I need to read the last file for a particular day, such as, "Jun 13" because the CSV file is cumulative for the entire day, so I don't want all the previous files, I just want the last file, for that day.

I ran an 'ls -al | grep "June 13" > myLs.txt' (simplified) to list all files from that day.

So.. instead of using a bunch of greps and cuts and then expr comparisons to find the last file for that day based on date/time.
I'm just assuming that the last line of each myLs.txt will be the last file for that day.

Does awk have a "last line" variable?
I was looking for it and I only found ones for number of "fields per record" and "current record number", but no "number of records total".

Hmm... I guess it would be better if i just, wc -l the textfile and then redirect that into a variable which I will then use in my awk command to echo the last line..... That would work wouldn't it.. hmm..

firstVar=`wc -l < ls.txt` # get wc -l
firstVar=`echo $firstVar| tr -d ' '` # remove extra spaces
echo "firstVar:${firstVar}" # echo test to see if var worked
secondVar=`awk "END {print $$firstVar}" ls.txt` # awk to print last line of file

Hm... I think I'm complicating things.
I'm getting closer.

ah hah! There's a tail command.

# print the last line of a file (emulates "tail -1")
sed '$!d' # method 1
sed -n '$p' # method 2

# print first line of file (emulates "head -1")
sed q

sed1liners