CODEWITHSUNDEEP
regexgrep
Charlie
Charlie

Reputation: 252

How to create a number range regexp grep filter from input variables

I have a relatively large amount of files inside a certain directory. There are always 4 types of files and each file type is numbered like so:

FileType<1-4>_<0000-9999>.ext

So, in the worst case I´d have 40000 files. Now, Im trying to write a simple bash script that filters out all files which are not contained in a range defined by user input, like for instance:

$ ./FilterScript 0 10

Which should output a list of all of the 4 filetypes going from the 0000 to the 0010. I have been working with grep and regexp and gotten some success. This is the example command I´m using to filter everything out (outputs from 0012 to 0023):

ls *.ext | grep -E '00[1][2-8]|00[2][0-3]'

Now the question is, how can I translate the user input to the script into the corresponding regexp expression, without manipulating the input variables as strings? If I want for example the range 90-120, the regexp expresion changes. Is this possible? I have tried using do loops, but in general I think the above mentioned command has a better performance.

Thanks in advance! Ch.

Upvotes: 1

Views: 60

Answers (1)

Ed Morton
Ed Morton

Reputation: 204015

Regarding ls *.ext | ... - don't do that, see https://mywiki.wooledge.org/ParsingLs.

Regarding using a regexp to match a range of numbers - don't do that, use a numeric comparison.

It sounds like you should be doing something like the following using any awk:

#!/usr/bin/env bash

awk -v beg="$1" -v end="$2" '
    BEGIN {
        for ( i=1; i<ARGC; i++) {
            n = split(ARGV[i],parts,/[_.]/)
            num = parts[n-1] + 0
            if ( (beg <= num) && (num <= end) ) {
                print ARGV[i]
            }
        }
        exit
    }
' ./*.ext

The above is obviously untested since you didn't provide any sample input/output for us to test a potential solution with.

Upvotes: 1

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