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Display lines not starts with #

 
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# 1  
Old 03-12-2010
Display lines not starts with #
hiiiii

$ grep ^"#" $file

Will give the lines , which starts with # .And I wanna get the lines which are not starting with #.
How to implement that.

Thanking you
KrishSmilie
# 2  
Old 03-12-2010
Code:
grep ^[^#] file

You try this.
Here the ^ inside the character class is similar to neglecting the # character.

---------- Post updated at 10:59 AM ---------- Previous update was at 10:54 AM ----------

Using sed you can also achieve this as following way.
Code:
sed '/^#/d' lines

Last edited by thillai_selvan; 03-12-2010 at 06:15 AM..
This User Gave Thanks to thillai_selvan For This Post:
figaro
# 3  
Old 03-12-2010
Code:
grep -v ^# file

# 4  
Old 03-12-2010
If you put the ^ symbol inside the character class, it will not negate the pattern. It means , match the patter starting with. So the following code is not at all work.

Code:
grep [^#] file

Instead of that you should use the following code
Code:
grep ^[^#] file

Here the symbol ^ outside of the character class negate the pattern. The symbol ^ Inside character class represent the starting.
# 5  
Old 03-12-2010
Actually, I think it should be:
Code:
grep "^[^#]*" file

otherwise you will filter out empty lines too
# 6  
Old 03-12-2010
Quote:
Originally Posted by Scrutinizer
Actually, I think it should be:
Code:
grep "^[^#]*" file

otherwise you will filter out empty lines too

No this is totally wrong.
It will match the lines that begins with #.
* matches the zero or more occurrence of the previous character.
So it will take it as "other than beginning with hash" it will consider as zero occurrence and will
match the lines that begins with #.
Last edited by thillai_selvan; 03-12-2010 at 06:59 AM..
# 7  
Old 03-12-2010
Here's a comparison of the various methods proposed so far, with line numbers where possible:
Code:
$ awk '{print NR,$0}' test.txt # Not a solution, but to show the original file
1 This is a line
2 This is another line
3
4 The previous line is empty
5
6 The previous line contains whitespaces
7 # This shouldn't show
8 This should show again
$ grep -n ^[^#] test.txt
1:This is a line
2:This is another line
4:The previous line is empty
5:
6:The previous line contains whitespaces
8:This should show again
$ sed '/^#/d' test.txt
This is a line
This is another line

The previous line is empty

The previous line contains whitespaces
This should show again
$ grep -nv ^# test.txt
1:This is a line
2:This is another line
3:
4:The previous line is empty
5:
6:The previous line contains whitespaces
8:This should show again
$ grep -n "^[^#]*" test.txt
1:This is a line
2:This is another line
3:
4:The previous line is empty
5:
6:The previous line contains whitespaces
7:# This shouldn't show
8:This should show again

 
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