Write a batch file to execute a program after 5 sec and show the remaining time

[whosyourdaddy]

How could I write a batch file to show the message like below and then run the program?
10:00:00 aaa.txt is waiting 5 seconds to start up...
10:00:01 aaa.txt is waiting 4 seconds to start up...
10:00:02 aaa.txt is waiting 3 seconds to start up...
10:00:03 aaa.txt is waiting 2 seconds to start up...
10:00:04 aaa.txt is waiting 1 seconds to start up...

Below is my script:

@ECHO OFF
setlocal
set seconds=%5
if "%seconds%"=="" set seconds=5
ECHO %TIME% aaa.txt is waiting %seconds% seconds to start up...
PING 127.0.0.1 -n %seconds% -w 1000  > NUL
START /B C:\aaa.txt
endlocal
exit

[Dusty]

Try this script, the timing is not precise, you may want to do some work on that.

Code: [Select]

@echo off
cls
setlocal enabledelayedexpansion
   
for /l %%1 in (5,-1,1) do (
    echo !time:~0,-3! aaa.txt is waiting %%1 second(s^) to startup...
    ping -n %%1 -w 100000 127.0.0.1 > nul
)

start /b "" c:\aaa.txt
exit



[whosyourdaddy]

Finally, I get a solution and want to share with all of you.

Below is my new scripts:
@echo off
set sec=5

:BEGIN
if %sec%==0 GOTO END
ECHO %TIME% aaa.txt is waiting %sec% seconds to start up...
PING 127.0.0.1 -n 2 -w 1000  > NUL
set /a sec -=1
CLS
goto BEGIN

:END
START /B C:\aaa.txt

[billrich]

C:\>type wait5.bat

Code: [Select]

@echo off


ECHO %TIME% aaa.txt is waiting 5 seconds to start up...

C:\batextra\sleep.exe 5
ECHO %TIME%

C:\aaa.txt
Output:
C:\>wait5.bat
 5:14:11.93 aaa.txt is waiting 5 seconds to start up...
 5:14:16.98