Question

在弹出窗口中，我显示的是html，我从另一个div复制并在弹出窗口中显示。在这里，我想验证此输入字段是否必需。并在输入框下方显示错误消息。的index.html × 载入中......

  <!-- html of change zip code -->

<div class="hidden" id="updateZipContent">
    <div class="zipContent">
        <div class="padding-bt-2">Please enter new zip code</div>
        <div class="row">
            <div class="text-left col-md-6 padding-bt-2">
                    <input ng-model="zipCode" type="text" class="form-control" maxlength="5" data-required="true" number-only>
            </div>  
            <div class="text-left col-md-4">
                <button class="btn btn-primary form-control">Update</button>
            </div>
        </div>
    </div>
</div>

更改邮政编码操作是用autoQuotectrl.js

编写的

$scope.changeZipCode = function()
                    {
                        $rootScope.myModal = true;
                        var firstDivContent = document.getElementById('updateZipContent');
                        var secondDivContent = document.getElementById('dynamicContect');
                        secondDivContent.innerHTML = firstDivContent.innerHTML;                        
                    }

为了保持其他动作分开，我编写了新的控制器utilityCtrl.js。在这里我写了隐藏这个弹出窗口的动作。

$scope.closePopup = function ()
        {
            console.log('here in utility');
            $rootScope.myModal = false;
            document.getElementById('dynamicContect').innerHTML = "";
        }

如何在此处设置验证？ https://plnkr.co/edit/aV65Nab9U9I6YlK2g4sY?p=preview

Answer 1

查看更新的plunker。

我们可以使用$compile指令。

$scope.changeZipCode = function()
                {
                    $rootScope.myModal = true;
                    var firstDivContent = document.getElementById('updateZipContent');
                    var secondDivContent = document.getElementById('dynamicContect');
                    var clonedElement = $compile(firstDivContent.innerHTML)($scope, function(clonedElement, scope) {
                      //attach the clone to DOM document at the right place
                      secondDivContent.innerHTML ="";
                      angular.element(secondDivContent).append(clonedElement);
                  });                        
                }

Answer 2

var app = angular.module('plunker', []);

app.controller('MainCtrl', function($scope) {
  $scope.name = 'World';
});

<html ng-app="plunker">

  <head>
    <meta charset="utf-8" />
    <title>AngularJS Plunker</title>
    <script>document.write('<base href="' + document.location + '" />');</script>
    <link rel="stylesheet" href="style.css" />
    <script data-require="angular.js@1.2.x" src="https://code.angularjs.org/1.2.25/angular.js" data-semver="1.2.25"></script>

  </head>

  <body ng-controller="MainCtrl">
    <div ng-form="myForm">
      <input type="text" required ng-model="user.name" placeholder="Username">
      
      <button ng-click="doSomething()" ng-disabled="myForm.$invalid">DO</button>
    </div>
  </body>

</html>

如何在角度js中验证没有表单标签的输入？

2 个答案: