我有一个包含一些预定义数据的数组
var data = [
{amount:20, speed:100},
{amount:40, speed:50}
];
然后我将数据添加到上述数组
function addMore() {
data = appendObjTo(data, {amount: 1500,speed:100});
}
function appendObjTo(thatArray, newObj) {
const frozenObj = Object.freeze(newObj);
return Object.freeze(thatArray.concat(frozenObj));
}
可以很好地添加数据,但是由于某些原因,我无法更改新数据的值
function runData() {
perSec = 0;
$.each(data, function( key, value ) {
perSecCalc = Math.round(value.speed/60);
perSec += perSecCalc;
// Below line works only for predefined objects, but not objects from "addMore()"
data[key].amount = value.amount-perSec;
});
setTimeout(function() {
runData();
},1000);
}
“ var data”中的预定义对象被更改时,“ addMore”中动态添加的数据不会更改。
为什么新数据没有变化?
更新: 看到这个fiddle
答案 0 :(得分:2)
您正在appendObjTo
函数中使用Object.freeze。 Object.freeze的定义-Object.freeze()。
由于使用Object.freeze()
创建了FrozenObj,因此不允许更改这些值。另外,您在控制台上没有任何错误。启用严格模式后,JS没有显示错误。我已经修改了小提琴以使其包含严格模式,并且您在执行data[key].amount = value.amount-perSec;
时可以看到它抛出错误。我还附加了一个小提琴来玩Object.freeze()方法,您可以自己尝试。
JS
(function () {
"use strict";
var data = [
{amount:20, speed:100}
];
function runData() {
var perSec = 0;
$.each(data, function( key, value ) {
var perSecCalc = Math.round(value.speed/60);
perSec += perSecCalc;
// Below line works only for predefined objects, but not objects from "addMore()"
data[key].amount = value.amount-perSec;
$('#test').prepend(data[key].amount+'<br>');
});
setTimeout(function() {
runData();
},1000);
}
function appendObjTo(thatArray, newObj) {
const frozenObj = Object.freeze(newObj);
return Object.freeze(thatArray.concat(frozenObj));
}
function addMore() {
data = appendObjTo(data, {amount: 1500,speed:100});
}
setTimeout(function() { addMore(); },1500);
runData();
})();
var arr = [10, 20, 30];
console.log(arr);
arr = Object.freeze(arr.concat([40, 50]));
console.log(arr);
arr[3] = 80;
console.log(arr); // doesn't change
// arr.push(60); // error, cannot add property 5, object is not extensible
arr = Object.freeze(arr.concat([{ x: 100 }]));
console.log(arr);
arr[5].x = 200;
console.log(arr); // changes, as Object.freeze only locks the first level values.