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# 1  
Old 07-16-2010
Question on regex with * and .
I have a basic question regarding * and . while using regex:

Code:
# echo 3 | grep ^[0123456789]*$
3

I think I understood why it outputs "3" here (because '*' matches zero or more of the previous character) but I don't understand the output of the following command:

Code:
# echo 3 | grep ^[0123456789].$
#

I thought I would see "3" here as well since '.' matches "one or more of previous characters". no?
Last edited by Scott; 07-19-2010 at 07:57 AM.. Reason: Code tags
# 2  
Old 07-16-2010
Quote:
Originally Posted by mirage
I have a basic question regarding * and . while using regex:

# echo 3 | grep ^[0123456789]*$
3

I think I understood why it outputs "3" here (because '*' matches zero or more of the previous character) but I don't understand the output of the following command:

# echo 3 | grep ^[0123456789].$
#

I thought I would see "3" here as well since '.' matches "one or more of previous characters". no?
Yes, but after the first character there is no more character to match.
Code:
^[0123456789].$

means 1 character begins with one of the characters between the square brackets and one single character at the end of the line.
This User Gave Thanks to Franklin52 For This Post:
mirage
# 3  
Old 07-16-2010
understood. Now I realize I also misinterpreted the output of `echo 3 | grep ^[0123456789]*$` even though it showed the result I expected.

I should be reading it as:
"Match characters
^ - starting with
[0123456789] - one of the digits from 0 thro' 9 and
*$ - zero or more of digits from 0 thro' 9 at the end. "

Thus it matches 3, 33, 34, 345 etc.

Thanks a lot!
# 4  
Old 07-16-2010
Quote:
Originally Posted by mirage
'.' matches "one or more of previous characters"
"." (dot) specifies "any single character"
"+" (plus) specifies "one or more of previous"

---------- Post updated at 01:04 AM ---------- Previous update was at 12:59 AM ----------
Quote:
Originally Posted by mirage
I should be reading it as:
"Match characters
^ - starting with
[0123456789] - one of the digits from 0 thro' 9 and
*$ - zero (none) or more of digits from 0 thro' 9 at the end. "

Thus it matches 3, 33, 34, 345 etc.

You should read as:

"Match characters
^ - starting with
[0123456789]* - zero or more digits from 0 thro' 9 and
$ - end."

---------- Post updated at 01:05 AM ---------- Previous update was at 01:04 AM ----------
Quote:
Originally Posted by mirage
I should be reading it as:
"Match characters
^ - starting with
[0123456789] - one of the digits from 0 thro' 9 and
*$ - zero or more of digits from 0 thro' 9 at the end. "

Thus it matches 3, 33, 34, 345 etc.

You should read as:

"Match characters
^ - starting with
[0123456789]* - zero (none) or more digits from 0 thro' 9 and
$ - end."
# 5  
Old 07-19-2010
Code:
# echo 3 | grep ^[123456789]+$
# echo 34 | grep ^[123456789]+$
#

but why not the above print "3" and "34" (respectively) if '+' should match one or more of the previous characters?
Last edited by Scott; 07-19-2010 at 07:56 AM.. Reason: Code tags
# 6  
Old 07-19-2010
Quote:
Originally Posted by mirage
Code:
# echo 3 | grep ^[123456789]+$
# echo 34 | grep ^[123456789]+$

Code:
[house@leonov] echo "3" | grep -E '^[123456789]+$'
3
[house@leonov] echo "34" | grep -E '^[123456789]+$'
34

# 7  
Old 07-19-2010
Hi.

What is printed?

Further to what dr.house said, + loses it's special meaning in basic regular expressions.

Try escaping it:

Code:
$ echo 34 | grep ^[123456789]\\+$
34

(or use extended regular expressions - egrep or grep -E)
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