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Top Forums Shell Programming and Scripting Extracting text between two constant strings Post 302554225 by shoaibjameel123 on Saturday 10th of September 2011 11:13:51 PM
Old 09-11-2011
Extracting text between two constant strings

Hi All,

I have a file whose common patter is like this:

Code:
.I 1
.U    
87049087
.S
Some text here too
.M
This is a text
.T
Some another text here
.P
Name of the book
.W
Some lines of more text. This text needs to be extracted.
.A
more text goes here too
.I 2
.U    
87049088
.S
Some text here too. More text from previous text
.M
This is a text
.T
Some another text here
.P
Name of the book
.W
Some lines of more text. This text needs to be extracted. This is text 2.
.A
more text goes here too

I want to extract text that is between .W and .A that is this text and store this text in 1.txt. The above pattern continues in the entire file. This means that I will start from 1.txt, then go to next pattern and

Code:
Some lines of more text. This text needs to be extracted.

and for the second pattern store in 2.txt

Code:
Some lines of more text. This text needs to be extracted. This is text 2.

As you can see the file numbers actually come from .I that is present in the above pattern.

I am using Linux with BASH and this is what I have done but seem it does not produce the desired results.

Code:
awk '/\.W/,/\.A/{c++}{print > c ".txt"}' FILE

 

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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)
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