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Combining gsub and substr in awk

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# 1  
Old 11-17-2010
Combining gsub and substr in awk
I have data a.txt:
Quote:
123456701234567890abcdefghijklmnopqrsccyy-mm-dd0000000001Pccyy-mm-ddABCDEFGH
123456701234567890abcdefghijklmnopqrsccyy-mm-dd0000000009Pccyy-mm-ddABCDEFGH
123456701234567890abcdefghijklmnopqrsccyy-mm-dd0000000050Pccyy-mm-ddABCDEFGH
I want to reformat file to look like this:

Quote:
1234567|01234567890abcdefghijklmnopqrs|1|ccyymmdd|
1234567|01234567890abcdefghijklmnopqrs|9|ccyymmdd|
1234567|01234567890abcdefghijklmnopqrs|50|ccyymmdd|
basically with the 3rd columns having leading zeros removed.

My code a.awk:

Code:
awk '{ v=substr($0, 48,10); print substr($0,1,7)"|"substr($0,8,30)"|"gsub("0*","
",v)"|"substr($0,59,4)substr($0,64,2)substr($0,67,2)"|"}' a.txt

always returns '2' in the third column and I'm wondering why and how i can strip out the leading zeroes while awk'ing:

Quote:
1234567|01234567890abcdefghijklmnopqrs|2|ccyymmdd|
1234567|01234567890abcdefghijklmnopqrs|2|ccyymmdd|
1234567|01234567890abcdefghijklmnopqrs|2|ccyymmdd|
I'm trying to avoid doing another while read file just to make that substitution.
I would use 'bc" if it is possible.

Thanks.
Gianni
# 2  
Old 11-17-2010
gsub dosn't return a string result, it updates variable in place so do gsub on your variable before the print.

Code:
awk '{ v=substr($0, 48,10); gsub("^0*","",v);
print substr($0,1,7)"|"substr($0,8,30)"|"v"|"substr($0,59,4)substr($0,64,2)substr($0,67,2)"|"}' a.txt

This User Gave Thanks to Chubler_XL For This Post:
giannicello
# 3  
Old 11-17-2010
Code:
$ awk '{ v=substr($0, 48,10); gsub("^0*","",v); print substr($0,1,7)"|"substr($0,8,30)"|" v "|"substr($0,59,4)substr($0,64,2)substr($0,67,2)"|"}' file 
1234567|01234567890abcdefghijklmnopqrs|1|ccyymmdd|
1234567|01234567890abcdefghijklmnopqrs|9|ccyymmdd|
1234567|01234567890abcdefghijklmnopqrs|50|ccyymmdd|
$ awk '{ print substr($0,1,7)"|"substr($0,8,30)"|" int(substr($0, 48,10)) "|"substr($0,59,4)substr($0,64,2)substr($0,67,2)"|"}' file
1234567|01234567890abcdefghijklmnopqrs|1|ccyymmdd|
1234567|01234567890abcdefghijklmnopqrs|9|ccyymmdd|
1234567|01234567890abcdefghijklmnopqrs|50|ccyymmdd|

This User Gave Thanks to anbu23 For This Post:
giannicello
# 4  
Old 11-17-2010
Code:
sed 's:0:|0:;s:ccyy-mm-dd0*:|:;s:P:|:;s:-::g;s:[A-Z]*$:|:' a.txt

Code:
$ cat a.txt
123456701234567890abcdefghijklmnopqrsccyy-mm-dd0000000001Pccyy-mm-ddABCDEFGH
123456701234567890abcdefghijklmnopqrsccyy-mm-dd0000000009Pccyy-mm-ddABCDEFGH
123456701234567890abcdefghijklmnopqrsccyy-mm-dd0000000050Pccyy-mm-ddABCDEFGH
$ sed 's:0:|0:;s:ccyy-mm-dd0*:|:;s:P:|:;s:-::g;s:[A-Z]*$:|:' a.txt
1234567|01234567890abcdefghijklmnopqrs|1|ccyymmdd|
1234567|01234567890abcdefghijklmnopqrs|9|ccyymmdd|
1234567|01234567890abcdefghijklmnopqrs|50|ccyymmdd|
$

Last edited by ctsgnb; 11-17-2010 at 03:45 PM..
This User Gave Thanks to ctsgnb For This Post:
giannicello
# 5  
Old 11-17-2010
Code:
awk '{ v=substr($0, 48,10)+0; print substr($0,1,7),substr($0,8,30),v,substr($0,59,4)substr($0,64,2)substr($0,67,2)OFS}' OFS='|' a.txt

Last edited by vgersh99; 11-17-2010 at 03:44 PM..
This User Gave Thanks to vgersh99 For This Post:
giannicello
# 6  
Old 11-17-2010
Wow. Thank you so much everyone!! I like the +0 solution for it's simplicity.

-Gianni
# 7  
Old 11-17-2010
Code:
sed 's/\(.\{7\}\)\(.\{30\}\).*00*\(.*\)P\(....\)-\(..\)-\(..\).*/\1|\2|\3|\4\5\6|/' file

GNU sed:
Code:
sed -r 's/(.{7})(.{30}).*00*(.*)P(....)-(..)-(..).*/\1|\2|\3|\4\5\6|/' file
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