Hi, Please show me how to print out the below line in batch programming: perl -e "sleep (rand 30)" I tried: set random_sec=perl -e "sleep (rand 30)" echo %random_sec" and it does not work. Thanks for helping. Codelearner

Codelearner, You need to enclose the random_sec variable with "%". Like this: echo %random_sec% -- kptech

I tried echo random_sec = %random_sec% and its output is nothing as shown below: random_sec=

No problems setting the var and echoing here, perhaps you should check the syntax you are using in the Set command.. You have a space before the = in this echo random_sec = %random_sec% but it's not shown on the output random_sec=

Here is my code: set number_of_loop=%1 set /a counter=0 rem Number of loop must be enter: if "%number_of_loop%"=="" ( goto error ) :loop set /a counter+=1 set random_sec=perl -e "sleep (rand 30)" echo random_sec=%random_sec% echo Number of loops=%counter% if "%counter%"=="%number_of_loop%" ( goto end ) else ( goto loop ) :error echo Need to enter number of loop echo Usage: Script_Name number_of_loop :end Result: C:\WorkingFolderTemp>set number_of_loop=1 C:\WorkingFolderTemp>set /a counter=0 C:\WorkingFolderTemp>rem Number of loop must be enter: C:\WorkingFolderTemp>if "1" == "" (goto error ) C:\WorkingFolderTemp>set /a counter+=1 C:\WorkingFolderTemp>set random_sec=perl -e "sleep (rand 30)" C:\WorkingFolderTemp>echo random_sec=perl -e "sleep (rand 30)" random_sec=perl -e "sleep (rand 30)" C:\WorkingFolderTemp>echo Number of loops=1 Number of loops=1 Problem: cannot print out the random second