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files indexed by production date

开发者 https://www.devze.com 2022-12-29 22:22 出处:网络
Each day an application creates a file called file_YYYYMMDD.csv where YYYYMMDD is the production date. Bu开发者_高级运维t sometimes the generation fails and no files are generated for a couple of days

Each day an application creates a file called file_YYYYMMDD.csv where YYYYMMDD is the production date. Bu开发者_高级运维t sometimes the generation fails and no files are generated for a couple of days.

I'd like an easy way in a bash or sh script to find the filename of the most recent file, which has been produced before a given reference date.

Typical usage: find the last generated file, disregarding those produced after the May 1st.

Thanks for your help


This script avoids:

  • Using sed repeatedly in a loop
  • Parsing ls
  • Creating a subshell in the while loop
  • Processing files that don't match the file_*.csv name pattern

Here's the script:

#!/bin/bash
while read -r file
do
    date=${file#*_}    # strip off everything up to and including the underscore
    date=${date%.*}    # strip off the dot and everything after
    if [[ $date < $1 ]]
    then
        break
    fi
done < <(find -name "file_*.csv" | sort -r)

# do something with $file, such as:
echo "$file"

Edit:

With Bash >= 3.2, you can do this using a regular expression:

#!/bin/bash
regex='file_([[:digit:]]+).csv'
while read -r file
do
    [[ $file =~ $regex ]]
    date=${BASH_REMATCH[1]}
    if [[ $date < $1 ]]
    then
        break
    fi
done < <(find -name "file_*.csv" | sort -r)

# do something with $file, such as:
echo "$file"


Try this:

#!/bin/bash

ls -r | while read fn; do
    date=`echo $fn | sed -e 's/^file_\([0-9]*\)\.csv$/\1/'` || continue
    if [ $date -lt $1 ]; then
        echo $fn
        exit
    fi
done

Just call this script with the reference date you want to compare with. Replace -lt with -le if you want to include the reference date.

Edit: An alternate solution, without piping an echoed variable. Note that I didn't test it, but it should work, too.

#!/bin/bash

ls -r | sed -e 's/^file_\([0-9]*\)\.csv$/\1/' | while read date; do
    if [ $date -lt $1 ]; then
        echo "file_${date}.csv"
        exit
    fi
done


Sorting file names with man 1 sort will fail if there's a newline character in a file name.

Instead we should use something like:

touch $'filename\nwith\777pesky\177chars.txt'  # create a test file

ls -1db * 

find ... -print0 | LC_ALL=C sort0 ... 

see:

Find all used extensions in subdirectories,

http://codesnippets.joyent.com/posts/show/2300

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