The string is setup like so:
href="PART I WANT TO EXTRACT">[link]
views:
270answers:
4
A:
grep -o "PART I WANT TO EXTRACT" foo
Edit: "PART I WANT TO EXTRACT" can be a regex, i.e.:
grep -o http://[a-z/.]* foo
ctd
2009-11-09 22:11:09
I'm using:> `grep -o 'http://[a-z/.A-Z0-9]*">\[link\]'`but it's giving me:> `http://img18.imageshack.us/img18/8079/69903601.jpg">[link]`I don't want the stuff on the end.
Josh
2009-11-09 22:32:06
Well, you could grep to find, and then use grep -o to just get the expression: >grep 'http://[a-z/.A-Z0-9]*">\[link\]' | grep -o 'http://[a-z/.A-Z0-9]*
ctd
2009-11-10 00:27:27
+2
A:
use awk
$ echo "href="PART I WANT TO EXTRACT">[link]" | awk -F""" '{print $2}'
PART I WANT TO EXTRACT
Or using shell itself
$ a="href="PART I WANT TO EXTRACT">[link]"
$ a=${a//"/}
$ echo ${a/&*/}
PART I WANT TO EXTRACT
ghostdog74
2009-11-09 23:34:30
A:
Here's another way in Bash:
$ string="href="PART I WANT TO EXTRACT">[link]"
$ entity="""
$ string=${string#*${entity}*}
$ string=${string%*${entity}*}
$ echo $string
PART I WANT TO EXTRACT
This illustrates two features: Remove matching prefix/suffix pattern and the use of a variable to hold the pattern (you could use a literal instead).
Dennis Williamson
2009-11-10 01:11:15