jQuery variable made of a variable and selector

Question

I am trying to construct a variable like the following:

var feed = new Instafeed({
    get: 'tagged',
    tagName: 'Breakfast',
    clientId: 'bff2c0f300c841a298a9198e499eee16',
    limit: 50,
    target:'instagramFeedBottom',
    template: '<li class="panel"><a href="{{link}}"><img class="front card" src="{{image}}" /></a><a href="{{link}}"><div class="back card"><p>{{model.user.full_name}}</p></div></a></li>',
    after: function() {
        RandomImg();
    },
    resolution: 'low_resolution'
});


    $('.front').on({
        mouseenter: function () {
            var x = $(this).parent().parent();
            console.log(x + $('.back'));
        }
        // mouseleave: function () {
        //    $(this).fadeIn(medium);
        // }
    });

I am having the following error in the console: [object Object][object Object]. I need to access the '.back' based on the '.front'.parent().parent()

console.log(x.find('.back'));

Arun P Johny
– Arun P Johny

2014-02-25 12:03:07 +00:00
Commented Feb 25, 2014 at 12:03 — Arun P Johny
– Arun P Johny, Commented Feb 25, 2014 at 12:03

Arun P Johny · Accepted Answer · 2014-02-25 12:03:42Z

1

Since x is a jQuery object use .find() like

console.log(x.find('.back'));

answered Feb 25, 2014 at 12:03

Arun P Johny

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Comments

Satpal · Accepted Answer · 2014-02-25 12:09:04Z

1

I think you need The context selector

$('.back', x)

OR

You can also use

x.find('.back')

Note: jQuery internally converts $('.back', x) to x.find('.back')

edited Feb 25, 2014 at 12:09

answered Feb 25, 2014 at 12:02

Satpal

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2 Comments

Aessandro Over a year ago

Why x is after as x suppose to be .panel?

Aessandro Over a year ago

@ArunPJohny thanks for that , although I was looking for an explanation to the fact my (wrong) solution didnt work console.log(x + $('.back'));.

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