在指令Angularjs

Question

有一些代码

app.directive('currencyInput', function ($filter, myFactory) {
return {
    require: '?ngModel',
    link: function (scope, elem, attrs, ctrl) {
        ctrl.$formatters.unshift(function () {
            return $filter('number')(ctrl.$modelValue);

        });

        ctrl.$parsers.unshift(function (viewValue) {
            let plainNumber = viewValue.replace(/[\,\.]/g, ''),
                b = $filter('number')(plainNumber);

            elem.val(b);
            //here I need to add listener if keyCode==13
            return plainNumber;
        });
    }
};
});

所以，我需要抓住keyCode == 13。我怎么能在我添加评论的地方做到这一点？我需要采取一些行动：

 elem.bind('keydown keypress', ($event) => {
  if ($event.which === 13) {
     let val=$element.val();
     if(attrs['param']=="amount") myFactory.process[attrs['param']]=val*24;
     else myFactory.process[attrs['param']]=$element.val();
     let i=0;
     for(let key in myFactory.process){
         if(myFactory.process[key]===""){
               scope.dashboard.currParam=i;//this doesn't works until user make one more action.
               break;
         }
         i++;
     }
  }
 })

但并非所有行动都是＆＃34;行为＆＃34;用户点击进入。 myFactory.process看起来像

process: {
        cost:"",
        amount:"",
        wrapping:"",
        risk:"",
        limit:"",
        franchise:""
    },

此字符串scope.dashboard.currParam=i;必须更新视图（DOM）。但是在我点击输入之前它不会更新。

Answer 1

您可以使用以下代码执行此操作：

elem.bind('keydown keypress', ($event) => {
  if ($event.which === 13) {
    scope.$apply(() => scope.$eval(attrs.enter, { $event });
    // enter key pressed here
  }
});

Answer 2

问题是焦点。所以，你需要做出类似的事情：

 elem.bind('keydown keypress', ($event) => {
 if ($event.which === 13) {
 let val=$element.val();
 if(attrs['param']=="amount") myFactory.process[attrs['param']]=val*24;
 else myFactory.process[attrs['param']]=$element.val();
 let i=0;
 for(let key in myFactory.process){
     if(myFactory.process[key]===""){
           scope.dashboard.currParam=i;
           let target = $event.target;
           target.blur();//this helps
           break;
     }
     i++;
 }
 }
 })