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Diego
Asked: 2020-04-28 22:37:03 +0800 CST 2020-04-28 22:37:03 +0800 CST

Why does this method only show me one object in the view and in the browser all?

  • 772

My problem is that it only shows me the first object of the json array but all of them appear in the browser console.

How could I show all those elements that the for returns in text?

<div id="success">
<div id="artistName"></div>

</div>
<script>
  $.ajax({ 
      type : 'POST', 
      url : 'http://ws.audioscrobbler.com/2.0/', 
      data :'method=artist.gettopalbums&'+'artist=cooldplay&'+api_key=57ee3318536b23ee81d6b27e36997cde&' + 'format=json',
      dataType : 'jsonp', 
      success : function(data) {
          for( var i=0 in data.topalbums.album) {
              $('#success #artistName').html(data.topalbums.album[i].name);
          } 
      }
    });
</script>
javascript

2 Answers

  1. Best Answer

    What you are doing in it foris replacing the code htmlthat is in <div>each cycle of for, therefore it only prints the last value. To print all of them, you could do it in the following way.

    var str = "";
for( var i=0 in data.topalbums.album)
    str += data.topalbums.album[i].name+"<br>";
$('#success').html(str);

    Or you can also do it in the following way and you save lines of code.

    for( var i=0 in data.topalbums.album)
    $('#success').append(data.topalbums.album[i].name+"<br>");

    What it does .append()is that it adds to you, as opposed to .html()replacing you.

    • 4

  2. This might help you: 

    $.ajax({ 
          type : 'POST', 
          url : 'http://ws.audioscrobbler.com/2.0/', 
          data :'method=artist.gettopalbums&'+'artist=cooldplay&'+api_key=57ee3318536b23ee81d6b27e36997cde&' + 'format=json',
          dataType : 'jsonp', 
          success : function(data) {
              var name = data.topalbums.album[i].name;

              for( var i=0 in data.topalbums.album) {
                   $('#success').append("<div>"+name+"</div>");
              } 
          }
    });
    • 1