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

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.

Answer 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

Answer 2

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"

Answer 3

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

Answer 4

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)}'