javascript - 如何删除循环内嵌套的 JS 对象？

Question

这个问题已经有答案了:

Delete an element inside a JSON array by value with jQuery (4 个回答)

已关闭 6 年前。

当属性与我的参数匹配时(此处当轮子的名称是“米其林”时)，我试图删除特定对象，但我无法使其工作......

我该怎么做？

var car = {
  type: "Fiat",
  model: "500",
  color: "White",
  wheels: [{
    name: "Goodyear",
    color: "Red"
  }, {
    name: "Goodyear",
    color: "Yellow"
  }, {
    name: "Goodyear",
    color: "Black"
  }, {
    name: "Michelin",
    color: "Blue"
  }]
};

$.each(car.wheels, function() {
  if (this.name == "Michelin") {
    delete this;
  }
})
console.log(car);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer 1

$.each 方法的回调函数采用两个参数:

$.each(car.wheels, function(i,item) {
   if (item.name == "Michelin") {
       delete car.wheels[i];
   }
});

但这不是最佳解决方案。通常，delete运算符用于删除属性对象。如果在数组上使用它，删除将删除数组项，但不会重新索引数组或更新其长度。这使得它看起来好像未定义:

var car = {
      type: "Fiat",
      model: "500",
      color: "White",
      wheels: [{
        name: "Goodyear",
        color: "Red"
      }, {
        name: "Goodyear",
        color: "Yellow"
      }, {
        name: "Goodyear",
        color: "Black"
      }, {
        name: "Michelin",
        color: "Blue"
      }]
};
$.each(car.wheels, function(i,item) {
    if (item.name == "Michelin") {
       delete car.wheels[i];
    }
});
console.log(car);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

解决方案是使用 filter 方法，该方法接受应用于数组中每个项目的回调提供的函数。

var car = {
      type: "Fiat",
      model: "500",
      color: "White",
      wheels: [{
        name: "Goodyear",
        color: "Red"
      }, {
        name: "Goodyear",
        color: "Yellow"
      }, {
        name: "Goodyear",
        color: "Black"
      }, {
        name: "Michelin",
        color: "Blue"
      }]
 };
car.wheels=car.wheels.filter(function(wheel){
  return wheel.name!='Michelin';
});
console.log(car);