Replace the line with specific pattern

Shell Programming and Scripting
1

Hello All

I'm trying to change one string from a file contening this patern:
xxxx-xxxx

4 numbers - end 4 other numbers
This is a sample of the file:

LDR  00679 am a2200205   4500
=001  3617
=008  030219s2000\\\\xxx|||||\||||\00|\0\spa\d
=020  \\$a0211-1942
=041  \\$aCastell�
=093  \\$a19/02/2003
=095  \\$a12/11/2001
=100  \\$aGorrotxategi Anieto, Xabier$91930
=245  \\$aArte paleol�tico parietal de Bizcaia

as you see there is a line :=020 \\$a0211-1942
So I would like to detect deep in a big file with a lot of records like this. and every row with xxxx-xxxx replace the code =020 by =022

I know that sed can do this but I have no idea how... Some one can help ??

Thanks a lot

2 
 sed -i.bak 's:=020 \(\\\\$a[0-9]\{4\}-[0-9]\{4\}\):=022 \1:' file
3

Hi
thanks for this rapid answer... Whooaw
So I guess that the way to grab the xxxx-xxxx patern is:

(\\\\$a[0-9]\{4\}-[0-9]\{4\}\)

The start is parentesis write?
I do not anderstand the \\\\ before $a
Then I anderstand this:
[0-9]\{4\}-[0-9]\{4\
it means.. get numeric numbers length 4... isn't it?

Thanks again

---------- Post updated at 03:50 AM ---------- Previous update was at 03:43 AM ----------

Hello again

and...if I want to check de print before replace?
May this sintaxe ok?

	sed -p'\(\\\\$a[0-9]\{4\}-[0-9]\{4\}\) \1:' file

Thanks

4

Hello,

Following may help you in same.

awk '{match($0,/.*[0-9][0-9][0-9][0-9]\-[0-9][0-9][0-9][0-9]/); a=substr($0,RSTART+4,RLENGTH); {if(a){print "022" a}}}' check_data_check_rem_check12113

Output will be as follows.

022 \\$a0211-1942

Thanks,
R. Singh

1 Like
5

Hi

very usefull. Thank you very much !!
Does any one can paste url of a tutorial sed and awk for beginners??
jajaja This tools seem to be very powerfull.

Cheers

6

I advice for this links for you
Famous Sed One-Liners Explained, Part I: File Spacing, Numbering and Text Conversion and Substitution - good coders code, great reuse
Sed - An Introduction and Tutorial

Regards
ygemici