Windows Batch Scripting: Parse CSV file and extract data - windows

Question

I am new to windows batch scripting so pardon my ignorance. I have a CSV file that looks like this:
ColumnA,ColumnB,ColumnC
01/02/2015,ABC,111
01/03/2015,DEF,222
01/03/2015,HHH,333
01/05/2015,XYZ,767
The number of rows in this file will vary but the columns will remain the same. I need to extract the date column A, row 1 and date from column A, last row.
In this case I need to extract 01/02/2015 and 01/05/2015. Next I want to store both these dates in separate variable.
How can I achieve this? I have no idea where to begin. The only thing I have is this:
for /F "tokens=1 delims=," %%s in (IntFlow.csv) do #echo %%s
This one line of code cuts the first column and echo's it. I don't know what do next. Please advice.

Answer 1

Use a for /f loop to parse the csv file line-by-line. Use skip=1 to skip the header row. Only set the first variable if it isn't already defined. Then just set the last variable on every loop iteration. It's bound to be correct sooner or later, right? :)
#echo off
setlocal
set "first="
set "last="
for /f "skip=1 usebackq delims=," %%I in ("test.csv") do (
if not defined first set "first=%%~I"
set "last=%%~I"
)
echo %first%
echo %last%
tokens=1 is already the default behavior. usebackq allows you to enclose the filename in quotes, so it should work if your csv filename includes a space or some other unsavory character in the name.
I see in your comment you've decided to use PowerShell instead. Here's a PowerShell script that'll let you populate variables with the dates.
$csv = ipcsv test.csv
$first = ($csv | select -first 1 | %{ $_.ColumnA })
$last = ($csv | select -last 1 | %{ $_.ColumnA })
$first
$last