If i use the below command , it is giving me the output with "," in between two name. how ? and also i would like to know the reason for the space used in between ([a-z]*\) *\([a-z]*\) and \2, \1/' and what is the use of specifying two [a-z] and what *\ means?
Last edited by joeyg; 06-02-2015 at 02:50 PM..
Reason: Please wrap scripts/commands and data inside CodeTags
Ok, i've been trying to write some shell scripts. nothing challenging, but just automating
All of the tutorials i read say to start the file with
#!/bin/bash
or whatever your path to bash is.
So i do it, and all of my scripts error out saying ./nameofscript:command not found
when i... (4 Replies)
Hello,
can someone please explain me what the following commands do.. i know the output but i would like to understand the break down of what they do step by step.
1) sed -n "/ $(date +\%R -d "-1 min")/,$"p req.txt| wc -l
2)
awk '/19:00/,/22:00/' app.log |grep "mytesturl"|grep... (2 Replies)
Can someone help me understand why
sed '/host/ s/$/QQQ/g' junk
is taking the file 'junk' which is this:
host Example_1
filename-"gfnwd.cfg";
hardware-ethernet-00:13:11:fd:7a:88;
fixed-address-10.10.6.19;
}
host Example_2
filename-"gfnwd_280.cfg";... (2 Replies)
I am relatively new to Shell Scripting. I can't understand the following two scripts. Can someone please spare a minute to explain?
1) content s of file a are
(021) 654-1234
sed 's/(//g;s/)//g;s/ /-/g' a
021-654-1234
2)cut -d: -f1,3,7 /etc/passwd |sort -t: +1n gives error (3 Replies)
Hey forum,
I have a problem with a script what used to work, but suddenly is not working anymore. I have been trying different things for an hour now and I give up :D .
#!/bin/sh
asukoht=`pwd`
template=$1
for values in... (4 Replies)
I am learning SED and just following the shell scripting book, i have trouble understanding the grep and sed statement,
Question : 1
__________
/opt/oracle/work/antony>cat teledir.txt
jai sharma 25853670
chanchal singhvi 9831545629
anil aggarwal 9830263298
shyam saksena 23217847
lalit... (7 Replies)
#!/bin/awk -f
BEGIN {i=1;file="modified.txt"}
{
if ($0 !~ /^DS:/) {print $0 >> file} else {
if ($0 ~ /^DS:/) {print "DS: ",i >> file;if (i==8) {i=1} else {i++}};
}
}
END {gzip file}
Can someone explain to me how this above script works, I got it from a friend but not able... (3 Replies)
Discussion started by: Kamesh G
3 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)