javascript - 如何引用 JavaScript 函数内的不同位置

Question

我正在尝试从 JavaScript 中的其他函数中访问设置和值/函数，并且相信我可能缺少一个简单的规则。这是一个非常简化的代码示例:

function h(){
    // example settings I want to declare for h()
    this.settings = {
        critical: 400,
        readCritical: function(){
            return this.settings.critical; // /!\ but this does not work
        }
    }
    this.hsub = function(){
        return this.settings.critical; // /!\ this does not work either
    }

}
var x = new h();
console.log(x.settings.critical); // This works fine
console.log(x.settings.readCritical()); // I can *call* this fine, but it can't read outside itself
console.log(x.hsub()); // I can also call this but same problem
console.log(h.settings); // `undefined`; I understand I can't see directly into the function

您可以看到 this.settings.ritic 值，我试图从相应的函数 readCritical 和 hsub 内部访问该值。我怎样才能做到这一点？特别是来自实例化的x。

也是一个相关的次要问题，我更愿意声明 var settings = {} 而不是 this.settings。这是可能的还是更好的？

Answer 1

您需要做的唯一更改是从 readCritical 函数中删除 .settings。

hsub 按原样工作。不确定为什么您认为存在问题

function h(){
    this.settings = {
        critical: 400,
        readCritical: function(){
            return this.critical; // reference `critical` directly
        }
    }
    this.hsub = function(){
        return this.settings.critical; // this works just fine
    }

}

var x = new h();

console.log(x.settings.critical);       // 400
console.log(x.settings.readCritical()); // 400
console.log(x.hsub());                  // 400

// This is expected to be `undefined`
console.log(h.settings);