Question

鉴于一个集合，如何删除仅与条件匹配的第一个项目？

例如，给定此集合：

[
  { id: 1, name: "don" },
  { id: 2, name: "don" },
  { id: 3, name: "james" },
  { id: 4, name: "james" }
]

过滤掉与{ name: "james" }匹配的第一个结果。

结果：

[
  { id: 1, name: "don" },
  { id: 2, name: "don" },
  { id: 4, name: "james" }
]

Answer 1

您正在寻找这样的解决方案吗？

使用Array.prototype.splice迭代并更新数组。

＆＃13;

var arr = [
  { id: 1, name: "don" },
  { id: 2, name: "don" },
  { id: 3, name: "james" },
  { id: 4, name: "james" }
];
// loop and remove the first match from the above array
for (i = 0; i < arr.length; i++) {
    if (arr[i].name == "james"){
         arr.splice(i, 1);
         break;
      }
  } 
// write into the browser console
console.log(arr);

＆＃13;

Answer 2

使用underscore.js，您可以使用 _.without 和 findwhere

var myarray = [
  { id: 1, name: "don" },
  { id: 2, name: "don" },
  { id: 3, name: "james" },
  { id: 4, name: "james" }
];
var arr = _.without(myarray, _.findWhere(myarray, {
 name: "james"
}));
console.log(arr);

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>

使用Loadash ，

var myarray = [
  { id: 1, name: "don" },
  { id: 2, name: "don" },
  { id: 3, name: "james" },
  { id: 4, name: "james" }
];


var result =_.without(myarray, _.find(myarray, { name: "james" }));
console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.js"></script>

Answer 3

与Lodash

var newArray = _.without(array, _.find(array, { name: "james" }));

删除集合中的第一个匹配项

3 个答案: