你也可以尝试这样做:

var myArray = [{'name': 'test'}, {'name':'test2'}];
var myObject = {'name': 'test'};
myArray.splice(myArray.indexOf(myObject),1);

在数组上使用splice函数。指定开始元素的位置和要删除的子序列的长度。

someArray.splice(pos, 1);

使用javascript的splice()函数。

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

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"}]

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"
  }
]

如果你在数组中的对象上没有任何你知道的属性(或者可能是唯一的)，但是你有一个你想要删除的对象的引用，你可以执行下面unregisterObject方法中的操作:

let registeredObjects = []; function registerObject(someObject) { registeredObjects.push(someObject); } function unregisterObject(someObject) { registeredObjects = registeredObjects.filter(obj => obj !== someObject); } let myObject1 = {hidden: "someValue1"}; // Let's pretend we don't know the hidden attribute let myObject2 = {hidden: "someValue2"}; registerObject(myObject1); registerObject(myObject2); console.log(`There are ${registeredObjects.length} objects registered. They are: ${JSON.stringify(registeredObjects)}`); unregisterObject(myObject1); console.log(`There are ${registeredObjects.length} objects registered. They are: ${JSON.stringify(registeredObjects)}`);