I have a modal Bootstrap popup form that opens with a button click in AngularJS, I am wondering how it still displays, even if the button is disabled.
Please see the button code below:
<a class="btn btn-sm btn-info" data-toggle="modal" href="#modal-form-submit" data-backdrop="static" data-keyboard="false" ng-disabled="!ItemName || ItemDescription">
Submit
</a>
I have below model popup code:
<div id="modal-form-submit" class="modal">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">×</button>
<h4 class="blue bigger"> DEMO MODEL FORM </h4>
</div>
<div class="modal-body">
<div class="row">
<div class="col-sm-12 col-xs-12">
<div class="form-group">
<label class="control-label" for="toitems">To:</label>
<div class="tags full-width">
<span class="tag" ng-repeat="tag in tags">{{ tag.Description }}</span>
</div>
</div>
</div>
</div>
</div>
<hr />
<div class="modal-footer">
<button class="btn btn-sm" data-dismiss="modal">
<i class="ace-icon fa fa-times"></i>
Cancel
</button>
<button class="btn btn-sm btn-primary " ng-disabled="!ItemCode || !ItemDescription"
ng-click="SaveEntireFormData()" type="button">
<i class="ace-icon fa fa-check"></i>
Confirm
</button>
</div>
</div>
</div>
</div>
I don't want to use jquery js, I really want to enable this with angular js.
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