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sed or awk - removing part of line?

 
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# 8  
Old 05-05-2011
Quote:
Originally Posted by vgersh99
Code:
sed -n '/ 603-/s/ 603-/ /p' myFile | sort


this works thanks!
# 9  
Old 05-05-2011
Quote:
Originally Posted by alis
This works except that the column with the area code is $3 not $2 so when I substituted $2 for $3 - the command works !!
Trying to understand this tho...I know $3=substr means substitution in column 3 but where did the 5 come from? please explain?
How many characters are in 603-?
# 10  
Old 05-05-2011
Quote:
Originally Posted by vgersh99
How many characters are in 603-?
I see 4..counting a space 5 ..why would you count the space..there is no space in the original file after -
Sorry I am a newbie.
# 11  
Old 05-05-2011
Quote:
Originally Posted by alis
I see 4..counting a space 5 ..why would you count the space..there is no space in the original file after -
Sorry I am a newbie.
Quoting the strings in the forums can get confusing - let me try it again.
How many characters do you count in quoted string ' 603-' ?

There's a space AFTER the first name and the AREA CODE.
You want to remove the area code. How would you distinguish an area code '603' from the '603' appearing in the rest of the phone number?

Or alternatively:
Code:
nawk '$2~/^603-/&& sub("603-","",$2)' myFile | sort

Last edited by vgersh99; 05-05-2011 at 06:33 PM..
# 12  
Old 05-06-2011
@vgersh99

Here is a caveat for you and for those readers that are not already aware of it (you can give it a try) :

your statement
Code:
# sed -n '/ 603-/s/ 603-/ /p' infile

can be written
Code:
# sed -n '/ 603-/s// /p' infile

because the empty // will then refer to the previous match Smilie

You will notice that the following does NOT work ... contrary to what we could have thought:
Code:
# sed -n '/ 603-/s/&/ /p' infile

Last edited by ctsgnb; 05-06-2011 at 05:52 AM..
 
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