1

实施细节：

我正在使用 ui-router 在 ui-view div 中加载表单页面。我提到了Andres Ekdahi的一个很好的例子，我如何使用相同的指令对多个表单进行 $dirty 检查？

表格1

 <form name="myForm" ng-controller="Controller" confirm-on-exit>

表格2

<form name="iForm" ng-controller="Controller" confirm-on-exit ng-model="myModel">

app.js（指令）

myApp.directive('confirmOnExit', function() {
        return {
            link: function($scope, elem, attrs) {
                // condition when back page is pressed 
                window.onbeforeunload = function(){
                    if ($scope.myForm.$dirty) {
                        return "The formI is dirty, do you want to stay on the page?";
                    }
                }
                // condition when user try to load other form (via icons )
                $scope.$on('$stateChangeStart', function(event, next, current) {
                    if ($scope.myForm.$dirty) {
                        if(!confirm("myForm. Do you want to continue ?")) {
                            event.preventDefault();
                        }
                    }

                    if ($scope.iForm.$dirty) {
                        if(!confirm("iform. Do you want to continue ?")) {
                            event.preventDefault();
                        }
                    }
                });
            }
        };
    });

错误：

第一次加载页面时，$dirty 值为 false。然后我填写表单详细信息并单击第三个图标（文件），我收到第二个表单脏检查if ($scope.iForm.$dirty)和$dirty警报错误。

angular.js:12520 TypeError: Cannot read property '$dirty' of undefined

和

<form name="iForm" ng-controller="Controller" confirm-on-exit="" ng-model="myModel" class="ng-pristine ng-untouched ng-valid ng-scope">

演示：Plunker

4

3 回答 3

2

有一个更简单的方法。

只需将您的指令放在表单元素上并通过链接功能访问表单控制器：

myApp.directive('confirmOnExit', function() {
        return {
            require: 'form',
            link: function($scope, elem, attrs, formController) {
                // condition when back page is pressed 
                window.onbeforeunload = function(){
                    if (formController.$dirty) {
                        return "The form '" + formController.$name + "' is dirty, do you want to stay on the page?";
                    }
                }
                // condition when user try to load other form (via icons )
                $scope.$on('$stateChangeStart', function(event, next, current) {
                    if (formController.$dirty) {
                        if(!confirm(formController.$name + ". Do you want to continue ?")) {
                            event.preventDefault();
                        }
                    }
                });
            }
        };
    });
于 2016-03-28T22:24:50.857 回答
1

在检查表格是否脏之前检查表格是否存在。做类似的事情

if ($scope.myForm && $scope.myForm.$dirty) {
    if(!confirm("myForm. Do you want to continue ?")) {
        event.preventDefault();
    }
}

if ($scope.iForm && $scope.iForm.$dirty) {
    if(!confirm("iform. Do you want to continue ?")) {
        event.preventDefault();
    }
}
于 2016-03-28T21:58:25.517 回答
1

name通过从 的属性中获取它，使您的指令更简单form，因此指令中只有一个条件，并且可以在许多地方重用。

<form name="myForm" ng-controller="Controller" confirm-on-exit>

代码

var form = $scope[attrs.name]; //which will take out form name & do search that inside scope
if (form.$dirty) { //form object would always exist, to make sure you could also add one more check `form &&`
    if(!confirm("myForm. Do you want to continue ?")) {
        event.preventDefault();
    }
}

演示在这里

于 2016-03-28T22:04:48.380 回答