Question

所以我试图编写一个经过并选择专长的应用程序并将其添加到html文件中。当我有一些寻找特定元素的东西时，它会相应地通过这些陈述。但是，当说，它没有提出要求时，它会再次通过该过程（它应该），但是然后不会将它添加到html中。事实上，它只是说＆＃34; undefined＆＃34;。

这是有问题的代码。（我已经包含了整个代码的截断版本。）

function feats(){
    var ft = prompt("Write your feat here.");
    var feat;
    switch (ft) {
        case 'alert':
            feat = "<b>Alert:</b> +5 initative. You cannot be surprised while you are conscious. Other creatures don't gain advantage on attack rolls against you as a result of being hidden from you.";
            initiative = initiative + 5;
            break;  
case 'defensive duelist':
        if (dex >=13) {
            feat = "<b>Defensive Duelist:</b> When you are wielding a finesse weapon with which you are proficient and another creature hits you with a melee attack, you can use your reaction to add your proficiency bonus to your AC for that attack, potentially causing the attack to miss you."
        } else {
            alert("You do not have sufficient dexterity for this feat. Try another one.");
            feats();
        }
        break;
        default:
            feat = " ";
    } 
    return feat;
}

feats（）是此开关/案例语句所在的函数。它应该在函数结束时返回专长，但是当dex不大于或等于13时它不会返回。那么当再次调用该函数时，如何让它返回另一个专长。

Answer 1

您的代码存在的问题是您没有返回函数调用feats()。这是必要的原因，因为否则逻辑继续，打破案例结构和returns feat。由于变量feat永远不会赋值，因此您获得undefined的值。要解决这个问题，你需要返回函数，这意味着函数将返回回调的值，在这种情况下是相同的函数 - 递归函数。

示例：我手动设置变量dex，因为代码示例中未包含该变量。

var dex = 12;

function feats() {
    var ft = prompt("Write your feat here.");
    var feat;
    switch (ft) {
        case 'alert':
            feat = "<b>Alert:</b> +5 initative. You cannot be surprised while you are conscious. Other creatures don't gain advantage on attack rolls against you as a result of being hidden from you.";
            initiative = initiative + 5;
            break;
        case 'defensive duelist':
            if (dex >= 13) {
                feat = "<b>Defensive Duelist:</b> When you are wielding a finesse weapon with which you are proficient and another creature hits you with a melee attack, you can use your reaction to add your proficiency bonus to your AC for that attack, potentially causing the attack to miss you."
            } else {
                alert("You do not have sufficient dexterity for this feat. Try another one.");
                return feats();
            }
            break;
        default:
            feat = " ";
    }
    return feat;
}

Answer 2

您只需更改行

即可

feats()

在else语句中

return feats()

但重新组织可能更简单，这取决于你对函数中多个return语句的感受。

function feats(){
    var ft = prompt("Write your feat here.");
    switch (ft) {
        case 'alert':
            initiative = initiative + 5;
            return "<b>Alert:</b>...";
        case 'defensive duelist':
            if (dex >=13) {
                return "<b>Defensive Duelist:</b>...";
            } else {
                alert("You do not have...");
                return feats();
            }
        default:
            return " ";
    } 
}

对我来说，至少，这读起来要好得多。

Answer 3

将feats();放在ifelse声明之外。

似乎无法做到这一点（函数调用） - Javascript

3 个答案: