China DOS Union

As the question says..
For example set /p a=Number or letter

How to write it?? (It's a randomly entered number)

I thought that using the regular expression of findstr would easily complete the task, but unexpectedly there was still a bug: when the last character of the string is the escape character ^, an incorrect judgment result will appear:
@echo off :begin cls set input= set /p input= Please enter a string: if "%input%"=="" goto begin echo. echo %input%|findstr "^*$">nul && echo The string you entered is purely numeric||( echo %input%|findstr "^*$">nul && echo The string you entered is purely alphabetic||echo The string you entered is neither purely numeric nor purely alphabetic ) echo. pause goto begin

Thank you, moderator大哥. I just tried it and it works very well...

I just don't know what the code means, heh heh...

Can you take some time to explain it... Thanks.

Wonderful~~Learned from the moderator~~ :))))

The simplest way, compare variable with number. If the variable is a number, the comparison will succeed. If the variable is not a number, the comparison fails.

if %a% GTR 0 echo %a% is a number

zerocq Please write out all of them, okay??

I'm a noob, only this sentence I don't understand~~

Direct comparison with numbers can cause errors! Test code:
@echo off
set /p num=Please enter a number:
if %num% geq 1 (echo YES) else echo NO
pause>nul
When you enter 10a, it returns YES!
I think there are two solutions:
1. findstr + regular expression, the previous moderator has used it;
2. Use the particularity of set /a:
For example: set /a num=10a, then num cannot be defined correctly. If it starts with a letter, the return result of num is 0. If it is pure numbers, it returns the original number. Using this feature, it can be used to roughly detect the input situation.