Question

我写了颜色猜测javascript代码的代码。但问题是每次运行此代码时，它选择相同的颜色而不是选择随机颜色。我想知道是否有人帮助我。我将非常感激！

<head>
    <title>Color Guessing Game</title>
    <meta name="author" content="Wajeb Saab">
</head>

<body onload="do_game()">

    <script>
        var target_color;
        var target_index;
        var guessed_color;
        var guesses = 0;
        var color = ["red", "blue", "yellow", "crimson", "darkorange", "olive", "silver", "steelblue", "tomato", "green"];
        var finished = false;

        function do_game()
        {
            color.sort();
            var random_number = Math.random() * (color.length-1);
            target_index = Math.floor(random_number);
            target_color = color[target_index];
            alert("Hint: The target color is " + target_color);

            var text = "I am thinking of one of these colors:\n\n" + color.join(", ")
                        + "\n\n What color am I thinking of?";

            while(!finished)
            {
                guessed_color = prompt(text);
                guesses++;
                finished = check_guess();
            }

            myBody=document.getElementsByTagName("body")[0];
                myBody.style.background = target_color;
        }

        function check_guess()
        {
            if(color.indexOf(guessed_color) < 0) // check if guessed color is not included in array
            {
                alert("Sorry, I don't recognize your color.\n\n" + "Please try again.");
                return false;
            }
            else if(guessed_color > target_color)
            {
                alert("Sorry, your guess is not correct!\n\n" + "Hint: your color is alphabetically higher than mine.\n\n"
                      + "Please try again.");
                return false;
            }
            else if(guessed_color < target_color)
            {
                alert("Sorry, your guess is not correct!\n\n" + "Hint: your color is alphabetically lower than mine.\n\n"
                      + "Please try again.");
                return false;
            }
            else
            {

                alert("Congratulations! You have guessed the color!\n\n" + "It took you " + guesses + " guesses to finish the game.\n\n"
                      + "You can see the color in the background.");
                return true;
            }
        }

    </script>

</body>

Answer 1

问题是你需要在check_guesses（）函数中返回true时重置guessed_color和target_color，然后你需要将finished = false设置为do_game（）函数的第一行。像这样......

function do_game() {
    finished = false;
    color.sort();
    var random_number = Math.random() * (color.length - 1);
    target_index = Math.floor(random_number);
    console.log(target_index);
    target_color = color[target_index];
    var text = "I am thinking of one of these colors:\n\n" + color.join(", ") + "\n\n What color an I thinking of?";
    while(!finished) {
        guessed_color = prompt(text);
        guesses++;
        finished = check_guesses();
    }
}

function check_guesses() {
    if(color.indexOf(guessed_color) < 0) {
        console.log("Sorry I don't know what color that is");
        return false;
    } else if(guessed_color > target_color) {
        console.log("Incorrect!\n\n Your color is Alphabetically higher than mine");
        return false;
    } else if(guessed_color < target_color) {
        console.log("Incorrect\n\n Your color is lower than mine");
        return false;
    } else {
        console.log("You WIN!");
        guessed_color = null;
        target_color = null;
        return true;
    }
}

截至目前，你设置了颜色，然后检查它，但是当你返回true时没有重置它，游戏结束，颜色被设置，所以下次你运行它时它将是相同的颜色。

Answer 2

谢谢，但我的问题通过应用来解决。

颜色= color.sort（）;

每次在颜色猜谜游戏javascript中选择相同颜色而不是随机颜色

2 个答案: