Artee's answer解释了语法错误发生的位置和方式。让我建议用一种不同的方式来编写它，它可以被认为更“像Erlang-like”一点，并且顺便摆脱了语法错误。
首先，让Computer变量包含计算机的选择，其格式与玩家的相同：

Computer = element(rand:uniform(3), {"Rock", "Paper", "Scissors"}),

然后，创建一个函数，该函数接受玩家的选择和计算机的选择，并返回获胜者：

who_wins("Rock", "Scissors") ->
    player;
who_wins("Rock", "Paper") ->
    computer;
who_wins("Paper", "Scissors") ->
    computer;
who_wins("Paper", "Rock") ->
    player;
who_wins("Scissors", "Rock") ->
    computer;
who_wins("Scissors", "Paper") ->
    player;
who_wins(Same, Same) ->
    tie.

(In最后一个子句，因为我们对两个参数使用相同的变量名，所以只有当参数相同时，此子句才匹配。）
然后，回到rps函数，我们可以将嵌套的if表达式替换为：

case who_wins(Player, Computer) of
    player ->
        Win();
    computer ->
        Lose();
    tie ->
        io:fwrite("It's a tie!")
end.

这里有几个错误：

• 没有“end”的乐趣（x2）Win = fun（）-〉io：fwrite（“你赢了！”）end，
• if语法

正确代码：

-module(project).
-export([rps/0]).

rps() ->
    Computer = rand:uniform(3),
    {ok,Player} = io:read("Rock, Paper, or Scissors?"),
    Win = fun() -> io:fwrite("You Win!") end,
    Lose = fun() -> io:fwrite("You Lose.") end,

    % player chooses rock (win, lose)
    if
        Player == "Rock" ->
            if
                Computer == 3 ->
                    Win();
                Computer == 2 ->
                    Lose()
            end;
    % player chooses paper (win, lose)
        Player == "Paper" ->
            if
                Computer == 1 ->
                    Win();
                Computer == 3 ->
                    Lose()
            end;
    % player chooses scissors (win, lose)
        Player == "Scissors" ->
            if
                Computer == 2 ->
                    Win();
                Computer == 1 ->
                    Lose()
            end
    end.