我有一个页眉,页脚,侧边栏和内容的布局,应该用于某些页面。所有页面都应包含在内容部分内。
<div class="wrapper">
<header class="main-header" ui-view="header"></header>
<aside class="main-sidebar" ui-view="left"></aside>
<div class="content-wrapper" ui-view></div><!-- all pages should be included here -->
<footer class="main-footer" ui-view="footer"></footer>
<aside class="control-sidebar control-sidebar-dark" ui-view="right"></aside>
<div class="control-sidebar-bg"></div>
</div>
我的路线:
.state('main', {
views: {
'left': {
templateUrl: 'partials/design/left.html'
},
'header': {
templateUrl: 'partials/design/header.html'
},
'right': {
templateUrl: 'partials/design/right.html'
},
'footer': {
templateUrl: 'partials/design/footer.html'
}
},
url: '/'
})
.state('info', {
url: '/info',
templateUrl: "partials/info.html",
})
.state('foobar', {
url: '/foobar',
templateUrl: "partials/foobar.html",
})
因此,信息和foobar应该是主要的儿童,但我不知道如何做到这一点。
答案 0 :(得分:0)
而不是
.state('main.info', {
url: '/info',
templateUrl: "partials/info.html",
})
.state('main.foobar', {
url: '/foobar',
templateUrl: "partials/foobar.html",
})
这样做:
{{1}}
答案 1 :(得分:0)
我发现ng-include
<div class="wrapper">
<header class="main-header" ng-include src="'partials/design/header.html'"></header>
<aside class="main-sidebar" ng-include src="'partials/design/left.html'"></aside>
<div class="content-wrapper"><ui-view></ui-view></div><!-- all pages should be included here -->
<footer class="main-footer"ng-include src="'partials/design/footer.html'"></footer>
<aside class="control-sidebar control-sidebar-dark" ng-include src="'partials/design/right.html'"></aside>
<div class="control-sidebar-bg"></div>
</div>
因此,将此模板作为父级的所有人都将显示在<ui-view></ui-view>
部分。