在Windows 7中从命令行打开目录中的最新文件

Question

I am using a device that generates an output file in .txt format. The file name is generated using tokens for the experiment name and an incremented token: <ExperimentName><IncrementedToken>.txt. The output directory is filled with a number of output files from an array of experiments, and it has become difficult to find the most recent output file.

I am trying to come up with a script to launch the most recent output file from the directory where these files are saved using the command line.

So far I have been able to use dir PathToOutputFolder /b /od | head -1 dir PathToOutputFolder /b /od | head -1 to find the most recent file, but am having trouble launching the file from the command line.

The closest (I think) that I've gotten so far is something along the lines of: start "" notepad dir PathToOutputFolder /b /od | head -1 start "" notepad dir PathToOutputFolder /b /od | head -1 but this gives me a "system cannot find path specified" error.

I'm thinking that the output of the dir command can't be passed to the start command, but would like to find a workaround.

Answer 1

(Caveat: This will fail if your console is set to Raster fonts and your file names contain characters that won't fit in the OEM codepage.)

for /f "delims=" %%x in ('dir PathToOutputFolder /b /o-d') do if not set filename set "filename=%%x"
start "" "%filename%"

If you have the option of using PowerShell that is much easier, though:

$filename = (Get-ChildItem PathToOutputFolder | sort LastWriteTime)[-1]
Invoke-Item $filename