如果对象与新数组不匹配，则从前一个数组中删除对象

Question

我正在尝试将前一个数组中不匹配的值删除到新数组中的值。

var previous = [
    {m: "195/50 15"},
    {m: "195/50 16"},
    {m: "195/55 15"},
    {m: "195/55 16"},
    {m: "195/55 20"}
];
var newOne = [
    {m: "195/55 15"},
    {m: "195/55 16"},
    {m: "195/55 20"}
];

console.log( previous.filter((obj) => obj['m'] !== newOne['m']) );

如果新数组为空，则过滤器应返回一个空数组。

我上面尝试的方法不起作用。

Answer 1

尝试使用 some 来确定该值是否在另一个数组中：

 var previous = [ {m: "195/50 15"}, {m: "195/50 16"}, {m: "195/55 15"}, {m: "195/55 16"}, {m: "195/55 20"} ]; var newOne = [ {m: "195/55 15"}, {m: "195/55 16"}, {m: "195/55 20"} ]; const result = previous.filter(p => { return newOne.some(n => nm === pm); }) console.log(result);

Answer 2

对比应该是这样的

var previous = [
    {m: "195/50 15"},
    {m: "195/50 16"},
    {m: "195/55 15"},
    {m: "195/55 16"},
    {m: "195/55 20"}
];
var newOne = [
    {m: "195/55 15"},
    {m: "195/55 16"},
    {m: "195/55 20"}
];

console.log( previous.filter((obj, i) => newOne[i] && obj['m'] !== newOne[i]['m']) );

Answer 3

在您的代码中，您正在使用过滤器 function 运行前一个数组中的对象并将每个 object 值与返回undefined的 newOne['m'] 进行比较

    let newArray = []

    previous.forEach(pobj => {
      newOne.forEach(nobj => {
        if(pobj['m'] === nobj['m']){
          newArray.push(pobj)
        }
      })
    })

    console.log(newArray)