Shell Programming and Scripting

Hi,
I have to separate a pattern from a string. Here are the details:

Pattern: number1.number2.number3.number4.number5
where, number 3,4,5 can be sinngle or double digits
number 1,2 are single digits.

Examples:
7.5.1.10.10
6.0.11.6.34

Now here is the sample string:

Code:

<MyName platform="HP-UX" product="som" pattern="7.5.1.8.9"/>

And here is the sed command that I am using:

Code:

sed -n 's/\(.*\)\(.\)\.\(.\)\.\(.\{1,2\}\)\.\(.\{1,2\}\)\.\(.\{1,2\}\)\(.*\)/\2.\3.\4.\5.\6/p'

The problem in the above case is if number 5 is double digit, it works fine..
If number 5 is single digit, The quotes : " is also printed..
I am confused why..
Please guide..
The above sed command on above string will give:
7.5.1.8.9"
where as 7.5.1.8.9 was expected.

Why is \(.\{1,2\}\) always double digits.. when i have specified that the character can be one or two?


Do I have to specify something like this?
\(\[0-9\]\{1,2\}\)
The following command dosent work either:

Code:

sed -n 's/\(.*\)\(\[0-9\]\)\.\(\[0-9\]\)\.\(\[0-9\]\{1,2\}\)\.\(\[0-9\]\{1,2\}\)\.\(\[0-9\]\{1,2\}\)\(.*\)/\2.\3.\4.\5.\6/p'

What am I missing?
Please guide..
Thanks..

Try this, the command isolates the part between .*pattern= and the next double quote:

Code:

sed 's/.*pattern="\([^"]*\)".*/\1/'

Regards

Hi,
I am sorry but I did not get your point.
Another thing I want to ask is that, what if there is no double quote, and has some other character?
I just want the sed to pick up number4,5,6 as one or two digit numbers.
What do you suggest in this case?

Code:

$ echo '<MyName platform="HP-UX" product="som" pattern=7.5.1.8.9/>' | sed 's#.*pattern=[^0-9]*\([0-9.][0-9.]*\).*#\1#'
7.5.1.8.9

$ echo '<MyName platform="HP-UX" product="som" pattern="7.5.1.8.9"/>' | sed 's#.*pattern=[^0-9]*\([0-9.][0-9.]*\).*#\1#'
7.5.1.8.9

Brilliant..
Can you please explain this:

Code:

[^0-9]*\([0-9.][0-9.]*\)

Thanks a lot

---------- Post updated at 06:22 PM ---------- Previous update was at 06:13 PM ----------

Hi Vgersh,
In the second example you gave, how does it ignore the double quotes?
pattern=[^0-9]
This does not contain any double quotes..
Can you please explain?

Thanks..

'pattern=[^0-9]*\([0-9.][0-9.]*\)'

'pattern=' followed by any character BUT a number ( [^0-9]) - repeated 0 or more time ( * ).
Followed by a 'capture' \(...\). The 'capture' consists of any number OR a '.' repeated 1 or more times.
Anything else not matched by a capture, gets eaten up by the trailing greedy '.*'.