jquery getJson method for complex json how to get values using jquery each method

Question

Hi am newbie to JSON i dont know how to iterate JSON over $.each function i worked on it but it display always shows value as undefined am not sure why it is

here how my json looks saved it as new file data.json

{ "users":[
            {
                "firstName":"Ray",
                "lastName":"Villalobos",
                "joined": {
                    "month":"January",
                    "day":12,
                    "year":2012
                }
            },
            {
                "firstName":"John",
                "lastName":"Jones",
                "joined": {
                    "month":"April",
                    "day":28,
                    "year":2010
                }
            }
    ]}

here i need to iterate name with Joined date in JSon and i iterate it over Jquery by like this

$(function(){
    $(document).on('click', '.users', function(e){
        e.preventDefault();                                            
        $.getJSON( "js/data.json", function( data ) {
          var items = [];
          $.each( data.users, function( key, val ) {
            console.log(data.users);                           
            items.push( "<li id='" + data.users.firstName + "'><span><p>" + data.users.firstName + "</p></span></li>" );
          });
          $( "<ul/>", {
            "class": "users-lists",
            html: items.join( "" )
              }).appendTo( ".nlst" );
       });
    });
});

help me out

chandu · Accepted Answer · 2014-10-20 07:15:59Z

1

In your jquery code change this line

items.push( "<li id='" + data.users.firstName + "'><span><p>" + data.users.firstName + "</p></span></li>" );

to

items.push( "<li id='" + val.firstName + "'><span><p>" + val.firstName + "</p></span></li>" );

jquery doc Here

answered Oct 20, 2014 at 7:15

chandu

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2 Comments

Vikranth Over a year ago

thanks i got and a small doubt how do i iterate the nested object within the users named as joined?

chandu Over a year ago

you need to add another $.each loop inside $.each loop. pass the value as parameter for inside $.each loop

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