Vista - batch file / script to replace information

 
cd "the directory with .csv files"
 
set $wsep = ","
# Get a list of files
var str filelist ; find "*.csv" > $filelist
while ($filelist <> "")
do
    # Get the next file. Read it into a variable.
    var str file, data, altered_data ; lex "1" $filelist > $file ; cat $file > $data
 
    while ($data <> "")
    do
        # Get the next row of data.
        var str row ; lex -e "1" $data > $row
 
        # Get the first column.
        var str column ; wex -e "1" $row > $column
        # If a .tif is present, alter it to .pdf.
        sal "^.tif^" ".pdf" $column > null
        # Insert column 1 back into row
        set $row = $column + "," + $row ; echo ($row+"\n") >> $altered_data
    done
 
    # The altered data is in $altered_data. Write it back to file.
    echo $altered_data > { echo $file }
done

 
# Get today's time, mm, dd and yyyy
var str mm, dd, yyyy, time ; set $time = gettime()
set $yyyy = { chex "4]" $time }
set $mm = { chex "2]" $time }
set $dd = { chex "2]" $time }
find ($yyyy+"-"+$mm+"-"+$dd+".csv") > $filelist
# In this case the while loop for $filelist is unnecessary,
# since there is only one file in $filelist.