具有es6箭头功能

let someArray = [
                 {name:"Kristian", lines:"2,5,10"},
                 {name:"John", lines:"1,19,26,96"}
                ];
let arrayToRemove={name:"Kristian", lines:"2,5,10"};
someArray=someArray.filter((e)=>e.name !=arrayToRemove.name && e.lines!= arrayToRemove.lines)

虽然这可能不适合这种情况，我发现前几天，如果你不需要改变数组的大小，你也可以使用delete关键字从数组中删除一个项目。

var myArray = [1,2,3];

delete myArray[1];

console.log(myArray[1]); //undefined

console.log(myArray.length); //3 - doesn't actually shrink the array down

最简单的解决方案是创建一个映射，按名称存储每个对象的索引，如下所示:

//adding to array
var newPerson = {name:"Kristian", lines:"2,5,10"}
someMap[ newPerson.name ] = someArray.length;
someArray.push( newPerson );

//deleting from the array
var index = someMap[ 'Kristian' ];
someArray.splice( index, 1 );

const someArray = [{name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}];

我们得到对象的索引它的name属性值为"Kristian"

const index = someArray.findIndex(key => key.name === "Kristian");
console.log(index); // 0

通过使用拼接函数我们删除了name属性值为“Kristian”的对象

someArray.splice(index,1);
console.log(someArray); // [{name:"John", lines:"1,19,26,96"}]

使用javascript的splice()函数。

这可能会有帮助:http://www.w3schools.com/jsref/jsref_splice.asp

这里有一个映射和拼接的例子

const arrayObject = [ {name: "name1"， value: "value1"}， {name: "name2"， value: "value2"}， {name: "name3"， value: "value3"}， ]; let index = arrayObject.map((item) => item.name).indexOf("name1"); If(索引> -1){ arrayObject。拼接(指数(1); console.log(“结果”,arrayObject); ｝

输出

Result [
  {
    "name": "name2",
    "value": "value2"
  },
  {
    "name": "name3",
    "value": "value3"
  }
]