How do you test whether a tab is active with dojo tabs

Question

How do you test whether a tab is active or not with a dojo tab container? (In JQuery this is simple... you can use something like this

if($("#accordion").accordion('option', 'active') == mytabNumber){

With dojo's, dijit.layout.TabContainer there must be a similiar way to do it without having to write a litener function and all that jazz.

perhaps something like...

if( dojo.byId("tab2"), {selected:true} ){

Thanks in advance!

score 6 · Accepted Answer · answered Jun 29 '11 at 21:30

6

You can compare the widget for the tab with the tab container's selectedChildWidget property, i.e.:

dijit.byId('tabContainer').selectedChildWidget == dijit.byId('tab2')

answered Jun 29 '11 at 21:30

Mara Morton

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Looks like .id is actually a property of the .selectedchildwidget object... so you can simplify this as... dijit.byId('tabContainer').selectedChildWidget.id = "tab2" – lance Jun 29 '11 at 22:59

score 2 · Answer 2 · answered Jun 30 '11 at 02:41

If you have a reference to the tab already, you can simply just check its 'selected' property to see if it's selected, regardless of the container it is in.

var tab2 = dijit.byId('tab2');
if (tab2.get('selected')) { /* do stuff */ }

I've created a more detailed example at http://jsfiddle.net/brianarn/ws28T/

score 0 · Answer 3 · answered Feb 21 '13 at 07:41

Here's more complete code answer that gives the surrounding code for Dojo 1.8:

require(["dijit/registry",  "dojo/ready", "dojo/domReady!"], function (registry, ready) {
    ready(function () { //wait till dom is parsed into dijits
        if (dijit.byId('tabContainer').selectedChildWidget == dijit.byId('tab2'))
            alert('Yes, we found it!');
    });
});

How do you test whether a tab is active with dojo tabs

3 Answers3