angularjs - 根据条件禁用 ui-sref

Question

你能告诉我一种禁用提交按钮的方法吗，该按钮会通过以下方式更改为新状态:

<a ui-sref="state">Submit</a>

仅当表单有效时才应启用该按钮。

ng-disabled 与 ui-sref 不起作用:

<form name="tickets">
  <button ng-disabled="!canSave()"><a ui-sref="view">Submit</a></button>
</form>

app.js 中的 canSave 函数是:

$scope.canSave = function(){
  return $scope.tickets.$dirty && $scope.tickets.$valid;
};

Answer 1

您可以简单地将其与 ng-click 配对，以便 ng-disabled 正常工作。

.controller('myCtrl', function($scope, $state) {
    // so that you can call `$state.go()` from your ng-click
    $scope.go = $state.go.bind($state);
})

<!-- call `go()` and pass the state you want to go to -->
<button ng-disabled="!canSave()" ng-click="go('view')>Submit</button>

这是使用自定义指令的更奇特的方法:

angular.module('myApp', ['ui.router'])
.config(function($stateProvider) {
  $stateProvider.state('home', {
    url: '/'
  });
})
.controller('myCtrl', function() {
  
})
.directive('uiSrefIf', function($compile) {
  return {
    scope: {
      val: '@uiSrefVal',
      if: '=uiSrefIf'
    },
    link: function($scope, $element, $attrs) {
      $element.removeAttr('ui-sref-if');
      $compile($element)($scope);
      
      $scope.$watch('if', function(bool) {
        if (bool) {
          $element.attr('ui-sref', $scope.val);
        } else {
          $element.removeAttr('ui-sref');
          $element.removeAttr('href');
        }
        $compile($element)($scope);
      });
    }
  };
})
;

<script src="//ajax.googleapis.com/ajax/libs/angularjs/1.3.2/angular.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/angular-ui-router/0.2.13/angular-ui-router.min.js"></script>

<div ng-app="myApp" ng-controller="myCtrl">
  <form name="form">
    <input ng-model="foo" ng-required="true">
    <button ng-disabled="form.$invalid">
      <a ui-sref-if="!form.$invalid" ui-sref-val="home">test</a>
    </button>
  </form>
</div>