Question

我需要使用javascript基于键值合并两个json对象。

我有两个不同的变量g和c。

术语：所有值都需要合并。

var g = [ { id: 36, name: 'AAA', goal: 'yes' },
    { id: 40, name: 'BBB', goal: 'yes' },
    { id: 57, name: 'CCC', goal: 'yes' },
    { id: 4, name: 'DDD', goal: 'yes' },
    { id: 39, name: 'EEE', goal: 'yes' },
    { id: 37, name: 'FFF', goal: 'yes' },
    { id: 59, name: 'GGG', goal: 'yes' },
    { id: 50, name: 'III', goal: 'yes' },
    { id: 43, name: 'HHH', goal: 'yes' },
    { id: 35, name: 'JJJ', goal: 'yes' } ]


 var c = [ { id: 36, name: 'AAA', circle: 'yes' },
    { id: 40, name: 'BBB', circle: 'yes' },
    { id: 57, name: 'CCC', circle: 'yes' },
    { id: 42, name: 'ZZZ', circle: 'yes' },
    { id: 4, name: 'DDD', circle: 'yes' },
    { id: 39, name: 'EEE', circle: 'yes' },
    { id: 37, name: 'FFF', circle: 'yes' },
    { id: 59, name: 'GGG', circle: 'yes' },
    { id: 43, name: 'HHH', circle: 'yes' },
    { id: 35, name: 'JJJ', circle: 'yes' },
    { id: 100, name: 'JJJ', circle: 'yes' } ]

我尝试了以下代码：但我合并了'c'变量中具有相同id的内容。但我需要合并比较'g'和'c'。

   var arrayList = [];
    for(var i in g) {
        var getid = g[i].id;
        var getname = g[i].name;
        var getgoal = g[i].goal;
        for(var j in c){
            var compareid = c[j].id;
            if(getid == compareid){
                var obj = {};
                obj.id = getid;
                obj.name = getname;
                obj.goal =  'yes';
                obj.circle = 'yes';
                console.log(obj);
                arrayList.push(obj);
            }
         }

    }
    console.log(arrayList);

预期产出：

[ { id: 36, name: 'AAA', goal: 'yes',circle: 'yes' },
    { id: 40, name: 'BBB', goal: 'yes',circle: 'yes' },
    { id: 57, name: 'CCC', goal: 'yes',circle: 'yes' },
    { id: 4, name: 'DDD', goal: 'yes' ,circle: 'yes' },
    { id: 39, name: 'EEE', goal: 'yes' ,circle: 'yes' },
    { id: 37, name: 'FFF', goal: 'yes' ,circle: 'yes'},
    { id: 59, name: 'GGG', goal: 'yes' ,circle: 'yes'},
    { id: 50, name: 'III', goal: 'yes' ,circle: 'no'},
    { id: 43, name: 'HHH', goal: 'yes' ,circle: 'yes'},
    { id: 35, name: 'JJJ', goal: 'yes' ,circle: 'yes'} ,
    { id: 42, name: 'ZZZ', goal: 'no' , circle: 'yes' },
    { id: 100, name: 'JJJ',goal: 'no' , circle: 'yes' }]

