按键组合对象数组

我试图在我的case id中按键组合/合并2个对象数组。

目的：

1. 我期待一个结果，我将有数组包含所有具有ID 1,2,3,4的对象
2. 合并顺序不应影响结果中的对象数量，例如combine(arr1,arr2)或combine(arr2,arr1)应该具有相同数量的对象的数组
3. 合并顺序只能影响结果对象，例如在combine(arr1,arr2) arr2键的情况下，值对可以覆盖arr1键，值就像深度jquery extend $ .extend（true，arr1ObJ，arr2ObJ）;

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

样本数据：

 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

怎么样的东西：

 function combineArrays(arr1, arr2, keyFunc) { var combined = [], keys1 = arr1.map(keyFunc), keys2 = arr2.map(keyFunc), pos1 = keys1.map(function (id) { return keys2.indexOf(id); }), pos2 = keys2.map(function (id) { return keys1.indexOf(id); }); arr1.forEach(function (item, i) { combined.push( $.extend(item, arr2[pos1[i]]) ); }); arr2.forEach(function (item, i) { if (pos2[i] === -1) combined.push( item ); }); return combined; } 

用作

 var combine = combineArrays(arr1, arr2, function (item) { return item.id; }); 

 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, keyFunc) { var combined = [], keys1 = arr1.map(keyFunc), keys2 = arr2.map(keyFunc), pos1 = keys1.map(function (id) { return keys2.indexOf(id); }), pos2 = keys2.map(function (id) { return keys1.indexOf(id); }); arr1.forEach(function (item, i) { combined.push( $.extend(item, arr2[pos1[i]]) ); }); arr2.forEach(function (item, i) { if (pos2[i] === -1) combined.push( item ); }); return combined; } var combine = combineArrays(arr1, arr2, function (item) { return item.id; }); output(combine); // // // /* pretty Print */ function output(inp) { var str = JSON.stringify(inp, undefined, 4); $('body').append($('

').html(str)); }  
  

Answer 2

这是一个解决方案。 它基本上遍历arr2的每个元素，并检查是否有一个匹配ID为arr1的元素。 如果是这样，它会使用arr2的值更新arr1中的匹配元素。 如果没有匹配，它只是将arr2中的元素推送到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); $('body').append($('

').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); $('body').append($('

').html(str)); }