Windows Batch 中的 Double If 语句

我将从代码开始：

@echo off
set number=0

echo What would you like to add to %number%
echo 1. Add nothing
echo 2. Add 1
echo 3. Add number of your choosing

set /p test=

:: doesn't add any number
if "%test%" == "1" (
    echo You have choosen to not add anything
    pause
    goto end
)

::Adds number 1 to 0
if "%test%" == "2" (
    set /a number=number+1
    echo You have added 1 to the number
    goto end
)


::Gives choice to add 2 or 3 to number 0
if "%test%" == "3" (
    echo Which number would you like to add
    echo 1. Number 2
    echo 2. Number 3

    set /p option=

    ::Adds 2 to 0
    if "!option!" == "1"(
        echo You have added 2 to the number
        set /a number=number+2
        pause
    )

    ::Adds 3 to 0
    if "!option!" == "2"(
        echo You have added 3 to the number
        set /a number=number+3
        pause
    )
    goto end
)

:end
echo Your number is %number%
pause

这将数字声明为 0，然后给你一个选项，将 0、1、2 或 3 添加到数字 0。我已经这样做了，所以我可以测试在 if 语句中是否有一个 If 语句，该语句会询问对于输入，但我迷路了。

将 0 或 1 添加到数字可以正常工作，但是一旦您选择选项 3（想要添加 2 或 3），它就会崩溃。

我收到错误：命令的语法不正确

我猜这是一个愚蠢的错误，如果你能指出它，那就完美了。