Finding most recent file in a directory

Hi,
Is there a simple and easy way in labview to write a program that goes
into a directory and looks for the most recent file. I can think of one
way where I use the list directory vi to obtain an array of the file
names in the directory and then use the file/directory info vi to
obtain a time stamp for each file in the directory and afterwards sort
all the time stamps. I'm not sure if this is the simplest way to do
this, and wanted to ask if anyone on the discussion forums knows of any.
Thank you

unclebump wrote:
Try here. http://zone.ni.com/devzone/cda/epd/p/id/3175
This example is not that great and a few tweaks can greatly improve it!
Building the path with string operations inside the loop automatically breaks this VI on anything but windows. There is a reason we have the "build path" function! It should always be used. The "string to path" and "path to string" should only be used for OS specific tasks.
You probably want to skip files that are actually links. Newer versions of "File/Directory info" tell you if the path points to a link instead of a file with a boolean output.
WIy do we need to keep an array of all files and maintain the corret index in a shift register. It seems more intuitive and less wasteful to just keep the newest file path in the shift register and not worry about the index.
The image below shows a quick attempt at improving the example. Modify as needed.
Message Edited by altenbach on 04-12-2007 02:45 AM
LabVIEW Champion . Do more with less code and in less time .
Attachments:
find_newest_real_file.png ‏9 KB

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Message was edited by: w7ox

Finding most recent file in a directory

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