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我正在使用Angular允許我根據需要創建儘可能多的輸入,但我想要一個簡單的警報來顯示輸入的值。如果我不使用重複,但在重複之後沒有使用,我已經能夠獲得價值。使用角度中繼器的多個輸入的警報輸入值
<div ng-controller="MainCtrl">
<form id="quickPick-form1" class="list" ng-submit="submit()" >
<fieldset data-ng-repeat="choice in choices" class="clearfix">
<input type="text" placeholder="Item" class="pull-left" ng-model="item">
<button class="remove pull-left" ng-show="$last" ng-click="removeChoice()">X</button>
</fieldset>
<button id="quickPick-button1" style="font-weight:600;font-style:italic;" class="addfields button button-calm button-block button-outline " ng-click="addNewChoice()">Tap me to add an item!</button>
<div class="spacer" style="width: 300px; height: 40px;"></div>
<p>Have enough items added? Click Randomize</p>
<button ng-click="showChoice()" id="quickPick-button2" class=" button button-calm button-block ">Randomize</button>
<div class="spacer" style="width: 300px; height: 20px;"></div>
</form>
</div>
<script>
.controller('MainCtrl', function($scope) {
$scope.choices = [{id: 'choice1'}, {id: 'choice2'}];
$scope.addNewChoice = function() {
var newItemNo = $scope.choices.length+1;
$scope.choices.push({'id':'choice'+newItemNo});
};
$scope.removeChoice = function() {
var lastItem = $scope.choices.length-1;
$scope.choices.splice(lastItem);
};
$scope.item = null;
$scope.showChoice = function(){
alert($scope.item);
};
});
</script>
哎呀看起來像我的例子冷落項目,但它是[NG-模型= 「項」。我已經更新了這個例子 – Rhino