Sponsored Content
Full Discussion: cut or nawk
Top Forums Shell Programming and Scripting cut or nawk Post 302675553 by capilla on Monday 23rd of July 2012 09:20:48 AM
Old 07-23-2012
Thanks but doesn't work
Code:
root@SERVER01:/opt/scripts\ > nawk '/busti/ && match($8,/(\/[^\/]*){5}/) {print substr($8,RSTART,RLENGTH)}' ruta2.txt
root@SERVER01:/opt/scripts\ >
root@SERVER01:/opt/scripts\>vi ruta2.txt
/SERVER01-01/=user/a3/85/=USER1
/SERVER01-01/=user/a3/85/=USER1/=+Eliminats
/SERVER01-01/=user/a3/85/=USER1/=+Enviats
/SERVER01-01/=user/a3/85/=USER1/=+Esborranys
/SERVER01-02/=user/c6/30/=USER2
/SERVER01-02/=user/c6/30/=USER2/=+Eliminats
/SERVER01-02/=user/c6/30/=USER2/=+Enviats
/SERVER01-02/=user/c6/30/=USER2/=+Esborranys
/SERVER01-02/=user/84/60/=USER3
/SERVER01-02/=user/84/60/=USER3/=+Eliminats
/SERVER01-02/=user/84/60/=USER3/=+Enviats
/SERVER01-02/=user/84/60/=USER3/=+Esborranys

 

6 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

how to access values of awk/nawk variables outside the awk/nawk block?

i'm new to shell scripting and have a problem please help me in the script i have a nawk block which has a variable count nawk{ . . . count=count+1 print count } now i want to access the value of the count variable outside the awk block,like.. s=`expr count / m` (m is... (5 Replies)
Discussion started by: saniya
5 Replies

2. Shell Programming and Scripting

Nesting - two nawk into one nawk

hi people; this is my two awk code: nawk '/cell+-/{r=(NF==8) ? $4FS$5FS$6 : NF==7 ? $4FS$5 : $4 ;c=split(r,rr);for (i=1;i<=c;i++){if(rr != "111111"){printf($3" %d ""\n",(i+3))}}printf("")}' /home/gc_sw/str.txt > /home/gc_sw/predwn.txt nawk -F'*' '{gsub(/ *$/,"")}$0=$1$($NF-2)'... (2 Replies)
Discussion started by: gc_sw
2 Replies

3. Shell Programming and Scripting

Cut Command error cut: Bad range

Hi Can anyone what I am doing wrong while using cut command. for f in *.log do logfilename=$f Log "Log file Name: $logfilename" logfile1=`basename $logfilename .log` flength=${#logfile1} Log "file length $flength" from_length=$(($flength - 15)) Log "from... (2 Replies)
Discussion started by: dgmm
2 Replies

4. Shell Programming and Scripting

Nawk Problem - nawk out of space in tostring on

Hi.. i am running nawk scripts on solaris system to get records of file1 not in file2 and find duplicate records in a while with the following scripts -compare nawk 'NR==FNR{a++;next;} !a {print"line"FNR $0}' file1 file2duplicate - nawk '{a++}END{for(i in a){if(a-1)print i,a}}' file1in the middle... (12 Replies)
Discussion started by: Abhiraj Singh
12 Replies

5. UNIX for Beginners Questions & Answers

Cut command: can't make it cut fields

I'm a complete beginner in UNIX (and not a computer science student either), just undergoing a tutoring course. Trying to replicate the instructions on my own I directed output of the ls listing command (lists all files of my home directory ) to My_dir.tsv file (see the screenshot) to make use of... (9 Replies)
Discussion started by: scrutinizerix
9 Replies

6. Shell Programming and Scripting

Using :<<cut / cut to comment out block of bash script

I am using : << cut / cut to comment out block of code. Works fine on few lines of script, then it gives me this cryptic error when I try to comment out about 80 lines. The "warning " is at last line of script. done < results 169 echo "END read all positioning parameters" 170... (8 Replies)
Discussion started by: annacreek
8 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)
All times are GMT -4. The time now is 01:15 AM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy