CODEWITHSUNDEEP
bashshellcsv
parmigiano
parmigiano

Reputation: 105

Loop Through List of Files in Bash

I have a list of files that I get by executing this: ls core_* | sort -n -t _ -k 2

which gives me something like this:

core_20171201142359.csv core_20171202131548.csv core_20171203141112.csv

The objective is to get a single file in which to append all the content of every single file in order.

So, I want to open every single file one by one, copy its content into another file, move the previous source file to another directory for safekeeping and move on.

To always get the very first file in order I use ls core_* | sort -n -t _ -k 2 | head -1, and I need to cycle all of those files.

How can I know when there are no more files that I need to process?

Upvotes: 3

Views: 880

Answers (3)

glenn jackman
glenn jackman

Reputation: 246764

For bash, you can store the filenames in an array:

files=(core_*)

Then the first entry is 

first="${files[0]}"

And you can iterate with this (the quotes are absolutely required)

for file in "${files[@]}"; do
    echo "$file"
done

Or, if you need to do something with all the files at once:

cat "${files[@]}" > core_all.csv

but if that's the case, you don't need to store them at all

cat core_* > core_all.csv

Upvotes: 3

cn0047
cn0047

Reputation: 17061

You can try this:

ls core_* | sort -n -t _ -k 2 | while read f; do cat $f >> total.csv; done

Also alongside with cat you can perform move etc.

Upvotes: 1

sahaquiel
sahaquiel

Reputation: 1838

for file in $(ls core_* | sort -n -t _ -k 2)
do
    cat ${file} >> one_big_file.csv
    mv ${file} /anywhere/you/want
done

Will read each file, copy all lines into one_big_file.csv, remove just read file to the /anywhere/you/want

Upvotes: 0

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