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Awk to replace a field

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# 1  
Old 04-25-2010
Awk to replace a field
Hi
I am using awk to replace the 4th feild of the input line. My code is below:

Code:
REP_LINE=$(echo $Line | awk -v var=$REPL_DT '{$4=var; print}')

It does replaces but all words/phrases after the 4th feild are erased. Is there a way to modify the 4th feild of the input line and to have all words/pharases that comes after the 4th feild.

For ex:
Code:
# ID          Owner       Last Reset Date
# JPN7386     Cindy       2010/03/17 (should be same as FIW-LR AP)

Here the script replace the Last reset date (4th feild) but all words following the 4th field, (should be same as FIW-LR AP) gets erased.

Thanks,
Sugan
# 2  
Old 04-26-2010
Can you post your input file and the desired output?

Regards
# 3  
Old 04-26-2010
The input line is given below:
Code:
# IN01379     Larry      2010/03/17    (should be the same as FIW-LR AM)

The desired output is:
Code:
# IN01379     Larry      2010/04/26    (should be the same as FIW-LR AM)

i.e I want to replace the 4th feild '2010/03/17' with 'current date'. The code I am using for this is:

Code:
Line="# IN01379     Larry      2010/03/17    (should be the same as FIW-LR AM)"
REPL_DT=`date +'%Y\/%m\/%d'`
echo $Line | awk -v var=$REPL_DT '{$4=var; print}'

This code gives me the following output:
Code:
# IN01379     Larry      2010/04/26

The words following the 4th feild, (should be the same as FIW-LR AM) gets erased.

I want to replace the 4th feild and at the same time retain the entire line including those words/phrases following the 4th feild.

Thanks,
Sugan
# 4  
Old 04-26-2010
hi,

I have executed your code, and its works fine as you expected.
I have changed the below code only.

Code:
REPL_DT=`date +'%Y/%m/%d'`

# 5  
Old 04-26-2010
You are right. It works. I am actually trying to format the output using awk and printf. That is where the words/phrases after 4th feild gets erased.

My code is:
Code:
 Line="# IN01379     Larry      2010/03/17    (should be the same as FIW-LR AM)"
 REPL_DT=`date +'%Y\/%m\/%d'`
 REP_LINE=$(echo $Line | awk -v var=$REPL_DT '{$4=var; print}')
 NEW_LINE=$(echo $REP_LINE |  awk '{printf "# %-7s %9s %16s\n", $2,$3,$4}')

The last line of the code fetches only till 4th feild. But in few lines I have more feilds after $4. I do not have to format the feilds after $4. But I have to format the first four feilds as shown above.

Any way to format only the first 4 feilds of the input line and retain the rest of line.

Thanks,
Sugan
# 6  
Old 04-26-2010
Hi,
Try this,

Code:
NEW_LINE=$(echo $REP_LINE |  awk '{printf "# %-7s %9s %16s %s\n", $2,$3,$4,substr($0,index($0,$5))}')

# 7  
Old 04-26-2010
Thank you very much. It works. It would be great if you can explain how it works.
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