前些天做项目, 小伙伴遇到一个js多对象数组, 求唯一问题, 思考了很久, 没有找到好的方案. 小伙伴使用了递归, 个人感觉不是最好的方案.
今天有空重写了一个.

原理分析

  • 遍历数组中的对象, 第一个对象和接下来所有的对象进行对比, 如果碰到一致的, 将其删除. 自身对象进行处理.

  • 问题主要处在删除后, 数组的索引发生变更, 之前没有考虑到. 一直不正确. 今天把代码贴出. 欢迎拍砖.! :)

  • es6 code 需要babeljs转换下

    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    13
    14
    15
    16
    17
    18
    19
    20
    21
    22
    23
    24
    25
    26
    27
    28
    29
    30
    31
    32
    33
    34
    35
    36
    37
    38
    39
    40
    41
    42
    43
    44
    45
    46
    47
    48
    49
    50
    51
    52
    53
    54
    55
    56
    57
    58
    59
    60
    61
    62
    63
    64
    65
    66
    67
    68
    69
    70
    71
    72
    73
    74
    75
    76
    77
    78
    79
    80
    81
    // es6 code
    const uniq = (source = [], fields = [], callback = ()=>{}) => {
      for(let i = 0; i < source.length; i++){
        for(let j = i+1; j < source.length; j++){
            /*
             * 对比对象中的字段(传入的参数)是否相等 返回一个和fileds数组相同数量的true, false数组,
             * 本列子返回[false, true], 对返回的true, false数组reduce计算, 看看对象是否相等, 如果相等
             * 调用callback, 删除原数组中相同的对象, 原数组相同的对象删除后, for循环需要回退.
             */
          if(fields.map((field) => {
            return source[i][field] == source[j][field];
          }).reduce((prev, next) => prev&&next)){
            source[i] = callback(source[i]);
            source.splice(j, 1);
            j--; // 回退for
          }
        }
      }
      return source;
    };

    let source = [
      {name: 'luo', age: 3, count:1}, 
      {name: 'v', age: 2, count:1}, 
      {name: 'luo', age: 3, count:1},
      {name: 'luo', age: 3, count:1},
      {name: 'luo', age: 3, count:1},
      {name: 'v', age: 2, count:1}, 
      {name: 'luo', age: 3, count:1}, 
      {name: 'v', age: 3, count:1}, 
      {name: 'v', age: 2, count:1}, 
      {name: 'v', age: 42, count:1}, 
    ];

    let ret = uniq(source, ['name', 'age'], (item) => {
      item.count += 1;
      return item;
    });

    * es5 转换好的code

    // es5 code
    'use strict';

    var uniq = function uniq() {
      var source = arguments.length <= 0 || arguments[0] === undefined ? [] : arguments[0];
      var fields = arguments.length <= 1 || arguments[1] === undefined ? [] : arguments[1];
      var callback = arguments.length <= 2 || arguments[2] === undefined ? function () {} : arguments[2];

      var _loop = function _loop(i) {
        var _loop2 = function _loop2(_j) {
          if (fields.map(function (field) {
            return source[i][field] == source[_j][field];
          }).reduce(function (prev, next) {
            return prev && next;
          })) {
            source[i] = callback(source[i]);
            source.splice(_j, 1);
            _j--;
          }
          j = _j;
        };

        for (var j = i + 1; j < source.length; j++) {
          _loop2(j);
        }
      };

      for (var i = 0; i < source.length; i++) {
        _loop(i);
      }
      return source;
    };

    var source = [{ name: 'luo', age: 3, count: 1 }, { name: 'v', age: 2, count: 1 }, { name: 'luo', age: 3, count: 1 }, { name: 'luo', age: 3, count: 1 }, { name: 'luo', age: 3, count: 1 }, { name: 'v', age: 2, count: 1 }, { name: 'luo', age: 3, count: 1 }, { name: 'v', age: 3, count: 1 }, { name: 'v', age: 2, count: 1 }, { name: 'v', age: 42, count: 1 }];

    var ret = uniq(source, ['name', 'age'], function (item) {
      item.count += 1;
      return item;
    });
    //console.log(ret)