如何在linux中处理具有固定宽度列的文件

Question

I would like to process below file:

01234000000000000000000+000000000000000000+
02586000000000000000000+000000000000000000-
12345000000000000000000+000000000000000000-
12122000000000000000000+000000000000000000+

I want to convert above file to:

01234,000000000000000000+,000000000000000000+
02586,000000000000000000+,000000000000000000-
12345,000000000000000000+,000000000000000000-
12122,000000000000000000+,000000000000000000+

Input file have fixed width columns 5,19,19 respectively.

I would like to solve using linux command.

I tried below command, but it is not working :(

awk 'BEGIN{FIELDWIDTHS="5 19 19";OFS=",";}{$1="$1,$2,$3"}' data.txt

Executing above command on ubuntu 14.04 LTS desktop OS, the output was nothing(blank).

Answer 1

Your attempt was quite close, although you forgot to {print} :

awk 'BEGIN{FIELDWIDTHS="5 19 19";OFS=","}{$1=$1}1' file

{$1=$1} assigns the first field to itself, which is enough to make awk "touch" each record. I've used the shorthand 1 , which is the shortest true condition. The default action is {print} .

Note that FIELDWIDTHS is a GNU awk extension, so if you're using a different version, you will have to go with a different approach. For example:

awk 'BEGIN{OFS=","}{print substr($0,1,5),substr($0,6,19),substr($0,25)}' file

Answer 2

$ sed -r 's/(.{5})(.{19})/\1,\2,/' file
01234,000000000000000000+,000000000000000000+
02586,000000000000000000+,000000000000000000-
12345,000000000000000000+,000000000000000000-
12122,000000000000000000+,000000000000000000+

Answer 3

that would be very easy:

sed -n 's/\(.\{5\}\)\(.\{19\}\)\(.\{19\}\)/\1,\2,\3/p' your_file

what it does, is to capture each line by 5, 19, 19 then print it out with , in between.

$ echo 01234000000000000000000+000000000000000000+ | sed -n 's/\(.\{5\}\)\(.\{19\}\)\(.\{19\}\)/\1,\2,\3/p'
01234,000000000000000000+,000000000000000000+

Answer 4

Perl救援：

perl -pe 'for $p (5, 25) { substr $_, $p, 0, "," }' data.txt

Answer 5

this is suitable task for cut as well

$ cut --output-delimiter=',' -c1-5,6-24,25- data.txt
01234,000000000000000000+,000000000000000000+
02586,000000000000000000+,000000000000000000-
12345,000000000000000000+,000000000000000000-
12122,000000000000000000+,000000000000000000+

• --output-delimiter=',' specify output field separator
• -c to select specified character(s)
• 1-5 first field
• 6-24 second field
• 25- rest of the line

Answer 6

awk '{sub(/.0/,",0")sub(/+/,"+,")}1' file

0123,000000000000000000+,000000000000000000+
0258,000000000000000000+,000000000000000000-
1234,000000000000000000+,000000000000000000-
1212,000000000000000000+,000000000000000000+