Question

我试图在属性与我的参数匹配时删除特定对象（此时轮子的名称是＆＃34;米其林＆＃34;）但我无法使其工作..

我怎么能这样做？

＆＃13;

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);

＆＃13;

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

＆＃13;

Answer 1

$.each方法的回调函数有两个参数：

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

但这最佳解决方案。通常，delete运算符用于从对象删除属性。如果你在数组上使用它，删除将删除数组项，但不会重新索引数组或更新其长度。这使它看起来好像是未定义的：

＆＃13;

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);

＆＃13;

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

＆＃13;

解决方案是使用filter方法接受回调提供的函数应用于数组中的每个项目。

＆＃13;

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);

＆＃13;

Answer 2

使用jQuery grep功能：

＆＃13;

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= $.grep(car.wheels,function(wheel){
  return wheel.name!='Michelin';
});
console.log(car);

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>

＆＃13;

如何在循环中删除嵌套的JS对象？

2 个答案: