Hi All,

Am curious to know how I'd format the %%i value to display two digits. (01, 02, 03, ..., 09)
I currently have ...Code: [Select]for /l %%i in (1,1,%MonitorLoops%) do (
if %%i LEQ 9 (set %%i=0%%i)
echo Check %%i - Performed at %TIME%
)And not surprisingly, not getting the result I was looking for.
Can ANYONE part with suggestion on how I'd answer this formatting query ??

Cheers,
CameronI'm sure this could be done in a very clever way, but down and dirty also works:

Code: [Select]@echo off
setlocal enabledelayedexpansion
for /l %%i in (1,1,%MonitorLoops%) do (
set num=%%i
if %%i LEQ 9 set num=0%%i
echo Check !num! - Performed at %TIME%
)

How's the project coming Cameron?

How about this...

using set /a avoids a string comparison

Code: [Select]@echo off
setlocal enabledelayedexpansion
set Monitorloops=20
for /l %%i in (1,1,%MonitorLoops%) do (
set /a num=%%i
set pad=
if !num! LEQ 9 set pad=0
set num=!pad!!num!
echo Check !num! - Performed at %TIME%
)

Or do the padding in a loop

Code: [Select]@echo off
setlocal enabledelayedexpansion
set Monitorloops=20
REM 1=no padding, 2 upwards=pad with zeros
set digits=2
for /l %%i in (1,1,%MonitorLoops%) do (
set /a num=%%i
set max=&set pad=
for /l %%p in (2,1,%digits%) do set max=9!max!&if !num! LEQ !max! set pad=0!pad!
set num=!pad!!num!
echo Check !num! - Performed at %TIME%
)
Many thanks SIDEWINDER & Dias ,

Took a snippet from each of your offerings to keep it short & sweet ...Code: [Select]for /l %%i in (1,1,%MonitorLoops%) do (
set /a num=%%i
if !num! LEQ 9 set num=0!num!
@echo Check !num! - Performed at %TIME% >> %OutputLog%
Quote from: Sidewinder on JUNE 03, 2008, 11:32:23 AM

How's the project coming Cameron?