84669 person learning
152542 person learning
20005 person learning
5487 person learning
7821 person learning
359900 person learning
3350 person learning
180660 person learning
48569 person learning
18603 person learning
40936 person learning
1549 person learning
1183 person learning
32909 person learning
业精于勤，荒于嬉;行成于思，毁于随。
This can be matched grep -o "ed2k[^"]*"
grep -o "ed2k[^"]*"
[root@localhost oldlog.md]# echo '你的地址' | grep -o "ed2k[^\"]*" ed2k://|file|%E7%A1%85%E8%B0%B7.Silicon.Valley.S01E01.%E4%B8%AD%E8%8B%B1%E5%AD%97%E5%B9%95.HDTVrip.1024X576.mkv|348759100|04dc44e4978ea2d4dc9bf41b08829a3e|h=rfe5smulijxlmz7mgtpmn2bt3vmofk56|/ ed2k://|file|silicon.valley.s01e01.720p.hdtv.x264-killers.mkv|843706966|3B4B08BD9C960B5B8F2372158F84B1AE|h=36632AROF43T63NSXFLIGSYEB4PNDEFN|/
It is recommended to use some HTML parser to extract the a tag separately for matching, which can avoid the problem of multiple a's in one line
By default grep doesn't support non-greedy modifiers, but you can use grep -P to use the Perl syntax.
How to do a non-greedy match in grep?
![[Web front-end] Node.js quick start](https://img.php.cn/upload/course/000/000/067/662b5d34ba7c0227.png)
This can be matched
grep -o "ed2k[^"]*"It is recommended to use some HTML parser to extract the a tag separately for matching, which can avoid the problem of multiple a's in one line
By default grep doesn't support non-greedy modifiers, but you can use grep -P to use the Perl syntax.
How to do a non-greedy match in grep?