How to create condition find number of file in batch file

Question

Can you help me please, I would like to create batch for count number of file start with key word "MM_" in this path.

D:\>dir /o | find "MM_"
27/06/2017  01:14 PM             1,592 MM_170613.csv
27/06/2017  01:14 PM             1,376 MM_170614.csv
27/06/2017  01:13 PM             2,223 MM_170615.csv
27/06/2017  01:12 PM             2,241 MM_170616.csv
27/06/2017  01:11 PM             2,497 MM_170617.csv

I have use command as Stephan provide as follow detail

@ECHO OFF C: cd "C:\MyDir" set cntAAA = dir /b *MM_*|find /c /v "" echo %cntAAA%

Result is show

0 echo off

Somebody can help?

find /? states a /c parameter to count... dir /b *MM_*|find /c /v "" — Stephan
– Stephan, Commented Jun 27, 2017 at 9:20
You need for /F to capture the output of a command (line)... — aschipfl
– aschipfl, Commented Jun 27, 2017 at 11:57

Compo · Accepted Answer · 2017-06-29 13:49:29Z

0

Try this instead:

@Echo Off
CD /D "C:\MyDir"
Set "cntAAA=Dir/B "MM_*.csv"|Find /C /V """ 
%cntAAA%
Pause

answered Jun 29, 2017 at 13:49

Compo

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