。

使用-在vimdo中从管道中输入。

而大多数语法是相同的。Vim具有更高级的特性，并且使用\{-}是非贪婪匹配的标准。参见帮助regexp。

在这种特定情况下，您可以在不使用非贪婪正则表达式的情况下完成工作。

试试这个非贪婪的正则表达式[^/]*来代替。*?：

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

echo "/home/one/two/three/myfile.txt" | sed 's|\(.*\)/.*|\1|'

别麻烦了，我在另一个论坛上看到的:)

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。

Sed当然有它的位置，但这不是其中之一!

正如迪伊所指出的:用切就可以了。在这种情况下，它要简单得多，也安全得多。下面是一个使用Bash语法从URL中提取各种组件的示例:

url="http://www.suepearson.co.uk/product/174/71/3816/"

protocol=$(echo "$url" | cut -d':' -f1)
host=$(echo "$url" | cut -d'/' -f3)
urlhost=$(echo "$url" | cut -d'/' -f1-3)
urlpath=$(echo "$url" | cut -d'/' -f4-)

给你:

protocol = "http"
host = "www.suepearson.co.uk"
urlhost = "http://www.suepearson.co.uk"
urlpath = "product/174/71/3816/"

正如你所看到的，这是一个更加灵活的方法。

(全部归功于Dee)

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="/"