Answer 1

如果obj中不存在id，则忘记在第一个循环中推送c，如果该对象的一个或多个id不存在，则在c循环存在于g。

var g = [
        { id: 36, name: 'AAA', goal: 'yes' },
        { id: 40, name: 'BBB', goal: 'yes' },
        { id: 57, name: 'CCC', goal: 'yes' },
        { id: 4, name: 'DDD', goal: 'yes' },
        { id: 39, name: 'EEE', goal: 'yes' },
        { id: 37, name: 'FFF', goal: 'yes' },
        { id: 59, name: 'GGG', goal: 'yes' },
        { id: 50, name: 'III', goal: 'yes' },
        { id: 43, name: 'HHH', goal: 'yes' },
        { id: 35, name: 'JJJ', goal: 'yes' }
    ],
    c = [
        { id: 36, name: 'AAA', circle: 'yes' },
        { id: 40, name: 'BBB', circle: 'yes' },
        { id: 57, name: 'CCC', circle: 'yes' },
        { id: 42, name: 'ZZZ', circle: 'yes' },
        { id: 4, name: 'DDD', circle: 'yes' },
        { id: 39, name: 'EEE', circle: 'yes' },
        { id: 37, name: 'FFF', circle: 'yes' },
        { id: 59, name: 'GGG', circle: 'yes' },
        { id: 43, name: 'HHH', circle: 'yes' },
        { id: 35, name: 'JJJ', circle: 'yes' },
        { id: 100, name: 'JJJ', circle: 'yes' }
    ],
    arrayList = [], obj_c_processed = [];

for (var i in g) {
    var obj = {id: g[i].id, name: g[i].name, goal: g[i].goal};

    for (var j in c) {
        if (g[i].id == c[j].id) {
            obj.circle = c[j].circle;
            obj_c_processed[c[j].id] = true;
        }
    }

    obj.circle = obj.circle || 'no';
    arrayList.push(obj);
}

for (var j in c){
    if (typeof obj_c_processed[c[j].id] == 'undefined') {
        arrayList.push({id: c[j].id, name: c[j].name, goal: 'no', circle: c[j].circle});
    }
}

console.log(arrayList);

Answer 2

使用unexcore.js，您可以编写如下函数：

var a = [ { id: 36, name: 'AAA', goal: 'yes' },
    { id: 40, name: 'BBB', goal: 'yes' },
    { id: 57, name: 'CCC', goal: 'yes' },
    { id: 4, name: 'DDD', goal: 'yes' },
    { id: 39, name: 'EEE', goal: 'yes' },
    { id: 37, name: 'FFF', goal: 'yes' },
    { id: 59, name: 'GGG', goal: 'yes' },
    { id: 50, name: 'III', goal: 'yes' },
    { id: 43, name: 'HHH', goal: 'yes' },
    { id: 35, name: 'JJJ', goal: 'yes' } ];

var b = [ { id: 36, name: 'AAA', circle: 'yes' },
    { id: 40, name: 'BBB', circle: 'yes' },
    { id: 57, name: 'CCC', circle: 'yes' },
    { id: 42, name: 'ZZZ', circle: 'yes' },
    { id: 4, name: 'DDD', circle: 'yes' },
    { id: 39, name: 'EEE', circle: 'yes' },
    { id: 37, name: 'FFF', circle: 'yes' },
    { id: 59, name: 'GGG', circle: 'yes' },
    { id: 43, name: 'HHH', circle: 'yes' },
    { id: 35, name: 'JJJ', circle: 'yes' },
    { id: 100, name: 'JJJ', circle: 'yes' } ];


function merge_object_arrays (arr1, arr2, match) {
  return _.union(
    _.map(arr1, function (obj1) {
      var same = _.find(arr2, function (obj2) {
        return obj1[match] === obj2[match];
      });
      return same ? _.extend(obj1, same) : obj1;
    }),
    _.reject(arr2, function (obj2) {
      return _.find(arr1, function(obj1) {
        return obj2[match] === obj1[match];
      });
    })
  );
}

document.getElementsByTagName('pre')[0].innerHTML = JSON.stringify(
  merge_object_arrays(a, b, 'id'), null, 2
);

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

尝试在此处运行。

Answer 3

尝试使用jquery。试试这个：

var g= [ 
    { id: 36, name: 'AAA', goal: 'yes' },
    { id: 40, name: 'BBB', goal: 'yes' },
    { id: 57, name: 'CCC', goal: 'yes' },
    { id: 4, name: 'DDD', goal: 'yes' },
    { id: 39, name: 'EEE', goal: 'yes' },
    { id: 37, name: 'FFF', goal: 'yes' },
    { id: 59, name: 'GGG', goal: 'yes' },
    { id: 50, name: 'III', goal: 'yes' },
    { id: 43, name: 'HHH', goal: 'yes' },
    { id: 35, name: 'JJJ', goal: 'yes' } ];


