这个问题在这里已有答案:
是否有可能避免“这个”赋值列表并以某种方式“传播”构造函数中指定的所有args?
class MyClass {
constructor(arg1, arg2, arg3, arg4, arg5) {
this.arg1 = arg1;
this.arg2 = arg2;
this.arg3 = arg3;
this.arg4 = arg4;
this.arg5 = arg5;
}
}
您可以使用一组键来迭代参数。
class MyClass {
constructor(...args) {
var keys = ['arg1', 'arg2', 'arg3', 'arg4', 'arg5']
keys.forEach((k, i) => this[k] = args[i]);
}
}
var instance = new MyClass('a', 'b');
console.log(instance);