How to use Bash script or awk to extract a file name

Question

I receive files with names constructed in the following format

[2letters e.g.AF][6-digit-number sequence][Date in ccyymmdd][Time in hhmmss]

For Example:

AF00010720120917144500.csv

I want to automate loading such files onto my database using the date part of the file.

something which may start like this:

#!/bin/bash
filename_datepart=$(echo `date -d "1 day ago" +"%d%m%Y"`)
filename="/home/hlosi/AF000107"$filename_datepart".csv"

But remember, the part 000107 changes with each new file.

How many files do you expect per day? Also, wouldn't you need the Time in hhmmsss? — beny23
– beny23, Commented Sep 17, 2012 at 12:58
I receive one file per day. I have tried reasoning with the data source about excluding the time part since it's only one file a day to no avail. — hlosukwakha
– hlosukwakha, Commented Sep 17, 2012 at 13:01

beny23 · Accepted Answer · 2012-09-17 13:09:03Z

1

You can use wildcards to fill in the unknown values

#!/bin/bash
file=/home/hlosi/AF??????`date -d "1 day ago" +"%d%m%Y"`??????.csv
echo $file

answered Sep 17, 2012 at 13:09

beny23

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1 Comment

hlosukwakha Over a year ago

This worked fine: file=/home/hlosi/AF*date -d "1 day ago" +"%d%m%Y"*.csv

Jeremy J Starcher · Accepted Answer · 2012-09-17 13:37:59Z

1

Here is a BASH solution:

#!/bin/bash

#The full name
fullname="/home/hlosi/AF00010720120917144500.csv"

#Rip off the directory
file=$(basename "$fullname")

#Now pull out just the characters that we want
extract=$(echo "$file" | cut -c3-8)
echo "You want: $extract"

edited Sep 17, 2012 at 13:37

answered Sep 17, 2012 at 13:01

Jeremy J Starcher

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5 Comments

hlosukwakha Over a year ago

Thanks, I have adapted this one to part of my problem. I will try it and get back to you. Thanks a lot.

tripleee Over a year ago

Bash has built-ins so you don't need cut either, but you have to do it in stages; cuttail=${file#????????}; cutc8=${file%$cuttail}; echo "${cutc8#??}". Note also the use of double quotes when you interpolate a variable, except in assignments.

Jeremy J Starcher Over a year ago

@tripleee -- I learned shell scripting before bash was common -- I tend to stick the old ways. You are right about the quotes though, I should have put those in.

tripleee Over a year ago

Actually the string substitutions go pretty far back; they are certainly not Bash-only.

chepner Over a year ago

Since the date part is at a fixed location, ${file:8:8} is one-step solution in bash (although not POSIX).

r2p2 · Accepted Answer · 2012-09-17 13:11:29Z

0

I think you want this. In case you have to handle multiple files.

#!/bin/bash
fpath=/home/hlosi/
filename_datepart=$(echo `date -d "1 day ago" +"%d%m%Y"`)
files=$(find $fpath -iname "*$filename_datepart.csv")
for file in $files
do
    echo "found file: " $file
done

answered Sep 17, 2012 at 13:11

r2p2

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1 Comment

hlosukwakha Over a year ago

Especially since these people take days to remove old files, my script may have to choose the correct file using the date part.

V H · Accepted Answer · 2012-09-17 19:20:57Z

forgive me for my ignorance there is -atime -ctime -mtime I think its -ctime

find  -ctime 1  -name \*.csv -print

-mtime match ending of csv and are exactly 1 day old, the trouble of this is it works in 24 hour period so files less than 24 hours but still yesterday would not show

this would be a simpler way of doing things since and would not care about changes in file name formatting for future proofing.

cd pathtcsv;
d=`date -d "1 day ago" +"%d"`
d=$d find . -type f -name \*.csv -ctime 1 -exec ls -l {} \;|awk '$7 ~ d'|awk '{print $NF}'|  awk '{ print substr( $0, length($0) - 1, length($0) ) }'

# D = $d which is set as yesterday's date, it finds files from yseterday that have csv ending it then does an ls, pipes into awk and checks out value 7 against the date of yesterday which $7 is the date value on ls -l, it finally prints the last field and pipes into a final awk which prints the string and splits from char position to char position which is what you wanted ? you need to figure out what chars you need here is another example of above for char positions of 0 to 10.

 d=$d find . -type f  -ctime 1 -name \*.csv -exec ls -l {} \;|awk '$7 ~ d'|awk '{print $NF}'|  awk '{ print substr( $0, 0, 10)}'

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