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Breaking long lines into (characters, newline, space) groups

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Top Forums Shell Programming and Scripting Breaking long lines into (characters, newline, space) groups
# 8  
Old 05-15-2009
Awesome! That works perfectly cfajohnson...thanks a bunch for your help with this. Awk is clearly the way to go, parses it in a split second Smilie

For anyone who comes across this, and happens to be dealing with ldif files, here is the final script for parsing the file:

Code:
#!/bin/ksh

echo "Where is the ldif file located that you would like to parse?"
read source_ldif
echo "Where would you like to output the parsed version to?"
read out_ldif

awk 'length > 78 {
    n=1
    while ( length($0) > 77 + n ) {
    printf "%s\n ", substr($0,1,77 + n)
    $0 = substr($0,78 + n)
    n=0
  }
  if (length) print
  next
}
{print}' "$source_ldif" > "$out_ldif"

# 9  
Old 05-15-2009

This version accepts an arbitrary line length:

Code:
## adjust length to taste
## or prompt for value
## or get value from environment
## or on the command line
## or wherever
length=66

awk -v x=${length:-79} 'length > x {
    n=1
    while ( length($0) > x - 1 + n ) {
    printf "%s\n ", substr($0,1,x - 1 + n)
    $0 = substr($0,x + n)
    n=0
  }
  if (length($0)) print
  next
}
{print}'

Last edited by cfajohnson; 05-15-2009 at 11:54 PM..
# 10  
Old 05-16-2009
I studied up a bit on awk, and generally I understand the script. There is one part I dont get though, maybe you could explain:

Code:
if (length ($0)) print
next

What does this do and why is it necessary? I get what the words mean, but I dont understand its purpose in the script. I tested various source text files with various text content, with lines of all lengths, but no matter what, if i remove this code, the script still seems to work perfectly.

In the variable length script, I didnt understand this either:

Code:
x=${length:-79}



What does the colon minus 79 mean? If you want to allow the user to set the length, could you "read x" and then set the variable as x=x-2? If youre setting the length to something else, why does the number 79 enter this version of the script?

Thanks for any explanation, Id like to understand awk better in general. I will be studying up on it.

# 11  
Old 05-16-2009
Quote:
Originally Posted by rowie718
I studied up a bit on awk, and generally I understand the script. There is one part I dont get though, maybe you could explain:

Code:
if (length ($0)) print
next

What does this do and why is it necessary? I get what the words mean, but I dont understand its purpose in the script. I tested various source text files with various text content, with lines of all lengths, but no matter what, if i remove this code, the script still seems to work perfectly.

It doesn't work perfectly is the line is missing. Given this text file:

Code:
abcdefghijklmnopqrstuvwxyz
ABCDEFGHIJKLMNOPQRSTUVWXY
abcdefghijklmnopqrstuvwxyz

If you call the script with a length of 13, this is the result:

Code:
abcdefghijklm
 nopqrstuvwxy
 z
ABCDEFGHIJKLM
 NOPQRSTUVWXY
abcdefghijklm
 nopqrstuvwxy
 z

If you remove that line, the result is:

Code:
abcdefghijklm
 nopqrstuvwxy
 ABCDEFGHIJKLM
 abcdefghijklm
 nopqrstuvwxy

It prints whatever is left after all lines of length have been removed.
Quote:
In the variable length script, I didnt understand this either:

Code:
x=${length:-79}

[FONT=monospace]

What does the colon minus 79 mean?

That's shell parameter expansion, not part of awk. If length isn't defined, it substitutes 79.
Quote:
If you want to allow the user to set the length, could you "read x" and then set the variable as x=x-2? If youre setting the length to something else, why does the number 79 enter this version of the script?

It's a default value.
# 12  
Old 05-17-2009
Thanks for the explanation. Im learning, but strangely, it seems that our systems are treating this code differently.

This script (without the if and next statements towards the end):

Code:
#!/bin/ksh

echo "How many characters per line?"
read length
echo "Where is the ldif file located that you would like to parse?"
read source_ldif
echo "Where would you like to output the parsed version to?"
read out_ldif

awk -v x=${length:-79} 'length > x {
    n=1
    while ( length($0) > x - 1 + n ) {
    printf "%s\n ", substr($0,1,x - 1 + n)
    $0 = substr($0,x + n)
    n=0

  }

}
{print}' "$source_ldif" > "$out_ldif"

applied to your source text:
Code:
abcdefghijklmnopqrstuvwxyz
ABCDEFGHIJKLMNOPQRSTUVWXY
abcdefghijklmnopqrstuvwxyz

with a length of 13, results in properly parsed text:
Code:
abcdefghijklm
 nopqrstuvwxy
 z
ABCDEFGHIJKLM
 NOPQRSTUVWXY
abcdefghijklm
 nopqrstuvwxy
 z

I tried to figure out which version of awk I have, but apparently it is not easy to do so. I am running os x client 10.4.11. On Apple's opensource distribution page, they list "awk-7" as an available download. I wonder how I can find out the version I am using, and if it makes sense that it's a different version of awk that accounts for the difference in output.
# 13  
Old 05-17-2009

You're right; it is unnecessary; this works as is:

Code:
awk -v x=${1:-79} 'length > x {
    n=1
    while ( length($0) > x - 1 + n ) {
    printf "%s\n ", substr($0,1,x - 1 + n)
    $0 = substr($0,x + n)
    n=0
  }
}
{print}' "$file"

The next exercise is to modify it so that the amount of indent can be specified.
# 14  
Old 05-17-2009
Quote:
Originally Posted by cfajohnson

The next exercise is to modify it so that the amount of indent can be specified.
Code:
if [ $# -eq 0 ]
then
  echo "USAGE: ${0##*/} FILE [width [ indent ]]"
  exit 1
fi

file=$1

awk -v width=${2:-79} -v indent=${3:-1} '
length > width {
    n = width
    while ( length($0) > n ) {
    printf "%s\n%" indent "s", substr($0,1, n), " "
    $0 = substr($0, n)
    n = width - indent
  }
}
{print}' "$file"

Last edited by cfajohnson; 05-17-2009 at 11:17 AM..
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