[英]cleanest way to make a modal dependent on a function? AngularJS
I have a modal that gets displayed on a click of a button, when that button is clicked it also triggers a couple of javascript functions. 我有一个模式,单击一个按钮后就会显示,单击该按钮时还会触发几个javascript函数。 is there a way to prevent the modal from displaying if one of the functions is false?
如果其中一个函数为假,是否有办法防止模式显示?
<form ng-submit="checkValues(); makeChange();" class="form-inline">
<div class="form-group">
<input type="text" ng-model="newChange.amount" maxlength="15" class="form-control"/>
<button type="submit" data-toggle="modal" data-target="#myModal" class="btn btn-default" >Go</button>
</div>
</form>
you should use $modal service of angular-ui-bootstrap. 您应该使用angular-ui-bootstrap的$ modal服务。 so you can check if conditions are met and programatically open your modal.
因此您可以检查条件是否满足,并以编程方式打开模态。
here is a sample 这是一个样本
<form ng-submit="checkValues();" class="form-inline">
<div class="form-group">
<input type="text" ng-model="newChange.amount" maxlength="15" class="form-control"/>
<button type="submit" class="btn btn-default" >Go</button>
</div>
</form>
scope.checkValues = {
if(!everything.ok){
//error message maybe
return;
}
var modalInstance = $modal.open({
templateUrl: 'myModalContent.html',
controller: 'ModalInstanceCtrl',
resolve: {
items: function () {
return items;
}
}
});
modalInstance.result.then(function (selectedItem) {
$scope.selected = selectedItem;
});
};
note: if you need programmatic check for validation. 注意:如果您需要通过程序检查进行验证。 i would recommend you to do validation on form element.
我建议您对表单元素进行验证。 using
使用
<form name="valueForm" ng-submit="valueForm.$valid && checkValues();" class="form-inline">
<div class="form-group">
<input type="text" name="amount" ng-model="newChange.amount" ng-maxlength="15" class="form-control"/>
<button type="submit" class="btn btn-default" >Go</button>
</div>
</form>
Since you are not using angular-ui-bootstrap, here is a way to do it using Bootstrap's Javascript. 由于您未使用angular-ui-bootstrap,因此这里是一种使用Bootstrap的Javascript进行操作的方法。
Since it's DOM related, it'd be best to put this in a directive like so. 由于它与DOM相关,因此最好将其放入类似的指令中。 This gives you a reusable button.
这为您提供了可重复使用的按钮。 All you need to do is pass it a function reference which it will call to check whether or not to open the modal.
您需要做的就是向它传递一个函数引用,它将调用该函数引用来检查是否打开模态。
HTML: HTML:
<div open-modal="#myModal" check="demo.check()">Go</div>
JS: JS:
.directive('openModal', function() {
return {
restrict: 'EA',
scope: {},
replace: true,
bindToController: {
check: '&',
openModal: '@'
},
transclude: true,
template: '<button type="button" class="btn btn-primary btn-lg" ng-transclude ng-click="vm.onClick()"></button>',
controllerAs: 'vm',
controller: function() {
var vm = this;
vm.onClick = function() {
if (vm.check()) {
$(vm.openModal).modal('show');
}
};
}
}
});
Plunker: http://plnkr.co/edit/ukDuPZOaGrYQSwCnhe51?p=preview 柱塞: http ://plnkr.co/edit/ukDuPZOaGrYQSwCnhe51?p=preview
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.