I have a function
awkvarrep() {
awk -F'|' '$1~/$1/{printf "%-10s %-30s %-15s %-30s %-15s\n", $2,$3,$4,$5,$6}' testfile
}
I'm calling it by this
VARREP=XYZ
awkvarrep $VARREP
since i'm passing $VARREP to the awkvarrep() function I want to use this with $1 but it dosen't seem to be... (5 Replies)
Hello,
I have a variable that displays the following results from a JVM....
1602100K->1578435K
I would like to collect the value of 1578435 which is the value after a garbage collection. I've tried the following command but it looks like I can't get the > to work. Any suggestions as... (4 Replies)
Hi all
I want to do a simple substitution in awk but I am getting unexpected output. My function accepts a time and then prints out a validation message if the time is valid. However some times may include a : and i want to strip this out if it exists before i get to the validation. I have shown... (4 Replies)
Hi,
Can some one please explain the following line please throw some light on the ones marked in red
awk '{print $9}' ${FTP_LOG} | awk -v start=${START_DATE} 'BEGIN { FS = "." } { old_line1=$0; gsub(/\-/,""); if ( $3 >= start ) print old_line1 }' | awk -v end=${END_DATE} 'BEGIN { FS="." } {... (3 Replies)
I want to replace comma with space and "*646#" with space.
I am using the following code:
nawk -F"|" '{gsub(","," ",$3); gsub(/\*646\#/"," ",$3);print}' OFS="|" file
I am getting following error:
Help is appreciated (5 Replies)
Would really appreciate it if someone could point out my mistake in this line of code, i've been staring blankly at it trying everything i can think of some time now and coming up with nothing.
#!/bin/bash
echo "Enter Username"
read Username
awk -F: -v var=${Username} '/^var:/... (9 Replies)
Hi, I want to print the first column with original value and without any double quotes
The output should look like
<original column>|<column without quotes>
$ cat a.txt
"20121023","19301229712","100397"
"20121023","19361629712","100778"
"20121030A","19361630412","100838"... (3 Replies)
Hello, I had some difficulty to understand the gsub function and maybe the regex in this script to remove all the punctuations:
awk 'gsub(//, " ", $0)' text.txtFile text.txt:
This is a test for gsub
I typed this random text file
which contains punctuation like ,.;!'"?/\ etc.
The script... (6 Replies)
Hello,
I have searched but failed to find what exactly im looking for,
I need to eliminate first "." in a output so i can use something like the following
echo "./abc/20141127" | nawk '{gsub("^.","");print}'
what i want is to use gsub result later on, how could i achieve it?
Let say... (4 Replies)
Discussion started by: EAGL€
4 Replies
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)