如何将值发送到Ionic中具有条件的其他页面

My app.js below:

我的app.js如下:

var app = angular.module('dbpjkApps', ['ionic'])

app.controller('kategoriCtrl', function($scope) {
    $scope.listKat = [
        {kat: 'math'}, {kat: 'physics'}, {kat: 'English'}, {kat: 'bahasa'},
    ]})
    .config(function($stateProvider, $urlRouterProvider) {
        $stateProvider

        .state('menu1', {
            url: '/menu1',
            templateUrl: 'xxx.html' //this is condition what i mean. 
        })

        .state('login1', {
            url: '/login1',
            templateUrl: 'login1.html'
         });

         $urlRouterProvider.otherwise('/login1');
    })
})

At the 1st step user are choosing category from select/combobox on login1.html.

在第一步,用户从login1.html上的select / combobox中选择类别。

How to write condition when user chooses math or physics, and then open page menu1.html or menu2.html accordingly? And how to get value that user chooses from login1.html on menu1.html or menu2.html?

当用户选择数学或物理时,如何写条件,然后打开页面menu1.html或menu2.html?如何从menu1.html或menu2.html上的login1.html中获取用户选择的价值?

Thank You :)

谢谢 :)

2 个解决方案

#1

You can use the ng-model on combobox which will give you selected value

您可以在组合框上使用ng-model,它将为您提供所选值

and just define the state in stateProvider

并只在stateProvider中定义状态

.state('menu1', {
            url: '/menu1',
            templateUrl: 'menu1.html' //this is condition what i mean. 
        })


.state('menu2', {
            url: '/menu2',
            templateUrl: 'menu2.html' //this is condition what i mean. 
        })

After that just check the ng-model value of combobox in Login Controller after that just add following line to redirect the page

之后只需在登录控制器中检查组合框的ng-model值,然后只需添加以下行即可重定向页面

if(condition1){
       $state.go('menu1'); 
}else if(condition2){
       $state.go('menu2'); 
}

2 个解决方案

#1

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