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By using "sed" command. Please help!!!

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# 1  
Old 05-07-2010
By using "sed" command. Please help!!!
Hello,

My input files are having header/body/footer. The first three digits finds whether the data is header/body/footer. For example,

Code:
010AAAAA 20100507 234KB BBBBBBBBBB_20100506.DAT
020CCCCC                     DDDDDDDDD         373983983           19750426           456.90               3983939EE
020FFFFF                     GGGGGGGGG         38938993H           19801102           783.33               DKJFDKJ983
030END OF FILE TOTAL 39839.22

From the above file, there is a date of birth in body (=020). I want to remove/substitute the dateofbirth with spaces. Like, 19750426 is the date of birth but want to substiute as spaces " ". Want to do this same in all records in the file. Can you please tell me how to do this in shell script?

The content of the file is,
Code:
From 1 to 3 = RECORD_TYPE
From 4 to 40 = NAME
From 41 to 80 = DESIGNATION

and so on.

Thanks,
VenkatSmilie
Last edited by vgersh99; 05-07-2010 at 06:39 AM.. Reason: code tags, please!
# 2  
Old 05-07-2010
like?

Code:
awk '/^02/ {$4=""}{print}' file

# 3  
Old 05-07-2010
Tried with the givien command. But it is eliminating the spaces from the remain fields also. But I want to substitute with only spaces in that DOB field. The result of the given command as:
Code:
010AAAAA 20100507 234KB BBBBBBBBBB_20100506.DAT
020CCCCC DDDDDDDDD 373983983          456.90 3983939EE
020FFFFF GGGGGGGGG 38938993H          783.33 DKJFDKJ98
030END OF FILE TOTAL 39839.22

Last edited by Scott; 05-07-2010 at 07:00 AM.. Reason: Code tags, PLEASE!
# 4  
Old 05-07-2010
How about ...

Code:
#! /bin/bash

cat in.file | while read LINE
do
  if [ "${LINE:0:3}" == "020" ]
  then
    echo "$LINE" | sed 's/ [[:digit:]]\{8\} /          /' >> out.file
  else
    echo "$LINE" >> out.file
  fi
done

more out.file
exit 0

Code:
[house@leonov] bash test.bash
010AAAAA 20100507 234KB BBBBBBBBBB_20100506.DAT
020CCCCC DDDDDDDDD 373983983          456.90 3983939EE
020FFFFF GGGGGGGGG 38938993H          783.33 DKJFDKJ983
030END OF FILE TOTAL 39839.22

# 5  
Old 05-07-2010
Hello,

My input files are having header/body/footer. The first three digits finds whether the data is header/body/footer. For example,

Code:
010AAAAA 20100507 234KB BBBBBBBBBB_20100506.DAT
020CCCCC                     DDDDDDDDD         373983983           19750426           456.90               3983939EE
020FFFFF                     GGGGGGGGG         38938993H           19801102           783.33               DKJFDKJ983
030END OF FILE TOTAL 39839.22

From the above file, there is a date of birth in body (=020). I want to remove/substitute the dateofbirth with spaces. Like, 19750426 is the date of birth but want to substiute as spaces "1#7#0#2#". Want to do this same in all records in the file. Can you please tell me how to do this in shell script?

Code:
The content of the file is,
From 1 to 3 = RECORD_TYPE
From 4 to 40 = NAME
From 41 to 80 = DESIGNATION
and so on.

A small change in my previous post. Instead of substitue the spaces, need to fill every 2nd character as # in the Date of Birth field.

---------- Post updated at 10:25 AM ---------- Previous update was at 06:00 AM ----------

dr.house! Thanks for your coding. But I tried and it is still giving the same input file content as result Smilie
Last edited by nvkuriseti; 05-07-2010 at 07:17 AM.. Reason: Code tags, PLEASE!
# 6  
Old 05-07-2010
Quote:
Originally Posted by nvkuriseti
...
From the above file, there is a date of birth in body (=020). I want to remove/substitute the dateofbirth with spaces. Like, 19750426 is the date of birth but want to substiute as spaces "1#7#0#2#". Want to do this same in all records in the file. Can you please tell me how to do this in shell script?
...
A small change in my previous post. Instead of substitue the spaces, need to fill every 2nd character as # in the Date of Birth field.
...
You haven't mentioned if the date of birth part begins at a specific column in your file. If it does, and if you know that column number, then you can do something like the following Perl one-liner:

Code:
$
$ cat f1
010AAAAA 20100507 234KB BBBBBBBBBB_20100506.DAT
020CCCCC                     DDDDDDDDD         373983983           19750426           456.90               3983939EE
020FFFFF                     GGGGGGGGG         38938993H           19801102           783.33               DKJFDKJ983
030END OF FILE TOTAL 39839.22
$
$
$ perl -plne 's/^(020.{64})(\d).(\d).(\d).(\d).(.*)$/$1$2#$3#$4#$5#$6/' f1
010AAAAA 20100507 234KB BBBBBBBBBB_20100506.DAT
020CCCCC                     DDDDDDDDD         373983983           1#7#0#2#           456.90               3983939EE
020FFFFF                     GGGGGGGGG         38938993H           1#8#1#0#           783.33               DKJFDKJ983
030END OF FILE TOTAL 39839.22
$
$

Over here I know that date of birth begins from column 68 in rows that start with "020". Hence I put "64" in the brace quantifiers so that total length of the part before it is 3 (length of "020") + 64 = 67.

HTH,
tyler_durden
# 7  
Old 05-07-2010
Quote:
Originally Posted by nvkuriseti
A small change in my previous post. Instead of substitue the spaces, need to fill every 2nd character as # in the Date of Birth field.
Correct me if I'm wrong but this requirement does not seem to bear any significance in the real world. Smilie
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