Get the 1st 99 characters and add new line feed at the end of the line


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# 1  
Get the 1st 99 characters and add new line feed at the end of the line

I have a file with varying record length in it. I need to reformat this file so that each line will have a length of 100 characters (99 characters + the line feed).

Code:
AU  *     A01        EXPENSE     6990370000       CWF SUBC TRAVEL & MISC 
MY  *     A02        RESALE      6990788000     Y PERMANENT PLACEMENT FEES-CWS
CZ  *     A04        RESALE      6990788000     Y PERMANENT PLACEMENT FEES-BUS SVCS 
GH  *     A06        EXPENSE     6990375009       TAXES EXPENSES-FROM WWER
CZ  *     A04        RESALE      6990788000     Y PERMANENT PLACEMENT FEES-BUS SVCS MAINTENANCE PAYMENT

I'm using this line of code to achieve this.
Code:
awk '{printf "%-99s\n",$0}' $stagedir/$1_SAVE > $workdir/$1_SAVE

However, we are having problems when the length of a record within the file exceeds 100 characters. The receiving program that processes my output file is unable to parse it correctly.

I was asked to truncate the records first, get the first 99-characters of each line, and then add the line feed character at the end of the line.

How do I do this? Please help me. Thank you.
# 2  
Try:

Code:
sed 's/\(.\{99\}\).*/\1/' < file >outfile

or

Code:
perl -lne 'print substr($_,0,99);' file

# 3  
Quote:
Originally Posted by udelalv
Code:
awk '{printf "%-99s\n",$0}' $stagedir/$1_SAVE > $workdir/$1_SAVE

However, we are having problems when the length of a record within the file exceeds 100 characters. The receiving program that processes my output file is unable to parse it correctly.

I was asked to truncate the records first, get the first 99-characters of each line, and then add the line feed character at the end of the line.
Code:
awk '{printf "%-99.99s\n",$0}' "$stagedir/$1_SAVE" > "$workdir/$1_SAVE"

# 4  
Thanks for the suggestions dennis.jacob and cfajohnson.

I went ahead and used
Code:
awk '{printf "%-99.99s\n",$0}' "$stagedir/$1_SAVE" > "$workdir/$1_SAVE"

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