Using More Than One Site Per Page

See Maxim Z's original question here.

The StackTack plugin doesn't need to select the entire document to work, you can feed it any selection you like.

$('div').stacktack();

The only requirement for an element to become a Tack is to have a valid StackTack ID. This affords you the ability to isolate Tacks to a certain portion of your page:

$('.my-tacks').stacktack();

You can also use multiple StackTack instances to target different APIs or specify different defaults. To target two different Stack sites on the same page, do the following:

$('.stackoverflow-tack').stacktack({site: 'stackoverflow.com'});
$('.serverfault-tack').stacktack({site: 'serverfault.com'});

<div id="stacktack-12345" class="stackoverflow-tack"></div>
<div id="stacktack-30303" class="serverfault-tack"></div>

Using More Than One Site Per Page

See Maxim Z's original question here.

The StackTack plugin doesn't need to select the entire document to work, you can feed it any selection you like.

$('div').stacktack();

The only requirement for an element to become a Tack is to have a valid StackTack ID. This affords you the ability to isolate Tacks to a certain portion of your page:

$('.my-tacks').stacktack();

You can also use multiple StackTack instances to target different APIs or specify different defaults. To target two different Stack sites on the same page, do the following:

$('.stackoverflow-tack').stacktack({site: 'stackoverflow.com'});
$('.serverfault-tack').stacktack({site: 'serverfault.com'});

<div id="stacktack-12345" class="stackoverflow-tack"></div>
<div id="stacktack-30303" class="serverfault-tack"></div>

The reason there is no class name override for the site parameter is because you can't represent a URL in a class name due to invalid characters.