How call variable outside function .each with Jquery?

Question

I want to be able to access the value of a nested each() function outside of the function.

I have code below or http://jsfiddle.net/z36UK/6/

HTML:

<div id="wrap">
  <ul data-role="listview" data-filter="true" data-filter-placeholder="Search..." id="ds-canho"> 
  </ul>
</div>

Javascript:

$(document).ready(function() {
  var urlx=[];  
  parseXml();
  alert(urlx);
  console.log(urlx);

  $.mobile.loading( "show" );
  $.ajax({
    type: "GET",
    url: 'https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20html%20where%20url%3D%27http://saigonpearl.info/home/notype/-0-trang-chu.html%27%20and%20xpath%3D%22%2F%2Fdiv%5B%40class%3D%27tindang%27%5D%22&diagnostics=true',
    dataType: "xml",
    success: parseXml
  });


  function parseXml(xml) {
    $(xml).find("div.tindang").each(function() {

      var url = $(this).find('a').attr("href"),
      urlx= url.replace('#menutop','');
      $("ul#ds-canho").append('<li>' + $(this).find('h3').text() + '</li>');
      $('ul#ds-canho').listview('refresh');
      $.mobile.loading( "hide" );
    });
  }

});

you want to use the value or the variable who'z having the value? in this case the `url` variable, right? — rockStar, Jun 06 '14 at 07:00
-1 because you didn't make any effort to format your code in a readable format. — iappwebdev, Jun 06 '14 at 07:05

score 0 · Answer 1 · edited May 23 '17 at 12:05

It's not about the each, is it? You can easily put the urlx variable there and assign to it from the inside:

function parseXml(xml) {
    var urlx;
    $(xml).find("div.tindang").each(function() {
        var url = $(this).find('a').attr("href"),
        urlx = url.replace('#menutop','');
        $("ul#ds-canho").append('<li>' + $(this).find('h3').text() + '</li>');
        $('ul#ds-canho').listview('refresh');
        $.mobile.loading( "hide" );
    });
    console.log(urlx);
    // if you wanted an array, use `urlx = [];` and `urlx.push(…);`
}

However, where you currently have placed the log and alert statements is

before the ajax request
outside of the ajax request callback

which means that it cannot work. You need to put them inside the callback, because you cannot return the data outside of it.

score 0 · Accepted Answer · edited Jun 06 '14 at 07:18

Please update below js code

$(document).ready(function() {                   
  $.mobile.loading( "show" );
  $.ajax({
    type: "GET",
    url: 'https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20html%20where%20url%3D%27http://saigonpearl.info/home/notype/-0-trang-chu.html%27%20and%20xpath%3D%22%2F%2Fdiv%5B%40class%3D%27tindang%27%5D%22&diagnostics=true',
    dataType: "xml",
    success: parseXml
  });
  $( document ).ajaxComplete(function() {
    parseXml();
  });
  function parseXml(xml) {
    window.urlx = '';
    $(xml).find("div.tindang").each(function() {

      var url = $(this).find('a').attr("href"),
          urlx = url.replace('#menutop','');

      $("ul#ds-canho").append('<li>' + $(this).find('h3').text() + '</li>');
      $('ul#ds-canho').listview('refresh');
      $.mobile.loading( "hide" );
    });
  }
});

EDIT

Add FSFIDDLE

How call variable outside function .each with Jquery?

2 Answers2