 var c= [ 
    { id: 36, name: 'AAA', circle: 'yes' },
    { id: 40, name: 'BBB', circle: 'yes' },
    { id: 57, name: 'CCC', circle: 'yes' },
    { id: 42, name: 'ZZZ', circle: 'yes' },
    { id: 4, name: 'DDD', circle: 'yes' },
    { id: 39, name: 'EEE', circle: 'yes' },
    { id: 37, name: 'FFF', circle: 'yes' },
    { id: 59, name: 'GGG', circle: 'yes' },
    { id: 43, name: 'HHH', circle: 'yes' },
    { id: 35, name: 'JJJ', circle: 'yes' },
    { id: 100, name: 'JJJ', circle: 'yes' } ];

var combine_obj={};
$.extend(combine_obj, g, c);

或使用简单的javascript

var combine_obj={};
for(var key in g)  combine_obj[key]=g[key];
for(var key in c)  combine_obj[key]=c[key];

Answer 4

你可以这样做，

var g = [ { id: 36, name: 'AAA', goal: 'yes' },
    { id: 40, name: 'BBB', goal: 'yes' },
    { id: 57, name: 'CCC', goal: 'yes' },
    { id: 4, name: 'DDD', goal: 'yes' },
    { id: 39, name: 'EEE', goal: 'yes' },
    { id: 37, name: 'FFF', goal: 'yes' },
    { id: 59, name: 'GGG', goal: 'yes' },
    { id: 50, name: 'III', goal: 'yes' },
    { id: 43, name: 'HHH', goal: 'yes' },
    { id: 35, name: 'JJJ', goal: 'yes' } ]


 var c = [ { id: 36, name: 'AAA', circle: 'yes' },
    { id: 40, name: 'BBB', circle: 'yes' },
    { id: 57, name: 'CCC', circle: 'yes' },
    { id: 42, name: 'ZZZ', circle: 'yes' },
    { id: 4, name: 'DDD', circle: 'yes' },
    { id: 39, name: 'EEE', circle: 'yes' },
    { id: 37, name: 'FFF', circle: 'yes' },
    { id: 59, name: 'GGG', circle: 'yes' },
    { id: 43, name: 'HHH', circle: 'yes' },
    { id: 35, name: 'JJJ', circle: 'yes' },
    { id: 100, name: 'JJJ', circle: 'yes' } ]

for (i = 0; i < g.length; i++) {        //Loop trough first array
    var curID = g[i].id;                //Get ID of current object
    var exists = false;                 
    for (j = 0; j < c.length; j++) {    //Loop trough second array
        exixts = false;
        if (curID == c[j].id){          //If id from array1 exists in array2
            exixts = true;
            tempObj = c[j];             //Get id,object from array 2
            break;
        }
    }
    if(exixts) {
        g[i]["circle"] = tempObj.circle;//If exists add circle from array2 to the record in array1
    }else{
        g[i]["circle"] = "no";          //If it doesn't add circle with value "no"
    }
}

for (i = 0; i < c.length; i++) {        //Loop trough array 2
    var curObj = c[i];
    var ex = true;
    g.forEach(function(row) {           //Loop to check if id form array2 exists in array1
        if (curObj.id == row.id){
            ex = false;
        }
    });
    if(ex){                             //If it doesn't exist add goal to object with value "no" and push it into array1
        curObj["goal"] = "no";
        g.push(curObj);
    }
}

console.debug(g);

我添加了一些注释来解释代码中发生了什么。

Psuedo代码，

//Loop trough g
//get id from g[i] and check if it exists in c
//if so
    //add circle from a2 to a1[i]
    //Add value of circle from c onto g[i]["circle"]
//otherwise
    //Add value of "no" onto g[i]["circle"]

//Loop trough c
    //If id isn't in g, add row with value c[i]["goal"] = "no" to g

基于javascript中的键值合并两个json对象

4 个答案: