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Question

关于在这个网站上比较两组JavaScript对象有很多问题，但我无法找到我想要实现的目标。

我有两组对象如下。

var players = [
                { "id" : "45783",
                  "r3" : "DED" },
                { "id" : "54762",
                  "r3" : "DEDCCCCDDDDC" },
                { "id" : "78903",
                  "r3" : "DEDCDBDDDDDC"}
              ]
var oldPlayers = [
                { "id" : "45783",
                  "r3" : "DE" },
                { "id" : "54762",
                  "r3" : "DEDCCCCDDDD" },
                { "id" : "78903",
                  "r3" : "DEDCDBDDDDD"}
              ]

我想比较上面的数据并找到var播放器中的确切位置，＆＃34; r3＆＃34;与oldPlayers不同的价值。

例如，

players ---- "r3" : "DED"
oldPlayers - "r3" : "DE"

上述数据的差异是玩家在[2]的字符位置有额外的D，假设数字从0开始。

我有每个循环，我打算获得这个位置，我被困在如何使这项工作。

var isOldPlayers = false;

// check if oldPlayers exists
 if(oldPlayers != undefined && oldPlayers != ''){
   isOldPlayers = true;
 }

$.each(players,function(i,v){

if(isOldPlayers){
  var oldScores = oldPlayers[i]['r3'];
  var newScores = this['r3'];

   if(newScores != oldScores){
      // how can I get the exact location of the difference?
    }

  }
});

Answer 1

它可以加速使用数组。如果字符串将包含与上面相同的字符，这将是一个很好的方法。这会对你有用吗？

var dIdx = [v['r3'], oldPlayers[i]['r3']].sort()[0].length;

// ["DED", "DE"].sort() --> ["DE", "DED"];
// ["DE", "DED"][0] --> "DE"
// "DE".length --> 2

// "DED" differs from "DE" at index 2

或者如果你自己需要差异，你可以这样做：

var scores = [v['r3'], oldPlayers[i]['r3']].sort(),
    diff = scores[1].substr(scores[0].length);

修改

要清楚，您不需要制作“players”对象数组。 点击图片进行演示。

var players = [
                { "id" : "45783",
                  "r3" : "DED" },
                { "id" : "54762",
                  "r3" : "DEDCCCCDDDDC" },
                { "id" : "78903",
                  "r3" : "DEDCDBDDDDDC"}
              ],
    oldPlayers = [
                { "id" : "45783",
                  "r3" : "DE" },
                { "id" : "54762",
                  "r3" : "DEDCCCCDDDD" },
                { "id" : "78903",
                  "r3" : "DEDCDBDDDDD"}
              ];
var desc = '';
$.each(players, function(i, v) {
  var dIdx = [v['r3'], oldPlayers[i]['r3']].sort()[0].length;

  desc += '<span class="h">' + v['r3'] + '</span> ' +
    'and <span class="h">' + oldPlayers[i]['r3'] + '</span> ' +
    'differ at index: <span class="hg">' + dIdx + '</span><br><br>';
});
$('body').html(desc);

Answer 2

如果我理解正确，这应该做你需要的：

var players = [
                { "id" : "45783",
                  "r3" : "DED" },
                { "id" : "54762",
                  "r3" : "DEDCCCCDDDDCDEWE" }, // I added chars here to show how to get the rest of the string that is different
                { "id" : "78903",
                  "r3" : "DEDCDBDDDDDC"}
              ]
var oldPlayers = [
                { "id" : "45783",
                  "r3" : "DE" },
                { "id" : "54762",
                  "r3" : "DEDCCCCDDDD" },
                { "id" : "78903",
                  "r3" : "DEDCDBDDDDD"}
              ]
$.each(players,function(index, element){
  var newR3= players[index]["r3"].split('');
  var oldR3= oldPlayers[index]["r3"].split('');
  var curId= players[index]["id"];
  
  $.each(newR3,function(i, el){
     if(el != oldR3[i]){
       $('#myDiv').append('Object '+index+' (id '+curId+') in players is different than object '+index+' (id '+curId+') in oldPlayers starting at index: '+i +'<br>');
       $('#myDiv').append('The rest of the string, from the fist difference to the end is: ' + newR3.slice(i).join('') +'<br><br>');
       
       return false;
       }


  });


});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<div id="myDiv"></div>

Answer 3

这是一个解决方案：

$.each(players,function(iNew,vNew){
    if(isOldPlayers){               
        $.each(oldPlayers,function(iOld,vOld){
            if(vNew.id === vOld.id && vNew.r3 !== vOld.r3){
                console.log('players ----- "r3" : ' + vNew.r3 + '\noldPlayers ----- "r3" : ' + vOld.r3);
            }
        }); 
    }
});

请注意，您需要两个内部循环，外部循环遍及玩家，对于每个循环，内部循环遍历oldPlayers数组以查找匹配。

在内部循环中，您可以使用this.id和this.r3代替vOld.id和vOld.r3。但是，此代码效率不高，因为必须为第一个数组的每个元素迭代第二个数组的所有元素。如果您知道所有ID都是唯一的，那么您可以使用传统的for循环进一步改进它，并在找到ID匹配后立即将其中断：

$.each(players,function(iNew,vNew){
    if(isOldPlayers){               
        for (var i = 0, len = oldPlayers.length ; i < len ; i += 1) {
            if(vNew.id === oldPlayers[i].id) {
                if(vNew.r3 !== oldPlayers[i].r3){
                    console.log('players ----- "r3" : ' + vNew.r3 + '\noldPlayers ----- "r3" : ' + oldPlayers[i].r3);
                }
            break;
            }
        }
    }
});

jQuery比较两组对象并获取值的差异位置

3 个答案:

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