javascript - 按键组合对象数组

Question

我正在尝试在我的案例 ID 中按键组合/合并 2 个对象数组。

目标:

1. 我期待一个结果，其中我将按照示例包含所有 ID 为 1、2、3、4 的对象
2. 合并顺序不应影响结果中对象的数量，例如 combine(arr1,arr2) 或 combine(arr2,arr1) 应该具有相同数量的数组对象
3. 合并的顺序只能影响结果对象，例如 combine(arr1,arr2) arr2 key,values pair can override arr1 key,values just like deep jquery extend $.extend( true , arr1ObJ,arr2ObJ );

JSFIDDLE:https://jsfiddle.net/bababalcksheep/u2c05nyj/

示例数据:

var arr1 = [{
  id: 1,
  name: "fred",
  title: "boss"
}, {
  id: 2,
  name: "jim",
  title: "nobody"
}, {
  id: 3,
  name: "bob",
  title: "dancer"
}];
var arr2 = [{
  id: 1,
  wage: "300",
  rate: "day"
}, {
  id: 2,
  wage: "10",
  rate: "hour"
}, {
  id: 4,
  wage: "500",
  rate: "week"
}];
var Result = [{
  "id": 1,
  "name": "fred",
  "title": "boss",
  "wage": "300",
  "rate": "day"
}, {
  "id": 2,
  "name": "jim",
  "title": "nobody",
  "wage": "10",
  "rate": "hour"
}, {
  id: 3,
  name: "bob",
  title: "dancer" 
}, {
  id: 4,
  wage: "500",
  rate: "week"
}];

Answer 1

这是一个解决方案。它基本上遍历 arr2 的每个元素并检查是否存在具有匹配 ID arr1 的元素。如果是这样，它将用 arr2 的值更新 arr1 中的匹配元素。如果没有匹配项，它只是将 arr2 中的元素插入 arr1。

var arr1 = [{id: 1,name: 'fred',title: 'boss'}, 
            {id: 2,name: 'jim',title: 'nobody'}, 
            {id: 3,name: 'bob',title: 'dancer'}];

var arr2 = [{id: 1,wage: '300',rate: 'day'}, 
            {id: 2,wage: '10',rate:'hour'},
            {id: 4,wage: '500',rate: 'week'}];

function combineArrays(arr1, arr2) {
  for(var i = 0; i < arr2.length; i++) {
    // check if current object exists in arr1
    var idIndex = hasID(arr2[i]['id'], arr1);
    if(idIndex >= 0){
      //update
      for(var key in arr2[i]){
        arr1[idIndex][key] = arr2[i][key];
      }
    } else {
      //insert
      arr1.push(arr2[i]);
    }
  }

  return arr1;
}

//Returns position in array that ID exists
function hasID(id, arr) {
  for(var i = 0; i < arr.length; i ++) {
    if(arr[i]['id'] === id)
    {
      return i;
    }
  }

  return -1;
}

var combine = combineArrays(arr1, arr2);
output(combine);

/* pretty Print */
function output(inp) {
  var str = JSON.stringify(inp, undefined, 4);
  $('body').append($('<pre/>').html(str));
}

var arr1 = [{
  id: 1,
  name: 'fred',
  title: 'boss'
}, {
  id: 2,
  name: 'jim',
  title: 'nobody'
}, {
  id: 3,
  name: 'bob',
  title: 'dancer'
}];

var arr2 = [{
  id: 1,
  wage: '300',
  rate: 'day'
}, {
  id: 2,
  wage: '10',
  rate: 'hour'
}, {
  id: 4,
  wage: '500',
  rate: 'week'
}];

function combineArrays(arr1, arr2) {
  for (var i = 0; i < arr2.length; i++) {
    var idIndex = hasID(arr2[i]['id'], arr1);
    if (idIndex >= 0) {
      for (var key in arr2[i]) {
        arr1[idIndex][key] = arr2[i][key];
      }
    } else {
      arr1.push(arr2[i]);
    }
  }

  return arr1;
}

function hasID(id, arr) {
  for (var i = 0; i < arr.length; i++) {
    if (arr[i]['id'] === id) {
      return i;
    }
  }

  return -1;
}

var combine = combineArrays(arr1, arr2);
output(combine);

/* pretty Print */
function output(inp) {
  var str = JSON.stringify(inp, undefined, 4);
  $('body').append($('<pre/>').html(str));
}

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