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在模板中绑定动画触发器名称时出现错误。[@triggerName] 似乎不起作用。我错过了什么？

零件

import {Component, ViewChild, ViewChildren, Input, ElementRef, Renderer} from '@angular/core';
import {trigger, state, style, transition, animate} from '@angular/core';
import 'rxjs/add/operator/map';

@Component({
    selector: 'Header',
    templateUrl: '/components/header/header.html',
    directives: [],

providers: [],
    host: {'(document:click)': 'closeLoginModal($event)'},
    animations: [
        trigger('showLoginTrigger', [
            state('true', style({opacity: '1',})),
            state('false', style({opacity: '0',})),
            transition('0 => 1', animate('400ms ease-in')),
            transition('1 => 0', animate('400ms ease-out'))
        ])
    ]
})
export class Header {
    static get parameters() {
        return [[ElementRef]];
    }
    constructor(ElementRef) {
        this.el = ElementRef.nativeElement;
        this.showLogin = true;
        this.showMenu = false;
        this.test = true;
    }
    closeLoginModal(e) {
        if (this.el.contains(e.target)) return;
        this.showLogin = false;
    }
}

模板翡翠

button.no-border((click)='test=!test') LOGIN
    .login-modal(*ngIf='showLogin', [@showLoginTrigger]='test')
        h2
            | Publisher
            br
            | Login

我得到的错误是：

无法绑定到“@showLoginTrigger”，因为它不是已知的本机属性

4

1 回答 1

3

问题在于语法 [@showLoginTrigger]='test'

RC4：应该是@showLoginTrigger='test'

RC5 及更高版本：应该是[@showLoginTrigger]='test'

于 2016-09-24T12:35:52.977 回答