As Jedi states there are various bugs, I'm no awk expert but I think the line:
should read:
so that the awk(1) command has something in the input file (called file1) to work on...
Hi
I have a variable which is a path ie:
UBERROR=/cobwrk/mchr/prodsup/ub/wrk/../error
is there anyway I can get the output of an echo to read:
#echo $UBERROR
/cobwrk/mchr/prodsup/ub/error
instead of
#echo $UBERROR
/cobwrk/mchr/prodsup/ub/wrk/../error
Many thanks! (2 Replies)
awk -F^ '{ if ((($1 != "M") && ($5 != "2")) || (($1 != "S") && ($5 != "7"))) print $0}' welcome > welcome1
The "&&" and "||" in the above command is not working with awk.
When I run the above command, the same content of welcome is copied to welcome1 without any difference.
Your reply is... (12 Replies)
Hi,
Does awk ever resolve params in the search pattern?..
The following awk doesnt know how to resolve ${tables}$ inside a loop.
k=`awk '/${tables}$/ ${graph}`
The search pattern has ${tables}$ and I am narrowing down my search with a $ at the end of string.
So...this leaves me with a... (13 Replies)
Hi friends,
I have a file list1 which has these 2 columns like
616449 0
434453 1
2151083 0
2226536 0
2132382 0
2136814 0
I have to put the result of col1 -col2 into another file list2 linewise.
e.g. It gives the below result if use the below code:
awk '{ print $1 - $2 }' list1 >... (2 Replies)
Hi all,
Hope someone can help me out here.
I have this BASH script (see below)
My problem lies with the variable path.
The output of the command find will give me several fields. The 9th field is the path. I want to captured that and the I want to filter this to a specific level.
The... (6 Replies)
I am tyring to resolve an environment variable that is part of a string I selected from our database.
Simply put, I want cd to this folder before checking if a file exists.
The variable $in_loc has the value '$PS_HOME/int/VSP' where $PS_HOME is the environment variable.
I am using cd... (6 Replies)
Hello Everyone,
I am trying to resolve a variable inside another variable.Let me go straight to the example.
Input:
Query=$Table_1 Join $Table_2
(Query itself is a variable here)
Now for two different cases I am assigning different values to Table_1 and Table_2
Case 1:... (14 Replies)
Hi All,
I have below variable,
xyz=\$AI_XFR
Now, if you will run the below command
=> echo $xyz
$AI_XFR
It is returning hardcoded string value.
Whereas in environment, there is value in it. Like below:
=> echo $AI_XFR
/home/aditya/sandbox/xfr/
I need to resolve this... (4 Replies)
My script
----------
for i in `cat n`;do
export k=`echo "CSN: "$i` //combining CSN: and value from n
echo "$k"
awk ''{print "CSN: "$0;}'{_=29}_&&_--' file1|tail -1 >> file2
done
In the above script i cannot able to resolve $k in awk command
file n contains
------------
0000
1111... (0 Replies)
I have the following script, and I want to assign the output ($10 and $5) from awk to N and L:
grdinfo data.grd | awk '{print $10,$5}'| read N L
output from gridinfo data.grd is: data.grd 50 100 41 82 -2796 6944 0.016 0.016 3001 2461. where N and L is suppose to be 3001 and 100. I use... (8 Replies)
Discussion started by: geomarine
8 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)