将输入值设置为空

时间:2016-03-14 10:21:36

标签: javascript angularjs input ionic-framework

我想在单击按钮后将输入值设置为null。单击按钮后,spremi()会执行它的操作,但我希望将podaci.tezina和podaci.mamac的值设置为起始值,以便用户输入新值。

 .controller('SpremiCtrl', function($scope, $localStorage, $ionicPopup, $location, $window) {


    $scope.ribe = ["Saran", "Stuka", "Som"];

    var ribice = [];

    $scope.spremi = function(t_tezina, m_mamac){

          console.log(t_tezina);
          console.log(m_mamac);
          var podaci = {Tezina: t_tezina, Mamac: m_mamac};  
          console.log(podaci);
          ribice.push(podaci);

          $localStorage.fish = ribice;
          console.log(ribice);

    var confirmPopup = $ionicPopup.alert({
    title: 'Spremljeno'

         });
       };

 html

<ion-view view-title="Browse">
      <ion-content>
      <select id="vrstaribe" ng-model="selekt" ng-options="r as r for r in ribe" selected>
       <option value="">Vrsta ribe</option> 
      </select>
       <label class="item item-input">
       <input id="tezina" type="number" placeholder="Tezina" ng-model="podaci.tezina">
       </label>
       <label class="item item-input">
       <input id="mamac" type="text" placeholder="Mamac" ng-model="podaci.mamac">
       </label>
       <button class="button button-positive" ng-click="spremi(podaci.tezina, podaci.mamac)">Spremi</button>
       <button class="button button-positive" ng-click="dohvati()">Prikaži</button>
       <button class="button button-positive" ng-click="isprazni()">Isprazni</button>

      </ion-content>
    </ion-view>

1 个答案:

答案 0 :(得分:0)

您应该在所有其他逻辑之后重置方法末尾的值,以便下次调用时它是新鲜的。

$scope.spremi = function(t_tezina, m_mamac){
    console.log(t_tezina);
    console.log(m_mamac);
    var podaci = {Tezina: t_tezina, Mamac: m_mamac};  
    console.log(podaci);
    ribice.push(podaci);

    $localStorage.fish = ribice;
    console.log(ribice);

    var confirmPopup = $ionicPopup.alert({
        title: 'Spremljeno'
    });


    // reset your variables here.
    podaci.Tezina = '';
    podaci.Mamac = '';
};

还应该注意,您应该使用$log.debug()(或朋友)而不是console.log(),因为您可以在构建版本时通知$logProvider禁用调试输出而不是调试。这可以让生活更轻松。

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