Creating a new object, not a reference
This question already has an answer here:
最简单的版本是使用JSON.parse/stringify
,最快的就是使用一个普通的克隆方法:
/* simplest */
var clone = JSON.parse(JSON.stringify(obj));
/* fastest */
function clone(obj) {
if (obj == null ||typeof obj != "object") return obj;
var copy = obj.constructor();
for (var attr in obj) {
if (obj.hasOwnProperty(attr)) copy[attr] = obj[attr];
}
return copy;
}
var clone2 = clone(obj);
You could write a deep clone Method, which copies every value of every property of your Object to a new one.
Note i extend Object.prototype to avoid type checking and for simplicities sake, this could be changed, if you feel unpleasent with it
Object.defineProperty(Object.prototype, "clone", {
enumerable : false,
value: function(deep) {
deep |= 0;
var type = typeof this;
if (type !== "object") {
return this.valueOf();
}
var clone = {};
if (0 === deep) {
for (var prop in this) {
clone[prop] = this[prop];
}
} else {
for (var prop in this) {
if ( typeof this[prop] !== "undefined" && this[prop] !== null)
clone[prop] = ( typeof this[prop] !== "object" ? this[prop] : this[prop].clone(deep - 1));
else
clone[prop] = "";
}
}
return clone;
}
});
Object.defineProperty(Array.prototype, "clone", {
enumerable : false,
value:function(deep) {
deep |= 0;
var clone = [];
if (0 === deep)
clone = this.concat();
else
this.forEach(function(e) {
if ( typeof e !== "undefined" && e !== null)
clone.push(( typeof e !== "object" ? e : e.clone(deep - 1)));
else
clone.push("");
});
return clone;
}
});
Example output and a Demo
var first = {
var1:0,
var2:0
var3:0
};
var second = first.clone(Infinity);
first.var1++;
console.log (first.var1,second.var1,second); //1 , 0
To apply this to your code, you just have to clone the Object app.Defaults = app.Variables.clone()
The first argument is the level of deepness. If omitted, only the first level is cloned, which would be enough in this case.
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