如何取出一行中的一段字符 (麻烦网中人老师看一下)

xinkeren

文本aa内容为<iframe name="lyric" frameborder="0" width="310" height="234" src="get_lyric.php?music_url=china%2Fboy%2Fyuquan%2F03%2F01.xtl"></iframe>
<object width="310" height="0" classid="clsid:CFCDAA03-8BE4-11cf-B84B-0020AFBBCCFA" id="RPlayer"><ARAM NAME="NOLABELS" VALUE="FALSE"><param name="AUTOSTART" value="TRUE"><param name="LOOP" value="true"><param name="SRC" value="http://music.cbbn.net/newmusic/china/boy/yuquan/03/01.xtl"><ARAM NAME="NOLOGO" VALUE="false"><ARAM NAME="RESET" VALUE="true"><ARAM NAME="OPEN" VALUE="FALSE"><ARAM NAME="END" VALUE="TRUE"><param name="MAINTAINASPECT" value="0"><param name="CONTROLS" value="ImageWindow"><ARAM NAME="CONSOLE" VALUE="clip1"></object></div>


要取出http://music.cbbn.net/newmusic/china/boy/yuquan/03/01.xtl
我的本意是从aa.txt 中取出 grep '\<httpxtl\>' aa.txt>bb.txt 其中 bb.txt的内容为
http://music.cbbn.net/newmusic/china/boy/yuquan/03/01.xtl
用grep sed awk均为能成功真是惭愧

xyb

cat a.txt | sed -n '/.*\(http:\/\/music.cbbn.net.*xtl\).*/{s//\1/;p}'

r2007

1. grep -o '\<http.*xtl\>' aa.txt>bb.txt

复制代码

xyb

原帖由 "r2007" 发表：

grep -o '\<http.*xtl\>' aa.txt>bb.txt

简洁，好！

whn3375

1. grep -o '\<http.*xtl\>' aa.txt>bb.txt
   
2. grep: illegal option -- o
   
3. Usage: grep -hblcnsviw pattern file . . .
   
4. 
   
5. Solaris8
   

复制代码

r2007

我的是gnu grep，请参考。

1. r2007@www r2007 $ grep --version
   
2. grep (GNU grep) 2.5.1
   
3. 
   
4. Copyright 1988, 1992-1999, 2000, 2001 Free Software Foundation, Inc.
   
5. This is free software; see the source for copying conditions. There is NO
   
6. warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
   
7. 
   
8.        -o, --only-matching
   
9.               Show only the part of a matching line that matches PATTERN.
   

复制代码

whn3375

又是版本的问题！

1. grep --version
   
2. grep: illegal option -- -
   
3. grep: illegal option -- e
   
4. grep: illegal option -- r
   
5. grep: illegal option -- o
   
6. Usage: grep -hblcnsviw pattern file . . .
   

复制代码

至今为止还不知道版本是什莫。 郁闷！