Question

假设我有一个数组：

    members = [
        {name: 'Anna', class: 'one'}, 
        {name: 'Bob', class: 'two'},  
        {name: 'Chuck', class: 'two'}];

    removed = members.myRemoveByClass('two');     //something like

    // removed is {name: 'Bob', class: 'two'} 
    // members is [{name: 'Anna', class: 'one'}, {name: 'Chuck', class: 'two'}]

我正在寻找myRemoveByClass的东西。 ES2015很好或使用Lodash。该阵列将已经订购。 None中的questions我seen match的用途looking。

Answer 1

您可以创建自己的Array类：

 class Members extends Array {
   removeByClass(className) {
     for(const [index, member] of this.entries())
        if(member.class === className) 
           return this.splice(index, 1)[0];
   }
}

用作

 const members = new Members([ {/*...*/}, {/*...*/} ]);
 members.removeByClass("...");

PS：“类”是一个非常糟糕的名称，因为它是保留关键字

Answer 2

这也可以

<!doctype html>
<html>
    <head>
        <meta charset="utf-8">
        <meta name="viewport" content="width=device-width, initial-scale=1">
        <meta name="csrf-token" content="{{ csrf_token() }}">
        <title>Laravel</title>
    </head>
    <body>
        <div class="content" id="app">
            <div class="title m-b-md">
                <hello-world-component></hello-world-component>
            </div>
        </div>
    </body>
<script src="{{ asset('js/app.js') }}"></script>
</html>

Answer 3

您可以找到第一次出现的索引，并检查该索引并从数组中拼接对象。

var members = [{ name: 'Anna', class: 'one' }, { name: 'Bob', class: 'two' }, { name: 'Chuck', class: 'two' }],
    index = members.findIndex(o => o.class === 'two'),
    removed = index !== -1 && members.splice(index, 1);

console.log(removed);
console.log(members);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 4

您可以使用Array.prototype.findIndex()：

findIndex()方法返回数组满足提供的测试功能的第一个元素的 index 。否则，它返回-1，表示没有元素通过测试。

然后通过该 index splice()对象。

var members = [
        {name: 'Anna', class: 'one'}, 
        {name: 'Bob', class: 'two'},  
        {name: 'Chuck', class: 'two'}];
var idx = members.findIndex(p => p.class=="two");
var removed = members.splice(idx,1);     
console.log(removed);
console.log(members);

Answer 5

您可以使用without和findWhere

var arr = _.without(members , _.findWhere(members , {
 class: "two"
}));

Answer 6

简单地：

function removeByClass(obj, prop) {
    for (let i = 0; i < obj.length; i++) {
        if (obj[i].class == prop) obj.splice(i, 1);
    }   
}

Answer 7

您可以使用some方法找到需要删除的member，然后返回true停止循环。

let members = [
  {name: 'Anna', class: 'one'}, 
  {name: 'Bob', class: 'two'},  
  {name: 'Chuck', class: 'two'}
];

function myRemoveByClass(members, className) {
  let foundMember = null;
  members.some((member, index) => {
    if (member.class == className) {
      members.splice(index, 1);
      foundMember = member;
      return true;
    }
  });
  
  return foundMember;
}

let removed = myRemoveByClass(members, 'two');
console.log(removed);
console.log(members);

如乔纳斯（Jonas）指出的here，您应该避免使用class，因为它是保留字。

Answer 8

您可以将Array.prototype.find()与Array.prototype.splice()组合使用

代码示例：

const members = [{ name: 'Anna', class: 'one' }, { name: 'Bob', class: 'two' }, { name: 'Chuck', class: 'two' }];

members.find((o, i) => {
  if (o.class === 'two') {
    members.splice(i, 1); // Find and remove first matching element
    return true;
  }
  return false;
});

console.log(members);

查找和删除Javascript对象数组中的第一个匹配元素

8 个答案: