Zusammenführen von zwei json-Objekt basierend auf Taste Wert in javascript

Brauche ich zum Zusammenführen von zwei json-Objekt basierend auf key-value mit javascript.

Habe ich zwei verschiedene Variablen g und c -.

Bedingungen: Alle Werte zusammengeführt werden müssen.

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' } ]

Habe ich versucht den folgenden code ein: aber ich Zusammenführen, was haben dieselbe id in 'c' variable. aber ich muss Zusammenführen, vergleichen 'g' und '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);

Erwartete Ausgabe:

[ { 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' }]

Was ist das problem dabei und was ist die Frage? So weit alles Euch erklärt habe, ist das, was Sie wollen ... nicht, was Sie getan haben, um zu versuchen, es zu tun
sorry, ich vergaß zu hinzufügen mein code.
Brauchen nur javascript, nicht jquery.
was ist Ihre erwartete result in this case var g = [ { id: 36, name: 'AAA', goal: 'yes' },....] var c = [ { id: 36, name: 'foo', circle: 'yes' },.....] ??
was wird erwartet, wenn zwei Objekte haben die gleiche id, aber unterschiedliche Namen?

InformationsquelleAutor RSKMR | 2015-05-07

javascript json

Vergessen Sie push - obj in der ersten Schleife in dem Fall die id existiert nicht in c und Durchlaufen c im Fall, dass eine oder mehrere id ' s für das Objekt existiert nicht in 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);

Der code funktioniert, danke. Es ist möglich, zu vereinfachen, dieser code ?
Ich bearbeitet meine Antwort. Sie können, weisen Sie mehrere Variablen mit nur einer var keyword, getrennt durch ein Komma. Ich vereinfachte das, die ich entfernt get* Variablen und ich entfernte die Verwendung von in_obj_c mit der || Kürzel (in diesem Fall: weisen Sie "Nein" zu obj.circle wenn es leer ist)
Hi, ich brauche noch eine Hilfe. Die gleiche Frage, aber ich habe 3 Variablen. (g,c und u). g und c sind die gleichen Werte. var a = [ { 'id': 101, 'name': 'TTT',"other":"ja" }, { 'id': 40, 'name': 'BBB' ,"other":"ja"}, { 'id': 50, 'name': 'III' ,"other":"ja"}, { 'id': 100, 'name': 'TTT' ,"other":"ja"} ]
Erwartet : [ { 'id': 36, 'name': 'AAA', 'Ziel': 'ja','Kreis': 'ja','sonstige':'Nein' }, { 'id': 40, 'name': 'BBB', 'Ziel': 'ja','Kreis': 'ja','sonstige':'ja'"}, { 'id': 57, 'name': 'CCC', 'Ziel': 'ja','Kreis': 'ja','sonstige':'Nein' }, { 'id': 4, 'name': 'DDD', 'Ziel': 'ja' ,'Kreis': 'ja','sonstige':'Nein' }, { 'id': 39, 'name': 'EEE', 'Ziel': 'ja' ,'Kreis': 'ja','anderen':"Nein"}, { 'id': 37, 'name': 'FFF', 'Ziel': 'ja' ,'Kreis': 'ja','sonstige':'Nein'}, { 'id': 59, 'name': 'GGG', 'Ziel': 'ja' ,'Kreis': 'ja','sonstige':'Nein'}, { 'id': 50, 'name': III, 'Ziel': 'ja' ,'Kreis': 'no','andere':'ja'},
{ 'id': 43, 'name': 'HHH', 'Ziel': 'ja' ,'Kreis': 'ja','sonstige':'Nein'}, { 'id': 35, 'name': JJJ, 'Ziel': 'ja' ,'Kreis': 'ja','sonstige':'Nein'} , { 'id': 42, 'name': 'ZZZ', 'Ziel': 'Nein' , 'Kreis': 'ja','sonstige':'Nein'}, { 'id': 100, 'name': JJJ,'Ziel': 'no' , 'Kreis': 'ja' ,'sonstige':'yes'}, { 'id': 101, 'name': TTT,'Ziel': 'no' , 'Kreis': 'no' ,'andere':'yes'}]

InformationsquelleAutor redelschaap

Verwenden undescore.js schreiben Sie eine Funktion wie diese:

JS:

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
);

HTML:

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

<pre>
</pre>

Versuchen Sie es hier.

Wie würden Sie diese chance, um Elemente aus arr2 überschreiben arr1 auf id = id und name !=name

InformationsquelleAutor Mustafa Dokumacı

Versuchen mit Hilfe von jquery. Versuchen Sie dies :

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);

ODER mit einfachen javascript -

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

Was sind g und c?

InformationsquelleAutor shweta chinchore

Könnte man es so machen,

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);

Habe ich einige Kommentare Hinzugefügt, um zu erklären, was Los ist in den code.

Pseudo-code,

//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

InformationsquelleAutor Maarten Peels

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