10 More Discussions You Might Find Interesting
1. Shell Programming and Scripting
I am outputting a line like this
print $2 "/" $4The last character though is a ":" and I want to remove it. Is there any neat way to remove it? Or am I forced to do something like this:
print $2 "/" substr($4, 1, length($4) - 1)Thanks. (6 Replies)
Discussion started by: benalt
6 Replies
2. Shell Programming and Scripting
In bash, how can one remove the last character of a string? In perl, the chop function would remove the last character. However, I do not know how to do the same job in bash.
Many thanks in advance. (12 Replies)
Discussion started by: LessNux
12 Replies
3. Shell Programming and Scripting
How can i remove the first and last character of strings like below:
"^^^613*"
"admt130"
"^^^613*"
"123456"
"adg8484"
"DQitYV09dh1C"
Means i wanna remove the quotes("").
Please help (17 Replies)
Discussion started by: proactiveaditya
17 Replies
4. Shell Programming and Scripting
Hi there,
i need some help to remove all occurrences of a certain character at the beginning of a string.
Example: my string is 00102030 and i want to remove all zeros from beginning of string so the result is 102030 (3 Replies)
Discussion started by: gigagigosu
3 Replies
5. Shell Programming and Scripting
Hello!
Please bare with me, I'm a total newbie to scripting. Here's the sudo code of what I'm trying to do:
Get file name
Does file exist?
If true
get length of file name
get network id (this will be the last 3 numbers of the file name)
loop x 2
If... (1 Reply)
Discussion started by: KatieV
1 Replies
6. Shell Programming and Scripting
Hello,
The last character is a comma ,
I have tried the following:
sed -e 's/\,$//' filename-to-read
however - there are still commas at the end of each line...:confused: (5 Replies)
Discussion started by: learning
5 Replies
7. Shell Programming and Scripting
Hi all,
Does anyone know how to code in ksh that will remove the first character in a string variable and replace that variable without the first character?
Example:
var1=ktest1 will become var1=test1
var2=rtest2 will become var2=test2
Need help please. (10 Replies)
Discussion started by: ryukishin_17
10 Replies
8. Shell Programming and Scripting
hi
I have a list of words in a text file. these words are appended by "." at their end. They look something like this.
word1.
word2.
word3.
word4.
word5.
I need to remove the last character "." from all the words. The output must look something like this.
word1
word2
word3... (7 Replies)
Discussion started by: ss3944
7 Replies
9. Shell Programming and Scripting
Hi All,
How to remove a box like special character which appears at the end of a string/line/record. I have no clue what this box like special character is. It is transparent square like box. This appears in a .DAT file at the end of header.
I'm to compare a value in header with a parameter.... (16 Replies)
Discussion started by: Qwerty123
16 Replies
10. Shell Programming and Scripting
Let's say I have a word "foobar23" in a file, and I want to pull the first "f" and last "3" character out of the world, how would I accomplish that?
# cat file
foobar23
I want the output to be:
f3 (3 Replies)
Discussion started by: LinuxRacr
3 Replies
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)