11-14-2011
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10 More Discussions You Might Find Interesting
1. Shell Programming and Scripting
I'm reading 2 input files but not getting expected value.
I should get an alpha value on file_1_data but not getting any.
Please help.
>cat test6.sh
awk '
FILENAME==ARGV { file_1_data=$0; print "----- 1 Line " NR " -----" $1; next }
FILENAME==ARGV { file_2_data=$0; print "----- 2... (1 Reply)
Discussion started by: pdtak
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2. Shell Programming and Scripting
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3. Shell Programming and Scripting
Hi guys,
I am new to AWK and unix scripting. Please see below my problem and let me know if anyone you can help.
I have 2 input files (example given below)
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4. Shell Programming and Scripting
Okay, so I've looked on here and found some similar things, but not exactly what I am looking for. I am working on creating a script that can back up some files, based on the contents of another file - the configuration file.
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5. Shell Programming and Scripting
Being new to this area .I have been assigned a task which i am unable to do . Can any one please help me .
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6. Shell Programming and Scripting
Hello,
Could somebody please give me an awk example on how to read from the standard input.
It means as the "read" function in Korn shell.
Thx in advance ... (3 Replies)
Discussion started by: rany1
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7. Shell Programming and Scripting
Can I do something like,
if($0==/^int.*$/) {
print "Declaration"
}
for an input like: int a=5;
If the syntax is right, it is not working for me, but I am not sure about the syntax. Please help.
Thanks,
Prasanna (1 Reply)
Discussion started by: prasanna1157
1 Replies
8. Shell Programming and Scripting
This is one of the strangest things that's happening to me.
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As a simple test, I wrote this:
#!/user/bin/perl -w
use strict;
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9. Shell Programming and Scripting
Hello,
I've been trying to come up with a solution for the following problem; I have an input file with two columns and I want to print as an output the first column without any changes but for the second column, I want to divide it by its last value. Example input:
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10. Open Source
I have a process that requires me to read data from huge log files and find the most recent entry on a per-user basis. The number of users may fluctuate wildly month to month, so I can't code for it with names or a set number of variables to capture the data, and the files are large so I don't... (7 Replies)
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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)