How to check if bootstrap modal is open, so i can use jquery validate

Question

i need to make a validation only if a modal is open, because if i open it, and then i close it, and the i press the button that opens the modal it doesn't work because it is making the jquery validation, but not showing because the modal was dismissed.

So i want to ad a jquery if modal is open so the i do validate, is this possible?

<script>
$(document).ready(function(){

var validator =$('#form1').validate(
 {
  ignore: "",
  rules: {

usu_login: {
  required: true
},
usu_password: {
  required: true
},
usu_email: {
  required: true
},
usu_nombre1: {
  required: true
},
usu_apellido1: {
  required: true
},
usu_fecha_nac: {
  required: true
},
usu_cedula: {
  required: true
},
usu_telefono1: {
  required: true
},
rol_id: {
  required: true
},
dependencia_id: {
  required: true
},
  },

  highlight: function(element) {
$(element).closest('.grupo').addClass('has-error');
        if($(".tab-content").find("div.tab-pane.active:has(div.has-error)").length == 0)
        {
            $(".tab-content").find("div.tab-pane:hidden:has(div.has-error)").each(function(index, tab)
            {
                var id = $(tab).attr("id");

                $('a[href="#' + id + '"]').tab('show');
            });
        }
  },
  unhighlight: function(element) {
$(element).closest('.grupo').removeClass('has-error');
  }
 });

}); // end document.ready

</script>

Answer 1

To avoid the race condition @GregPettit mentions, one can use:

($("element").data('bs.modal') || {})._isShown    // Bootstrap 4
($("element").data('bs.modal') || {}).isShown     // Bootstrap <= 3

as discussed in Twitter Bootstrap Modal - IsShown.

When the modal is not yet opened, .data('bs.modal') returns undefined, hence the || {} - which will make isShown the (falsy) value undefined. If you're into strictness one could do ($("element").data('bs.modal') || {isShown: false}).isShown