Question

我想使用JQuery将以下XML代码转换为对象这就是我的XML外观：

<league name="German 1st Bundesliga>
    <club name="FC Bayern München"          coach="Ancelotti"   stadium="Allianz Arena"     clubId="1000" />
    <club name="Borussia Dortmund"          coach="Tuchel"      stadium="Signal Iduna Park" clubId="1001" />
    <club name="Borussia Mönchengldabach"   coach="Schubert"    stadium="Borussia-Park"     clubId="1002" />
    <club name="FC Schlke 04"               coach="Weinzierl"   stadium="Veltins-Arena"     clubId="1003" />
 </league>

这是我创建对象的代码

var $leagueNode = $(response).find('league');   

var clubList = [];

for(var i=0; i < $leagueNode.children().length; i++){
    console.log($leagueNode.children()[i]);
    var $clubNode = $($leagueNode.children()[i]).find('club');              
    var club = {};
    club.clubid = $clubNode.attr('clubid');
    club.name = $clubNode.attr('name');
    club.coach = $clubNode.attr('coach');
    club.stadium= $clubNode.attr('stadium');
    console.log(club);
}

打印到控制台的对象显示所有属性的undefined。我的代码出了什么问题？

Answer 1

要执行您需要的操作，您可以使用map()创建一个对象数组。试试这个：

＆＃13;

var $leagueNode = $('<league name="German 1st Bundesliga"><club name="FC Bayern München" coach="Ancelotti" stadium="Allianz Arena" clubId="1000" /><club name="Borussia Dortmund" coach="Tuchel" stadium="Signal Iduna Park" clubId="1001" /><club name="Borussia Mönchengldabach" coach="Schubert" stadium="Borussia-Park" clubId="1002" /><club name="FC Schlke 04" coach="Weinzierl" stadium="Veltins-Arena" clubId="1003" /></league>');
var obj = $leagueNode.find('club').map(function() {
  var $club = $(this);
  return {
    clubid: $club.attr('clubid'),
    name: $club.attr('name'),
    coach: $club.attr('coach'),
    stadium: $club.attr('stadium'),
  };
}).get();

console.log(obj);

＆＃13;

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

＆＃13;

使用JQuery将Childnode转换为Object

1 个答案: