10 More Discussions You Might Find Interesting
1. Shell Programming and Scripting
Hi,
I want to print last column of 3 input files with awk.
cat file1.txt file2.txt file3.txt
file1.txt
1 ad 200
2 ss 300
file2.txt
1 mm 444
2 ee 555
file3.txt
1 kk 999
2 jj 555
My o/p should be :-
1 200 444 999
2 300 555 555 (3 Replies)
Discussion started by: satishmallidi
3 Replies
2. Shell Programming and Scripting
Hi All,
Do we know how to read input file within awk script and send output toanother log file. All this needs to be in awk script, not in command line. I am running this awk through crontab.
Cat my.awk
#!/bin/awk -f
function test(var){
some code}
{
}
END
{
print"test code"
} (5 Replies)
Discussion started by: random_thoughts
5 Replies
3. Shell Programming and Scripting
hello,
suppose, entered input is of 1-40 bytes, i need it to be converted to 40 bytes exactly.
example: if i have entered my name anywhere between 1-40 i want it to be stored with 40 bytes exactly.
enter your name:
donald duck (this is of 11 bytes)
expected is as below - display 11... (3 Replies)
Discussion started by: shravan.300
3 Replies
4. UNIX for Dummies Questions & Answers
Hi Friends,
Could you please tell me why i am getting the below eror while working with awk. I am confused :confused: what to do ?
awk: 0602-591 String 1,9,20,6,6 cannot be longer than 399 bytes. The source line is 1.
The error context is
>>> <<<
awk: 0602-591... (2 Replies)
Discussion started by: i150371485
2 Replies
5. UNIX for Dummies Questions & Answers
Hello guys. I really hope someone will help me with this one..
So, I have to write this script who:
- creates a file home/student/vmdisk of 10 mb
- formats that file to ext3
- mounts that partition to /mnt/partition
- creates a file /mnt/partition/data. In this file, there will... (1 Reply)
Discussion started by: razolo13
1 Replies
6. Shell Programming and Scripting
awk is failing because of input file is greater than 3000 bytes.
,'11 cannot be longer than 3,000 bytes.
The input line number is 1.13071e+06. The file is abc.txt.
The source line number is 1.
Is there any way to wrap the input lines to avoid this error ? (or) adding newline character at... (4 Replies)
Discussion started by: ysvsr1
4 Replies
7. Shell Programming and Scripting
Hi,
i have following line in my code.
eport.pl < $4 | dos2ux | head -2000 | paste -sd\| - | awk -v S="$1" '
Issue is, i get a message saying "awk:input line | found /file/path cannot be longer than 3000 bytes."
"source line number is 3"
Can someone help me with this please? (4 Replies)
Discussion started by: usustarr
4 Replies
8. Programming
Hello, everyone.
Could someone, please, tell me how to get the number of bytes in the terminal input queue without blocking and without consuming these bytes? I guess it could be called the peek functionality.
I've looked at termio tcgetattr() and tcsetattr() functions but could not find... (4 Replies)
Discussion started by: Lucy.Garfeld
4 Replies
9. Shell Programming and Scripting
I am creating ASCII file from Oracle procedure into Unix box.
I undertstand there is NO CRLF as I am writing it into one complete string .. but need to know what is best way to format the file with 80bytes per line only before handing over to another program.
Thanks in advance
regards... (14 Replies)
Discussion started by: u263066
14 Replies
10. News, Links, Events and Announcements
Surreal quote from the news link below:
http://www.washingtonpost.com/wp-dyn/articles/A44615-2002Nov12.html (0 Replies)
Discussion started by: Neo
0 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)