I need a command to get distinct extensions files inside a folder with the number of occurency of each extension.
So how can I do this?
You could do it with a PowerShell one-liner like this:
powershell "$ext = @{}; gci *.* | %{ $ext[$_.Extension]++ }; $ext"
If using this in a .bat script, double the %
sign.
In pure Batch, this is the simplest way I can think of:
@echo off & setlocal
for %%I in (*.*) do (
set /a ext[%%~xI] += 1
)
set ext[
Or condensed to a one-liner (still in a .bat script):
@setlocal & @(for %%I in (*.*) do @set /a ext[%%~xI] += 1) & set ext[
Or directly from the cmd console:
cmd /c ">NUL (@for %I in (*) do @set /a ext[%~xI] += 1) & set ext["
(The cmd /c
there behaves like setlocal
in a bat script, helping you avoid junking up your environment with useless variables.)
Here is solution from superuser.com . Save this code inside a .bat
file.
@echo off
set target=%~1
if "%target%"=="" set target=%cd%
setlocal EnableDelayedExpansion
set LF=^
rem Previous two lines deliberately left blank for LF to work.
for /f "tokens=*" %%i in ('dir /b /s /a:-d "%target%"') do (
set ext=%%~xi
if "!ext!"=="" set ext=FileWithNoExtension
echo !extlist! | find "!ext!:" > nul
if not !ERRORLEVEL! == 0 set extlist=!extlist!!ext!:
)
echo %extlist::=!LF!%
endlocal
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