如何检查bootstrap modal是否打开,所以我可以使用jquery validate?
我只需要在模式打开的情况下进行验证,因为如果我先打开它,然后将其关闭,则我按下打开模式的按钮将不起作用,因为它正在进行jquery验证,但未显示,因为该模式已被撤消.
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.
因此,如果模式已打开,我想广告一个jQuery,以便我进行验证,这可能吗?
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) {
$(element).closest('.grupo').removeClass('has-error');
}
});
}); // end document.ready
</script>
为避免@GregPettit提到的竞争条件,可以使用:
To avoid the race condition @GregPettit mentions, one can use:
($("element").data('bs.modal') || {})._isShown // Bootstrap 4
($("element").data('bs.modal') || {}).isShown // Bootstrap <= 3
,如 Twitter引导程序模式-IsShown 中所述.
当尚未打开模态时,.data('bs.modal')
返回undefined
,因此返回|| {}
-这将使isShown
成为(虚假的)值undefined
.如果您严格要求的话,可以做($("element").data('bs.modal') || {isShown: false}).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