topic batch file for finding full path of a file in Operating System - Microsoft

batch file for finding full path of a file

prajith — Wed, 30 May 2007 01:11:01 GMT

Hi experts,

I am new to batch file scripting.can any body help me to create a batch file which accepts a file name from the user.this file name has to be searched in the computer and the full path should be returned.if the file doesnt exist it should display an error.

awaiting your reply
thanx in advance

Re: batch file for finding full path of a file

Roger Faucher — Wed, 30 May 2007 05:54:13 GMT

I don't know how to convert this into a batch file but this is how it's done from the Command Prompt:

CD\
DIR /s

Make a great day!

Roger

Re: batch file for finding full path of a file

Edgar Zapata — Thu, 31 May 2007 04:28:35 GMT

Hi Prajith,

@echo off
cd \
set /p fileName="Insert file name to be searched.: "
title searching file
echo.
dir %filename% /s
pause

Note: if the file has blanks between words, users must use ""
Example:

Insert file name: "NCRM TaskForce Minutes 2006_11_15.doc"

hope that will do.

Tnks.

Re: batch file for finding full path of a file

Hugo Tigre — Thu, 31 May 2007 05:15:50 GMT

hi prajith,

You can run this bath like this example:

FilePath.bat filename

If the file file as spaces between you have to use "" like for example:

FilePath.bat "file name.txt"

Note: All files found will be outputted in the end, and the last or only file found will be saved in the variable %filepath" so you can used it latter or even used it on another script.

***copy the following code to a text file, and change the extension to .bat or .cmd.***

::----------------------------------------
@echo off
set filename=%1
set filenamePath="%userprofile%\desktop\filenamePath.txt"

c:
cd\

dir %filename% /s/b | findstr %filename% > %filenamePath%

for /f "usebackq tokens=1 delims=" %%i in (%filenamePath%) do (
set filePath=%%i
)

type %filenamepath%
del %filenamePath%
::------------------------------------------

I don't really know why you need this, since a "dir /s/a filename" will find the file you need, but here it is anyway.

Hope this helps,
best regards,
Hugo Tigre