Set Javascript object value at time of creation

Question

Is it possible to set one object's key $thumbnailContainer to the value of another object thumbnailImg within the same object gallerySelectors?

Example:

var gallerySelectors = {
    '$thumbnailContainer' : $('#thumb'),
    'thumbnailImg' : this.$thumbnailContainer.find('img')
};

Ry- · Accepted Answer · 2012-06-27 22:13:05Z

1

No, unfortunately, there isn't. Luckily, the alternative is just as simple:

var gallerySelectors = {
    $thumbnailContainer: $('#thumb')
};

gallerySelectors.thumbnailImg = gallerySelectors.$thumbnailContainer.find('img');

You could also make a function to do that for you if the issue is that you want to pass the object inline.

answered Jun 27, 2012 at 22:13

Ry-♦

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Comments

sandradev · Accepted Answer · 2012-06-27 22:17:56Z

0

No, because this will not be bound to the object in the situation you show.

I don't think there is a way to do this.

You could try

var gallerySelectors = {
    '$thumbnailContainer' : $('#thumb'),
    'thumbnailImg' : gallerySelectors.$thumbnailContainer.find('img')
};                     ^--------------- NOTE!!!

but I'm not sure whether I would trust this to work everywhere. I guess you'd have to look into the language spec to see whether the members are always guaranteed to be initialized in order.

I would just take the easy route:

var gallerySelectors = {
    '$thumbnailContainer' : $('#thumb'),
    'thumbnailImg' : $('#thumb').find('img')
};

edited Jun 27, 2012 at 22:17

answered Jun 27, 2012 at 22:12

sandradev

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2 Comments

nbrooks Over a year ago

I'm sure you didn't mean to type this: $('#thumb').$thumbnailContainer.find('img')

bob_cobb Over a year ago

Bummer. Thanks anyway. This is the method that I was probably going to end up doing.

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