Batch file IF statement only using first character - batch-file

Question

Not quite sure what I'm doing wrong. For the most part, this works. If I were to type just 1, I would go to opt1.
The problem is, if I type "11", "1111", or even "1234567" it always goes to opt1. The only time it seems to not select opt1 is when the first number is something other than 1.
Likewise, entering 21 chooses option 2. The only way I've been able to get this to work as intended, which is only entering 1, 2, or 3 selects the respective options, is to omit the IF NOT statement.
Could someone kindly point me in the right direction?
#ECHO OFF
CLS
:MAIN_MENU
ECHO Welcome Menu
ECHO.
ECHO 1 - Option 1
ECHO 2 - Option 2
ECHO 3 - Option 3
ECHO.
SET ans=
SET /P ans="Select your option and then press ENTER: "
IF NOT "%ans%" == "" SET ans=%ans:~0,1%
IF "%ans%" == "1" GOTO opt1
IF "%ans%" == "2" GOTO opt2
IF "%ans%" == "3" GOTO opt3
ECHO "%ans%" is not a valid option, please try again!
ECHO.
PAUSE
CLS
GOTO MAIN_MENU
:opt1
ECHO This is option 1
PAUSE
CLS
GOTO MAIN_MENU
:opt2
ECHO This is option 2
PAUSE
CLS
GOTO MAIN_MENU
:opt3
ECHO This is option 3
PAUSE
CLS
GOTO MAIN_MENU

Answer 1

With the code you snipped I can't replicate the bug, however, I do remember to face that very same problem long time ago.
The answer is the quotes; just add it and you will avoid more complex code:
:input
cls
set ans=:
set /p ans=
if "%ans%" equ "1" (
goto opt1
) else (
goto input
)
:opt1
echo success!
EDIT: Regarding the first ans that was set, it is to clean the shell memory. Sometimes a variable could be stored in Windows memory, and even if you do not input anything to the set /p variable, and only press return, the batch will check for the variable and it will already have a value previously set in the memory. Example:
set /p option=
The user input "1". As a result, Windows will store option=1, or, %option% resulting in 1.
So, the even if you do not input anything, the next code will move forward as the answer is previously stored in the memory:
set /p option=
The user does NOT input anything, only press return. The code will move on:
if "%option%" equ "1" do (
echo %option% is equal to 1 because it was previously stored.
)
EDIT: Here is the code, with comments and explanations:
#echo off
:main_menu
cls
echo Welcome menu
echo.
echo 1 - option 1
echo 2 - option 2
echo 3 - option 3
echo.
rem Blank "ans" will clean the memory for a previous set value.
set ans=
rem You can either quote the whole variable with its option, ot not quote it at all.
rem However, its always better to quote it, as follows.
set /p "ans=select your option and then press enter:"
if "%ans%" == "1" goto opt1
if "%ans%" == "2" goto opt2
if "%ans%" == "3" goto opt3
rem The following line will force a loop to main menu if only return is pressed; blank options will not display message.
if "%ans%" == "" goto main_menu
rem In case that another option rather than 1, 2 or 3 are pressed, the loop will warn you, and then return to main menu.
echo "%ans%" is not a valid option, press any key to try again!
pause > nul
goto main_menu
:opt1
cls
echo this is option 1
pause
goto main_menu
:opt2
cls
echo this is option 2
pause
goto main_menu
:opt3
cls
echo this is option 3
pause
goto main_menu
Regarding the %ans:~0,1%, it is to limit the string to a certain position.
So, for example, %any_variable:~0,5% is saying "Start the string since character 0, and go ahead until character 5". You can try that out in the following example. Notice that all echo lines will type out the very same text:
#echo off
set example=Batch files are kinda cool
echo %example%
echo %example:~0,11% are kinda cool
echo Batch files %example:~12%
pause

Answer 2

As it turns out, there were remnants in the code that needed to be removed.
IF NOT "%ans%" == "" SET ans=%ans:~0,1%
IF "%ans%" == "1" GOTO opt1
I'm not entirely sure what that first line was setting however, removing it resolved the issue.
Thanks for all the help, much appreciated!