2017-02-20 69 views
0

我多個菜單初學者到離子2如何根據條件顯示

下面是其中正在使用,使側面菜單使用離子框架

<ion-content> 
<ion-list *ngIf="!isUserLoggedIn()"> 
     <button menuClose ion-item *ngFor="let p of loginPages" (click)="openPage(p)"> 
     {{p.title}} 
     </button> 
    </ion-list> 
    <ion-list *ngIf="isUserLoggedIn()"> 
     <button menuClose ion-item *ngFor="let p of logoutpages" (click)="openPage(p)"> 
     {{p.title}} 
     </button> 
    </ion-list> 
    </ion-content> 

我的HTML代碼,這是我的應用程序組件

currentuser; 
    @ViewChild(Nav) nav: Nav; 
    auth:any; 
    rootPage: any = LoginPage; 
    loginPages:PageInterface[]=[ 
     { title: 'Login', component: LoginPage } 
    ]; 

    logoutpages :PageInterface[] = [ 

     { title: 'My Complaints', component: MycomplaintsPage }, 
     { title:'My Neighbours',component:NeighboursPage}, 
     { title:'Notifications',component:NotificationsPage}, 
     { title:'Directory',component:TabsPage}, 
     { title:'chat',component:ChatlistPage}, 
     { title:'Events',component:EventPage},  
     { title:'settings',component:SettingsPage}, 
     { title: 'LogOut',component:LoginPage,logsOut:true } 

    ]; 
isUserLoggedIn(): boolean { 
    let user = this.authservice.getcurrentuser(); 
return user !== null; 

    } 

已經顯示基於isUserLoggedIn的logoutpages。 我需要顯示基於UserRole的有條件的另一個菜單

請諮詢我, 感謝&問候

回答

0

您可以將菜單給ID並啓用或禁用根據你的病情菜單。 https://github.com/driftyco/ionic/tree/master/demos/src/menu

menu.enable(true/false, #menu);

+0

感謝烏爾Reply.I在所謂的新funtion isUserLoggedIn()後,無法安慰的價值。是我的功能檢查(){ VAR AUTH = localStorage.getItem( '電子郵件' ); if(auth!= null) { this.authservice.checkRole(); } } –

+0

嘗試window.localStorage.getItem('email'),甚至檢查undefined以及null(auth!='undefined')。 – Naveen

+0

我稱這個函數爲constructor.Check以下條件:ion-list * ngIf =「currentuser ==='6'」> 不能DISPLY菜單,但我的角色ID也是‘6’ –