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Full Discussion: awk prints only last line
Top Forums Shell Programming and Scripting awk prints only last line Post 302969640 by RavinderSingh13 on Friday 25th of March 2016 12:20:21 AM
Old 03-25-2016
Hello SkySmart,

Could you please try following and let me know if this helps.
Code:
echo "${NEWTEXTS}" | awk -F, '{a=$1; b=$2; c=$3; d=$4; e=$5; f=$6 } {printf("%s,%s,%s,%s,%s,%0.f\n", a,b,c,d,e,f)}'

Off course culprit there was END in your code. Also you could go through the manual page of awk by doing man awk, as follows mentioned in that for END.
Quote:
BEGIN and END are two special kinds of patterns which are not tested against the input. The action parts of all BEGIN patterns are merged as if all the statements had been written in a single BEGIN block. They are executed before any of the input is read. Similarly, all the END blocks are merged, and executed when all the input is exhausted (or when an exit statement is executed). BEGIN and END patterns cannot be combined with other patterns in pattern expressions. BEGIN and END patterns cannot have missing action parts.
Hope this helps.

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
SkySmart
 

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LEARN ABOUT DEBIAN

bup-margin

bup-margin(1)						      General Commands Manual						     bup-margin(1)

NAME

       bup-margin - figure out your deduplication safety margin

SYNOPSIS

       bup margin [options...]

DESCRIPTION

       bup margin  iterates  through  all  objects  in	your  bup repository, calculating the largest number of prefix bits shared between any two
       entries.  This number, n, identifies the longest subset of SHA-1 you could use and still encounter a collision between your object ids.

       For example, one system that was tested had a collection of 11 million objects (70 GB), and bup margin returned 45.  That  means  a  46-bit
       hash  would be sufficient to avoid all collisions among that set of objects; each object in that repository could be uniquely identified by
       its first 46 bits.

       The number of bits needed seems to increase by about 1 or 2 for every doubling of the number of objects.  Since SHA-1 hashes have 160 bits,
       that  leaves 115 bits of margin.  Of course, because SHA-1 hashes are essentially random, it's theoretically possible to use many more bits
       with far fewer objects.

       If you're paranoid about the possibility of SHA-1 collisions, you can monitor your repository by running bup margin occasionally to see	if
       you're getting dangerously close to 160 bits.

OPTIONS

       --predict
	      Guess  the offset into each index file where a particular object will appear, and report the maximum deviation of the correct answer
	      from the guess.  This is potentially useful for tuning an interpolation search algorithm.

       --ignore-midx
	      don't use .midx files, use only .idx files.  This is only really useful when used with --predict.

EXAMPLE

	      $ bup margin
	      Reading indexes: 100.00% (1612581/1612581), done.
	      40
	      40 matching prefix bits
	      1.94 bits per doubling
	      120 bits (61.86 doublings) remaining
	      4.19338e+18 times larger is possible

	      Everyone on earth could have 625878182 data sets
	      like yours, all in one repository, and we would
	      expect 1 object collision.

	      $ bup margin --predict
	      PackIdxList: using 1 index.
	      Reading indexes: 100.00% (1612581/1612581), done.
	      915 of 1612581 (0.057%)

SEE ALSO

       bup-midx(1), bup-save(1)

BUP

       Part of the bup(1) suite.

AUTHORS

       Avery Pennarun <apenwarr@gmail.com>.

Bup unknown-															     bup-margin(1)
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