我正在尝试从Angular中的ajax响应中提取div。我不想显示所有页面。我做过这样的事情,但注意到了。
$http({
method: 'GET',
url: '/angularJS/eg1.html'
})
.then(function (response) {
$scope.page = $sce.trustAsHtml(response.data)
$scope.result = $scope.page.document.getElementById('ajax');
});
我的观点是这样的
<div class="container">
<div class="row" ng-bind-html="result">
</div>
我要展示的div:
<div id="ajax">
<h2>Guess the Number !</h2>
<p class="well lead">Guess the computer generated random number between 1 and 1000.</p>
<label>Your Guess: </label><input type="number" ng-model="guess"/>
<button ng-click="verifyGuess()" class="btn btn-primary btn-sm">Verify</button>
<button ng-click="initializeGame()" class="btn btn-warning btn-sm">Restart</button>
<p>
</div>
我是棱角分明的新人,请耐心等待:)。
答案 0 :(得分:0)
Yo可以使用jqLite
按id
获取元素。
喜欢这个
var elem = angular.element('<div id='dt'> <span id='ajax'>312312</span></div>');
console.log(elem[0].querySelector('#ajax'));
jsfiddle上的实例。
angular.module('ExampleApp', [])
.controller('ExampleController', function($scope) {
$scope.name = "<div id='dt'> <span id='ajax'>312312</span></div>";
var elem = angular.element($scope.name);
console.log(elem[0].querySelector('#ajax'));
});
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
<div ng-app="ExampleApp">
<div ng-controller="ExampleController">
<form name="testForm">
<input name="myName" ng-model="name">
</form>
</div>
</div>