Javascript 迭代问题 - 可变变量

Question

不是旧 JS 的专家所以这里开始

我有

store1.baseParams.competition = null;
store2.baseParams.competition = null;
store3.baseParams.competition = null;

我想做的是

for (i=1; 1<=3; 1++) {
    store + i +.baseParams.competition = null;
}

希望我想做的事情有意义 - 是否可能

基本上通过添加来创建一个变量/对象

干杯

Answer 1

实现此目的的一种方法是通过 eval() - (通常是一个非常糟糕的主意)

for (var i=1; i<=3; i++) {
    eval("store" + i + ".baseParams.competition = null;");
}

另一种更复杂但相对有效的方法是创建一个函数，使您能够在运行时动态地改变任意深度的对象层次结构。这是一个这样的函数:

/*
   Usage: 
     Nested objects:
     nested_object_setter(object, ['property', 'propertyOfPreviousProperty'], someValue);
     Top-level objects:
     nested_object_setter(object, 'property', someValue);
 */

function dynamic_property_setter_base(obj, property, value, strict) {
   var shouldPerformMutation = !strict || (strict && obj.hasOwnProperty(property));
   if(shouldPerformMutation) {
      obj[property] = value;    
   }
   return value;
}

function dynamic_property_setter(obj, property, value) {
   return dynamic_property_setter_base(obj, property, value, false);
}

function nested_object_setter(obj, keys, value) {
    var isArray = function(o) {
      return Object.prototype.toString.call(o) === '[object Array]';
    };

    //Support nested keys.
    if(isArray(keys)) {
       if(keys.length === 1) {
          return nested_object_setter(obj, keys[0], value);
       }

       var o = obj[keys[0]];
       for(var i = 1, j = keys.length - 1; i < j; i++)
          o = o[keys[i]];
       return dynamic_property_setter(o, keys[keys.length - 1], value);
    }

    if(keys != null &&
       Object.prototype.toString.call(keys) === '[object String]' &&
       keys.length > 0) {
       return dynamic_property_setter(obj, keys, value);
    }

    return null;
}

您的代码如下所示:

for(var i = 1; i <= 3; i++) 
  nested_object_setter(this, ['store' + i, 'baseParams', 'competition'], null);

这是另一个例子，在 JS 控制台中运行:

> var x = {'y': {'a1': 'b'}};
> var i = 1;
> nested_object_setter(this, ['x','y','a' + i], "this is \"a\"");
> x.y.a1
"this is "a""

另一种方法，恕我直言，这是最简单但可扩展性最低的方法:

this['store' + i].baseParams.competition = null;