Question

我有一系列停车对象，每个停车场有4层，每层有10个车位

var parking = [
  {type:'car',plateNumber:'D555',parkingLevel:L1,parkingNumber:p1},
  {type:'car',plateNumber:'M225',parkingLevel:L2,parkingNumber:p9},
  {type:'car',plateNumber:'T765',parkingLevel:L2,parkingNumber:p2},
  {type:'car',plateNumber:'V366',parkingLevel:L4,parkingNumber:p3},
  {type:'car',plateNumber:'D121',parkingLevel:L1,parkingNumber:p5},
  {type:'car',plateNumber:'S909',parkingLevel:L3,parkingNumber:p1},
  {type:'car',plateNumber:'H551',parkingLevel:L4,parkingNumber:p5},
  {type:'car',plateNumber:'S305',parkingLevel:L3,parkingNumber:p1}
]

我想确定当我们将新物品推到停车场阵列时，避免重复停车场我的意思是我们不能将p1和p1推到L1

这就是我的尝试：

$.each(parking, function (i, item) {
    var newCar = {
        level: car.parkingLevel,
        number: car.parkingNumber
    };
    var jsonFL = $.map(filterParking, function(v,i) {
        return JSON.stringify(v);
    });
    jsonFL.indexOf( JSON.stringify(newCar) ) > -1 || filterParking.push(newcar);
});

Answer 1

您可以使用地图。

var map = {}

// when pushing on the array
map[car.plateNumber] = car

// when removing a car
delete map[car.plateNumber]

// check if a car is parked
"undefined" !== typeof map["D555"]

// bonus: you can now get cars by plate number
map["D555"]  // {type:'car',plateNumber:'D555',parkingLevel:L1,parkingNumber:p1}

下划线解决方案会增加不必要的依赖性，并且可能会很慢，具体取决于您添加汽车的频率以及阵列增长的大小。

Answer 2

var parking = [
  {type:'car',plateNumber:'D555',parkingLevel:L1,parkingNumber:p1},
  {type:'car',plateNumber:'M225',parkingLevel:L2,parkingNumber:p9},
  {type:'car',plateNumber:'T765',parkingLevel:L2,parkingNumber:p2},
  {type:'car',plateNumber:'V366',parkingLevel:L4,parkingNumber:p3},
  {type:'car',plateNumber:'D121',parkingLevel:L1,parkingNumber:p5},
  {type:'car',plateNumber:'S909',parkingLevel:L3,parkingNumber:p1},
  {type:'car',plateNumber:'H551',parkingLevel:L4,parkingNumber:p5},
  {type:'car',plateNumber:'S305',parkingLevel:L3,parkingNumber:p1}
]


var newCar = {
  parkingLevel: car.parkingLevel,
  parkingNumber: car.parkingNumber
};

for(var i=parking.length-1; i>=0; i--){
  if(parking[i].parkingLevel === newCar.parkingLevel && parking[i].parkingNumber === newCar.parkingNumber)
    break;

  if(i === 0){
    parking.push(newCar);
  }
}

Answer 3

Check whether a car is already parked in the given slot before adding to the collection. Assuming newcar is the new car object you are trying to add, then:

var parkedCars = parking.filter(function(c){ return c.parkingLevel === newcar.parkingLevel && c.parkingNumber === newcar.parkingNumber });
if(parkedCars.length === 0){ //this means there isnt any car parked in that level and slot
  parking.push(newcar);
}

Answer 4

我使用了underscore.js：

function containsObj(obj, objArray) {
   var obj= _.find(objArray, function(val){ return _.isEqual(obj, val);});
   return (_.isObject(obj))? true:false;
}

只需包含库文件并使用此功能。

将对象项目推入数组时避免重复

4 个答案: