Parse nested xml of recursive dir list with jQuery

Question

I have xml based recursive dir list. I want to get file list from specific directory for example .By XML I should get only img2.gif and img3.gif and img6.gif .But script also returns img4 and img5 which are actually in sub dir and i dont want to process subdirs. Objective is to get files from requested dir and ignore nested sub dirs. One way is to change xml structure from nested to normal but i dont want that. XML

<directory name=".">
   <file name="img1.gif" />
   <directory name="images">
      <file name="images/img2.gif" /> 
      <file name="images/img3.gif" /> 
      <directory name="images/delta">
         <file name="images/delta/img4.gif" /> 
         <file name="images/delta/img5.gif" /> 
      </directory>
      <file name="images/img6.gif" /> 
   </directory>
</directory>

JAVASCRIPT

function parse_files(path) {
    $.ajax({
        type: "GET",
        url: "list.xml",
        dataType: "xml",
        success: function(xml) {

            $(xml).find('directory').each(function(){
                if($(this).attr('name')==path) {

                    $(this).find('file').each(function(){
                        var list = $(this).attr('name');
                        alert(list);
                    });
                 }
            });
        }

    });
}

parse_files("images");
// returns img2,img3,img4,img5,img6
// should return img2,img3,img6

adeneo · Accepted Answer · 2013-03-30 15:30:16Z

1

Find only direct children with children(), or find(' > file') or with context $(' > file', this) etc, as just find() finds all descendents, even the ones that are nested further down in the next directory :

$(xml).find('directory').each(function(){
     if($(this).attr('name')==path) {
         $(this).children('file').each(function(){
             var list = $(this).attr('name');
             alert(list);
         });
     }
});

edited Mar 30, 2013 at 15:30

answered Mar 30, 2013 at 15:14

adeneo

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3 Comments

charlietfl Over a year ago

find() finds all descendents is even better explanation since children are only considered to be first level down

django Over a year ago

Thank you @adeneo.That does the trick. Is there a way/ syntax that i can reach directory with name directly so I dont have to .find('directory').each(function() ? I want to avoid iterate.

adeneo Over a year ago

$('directory[name="path"] > file', xml).each(function() { ... });

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