在网站文章列表中显示类别树 Odoo
Display in the list of website articles the tree of categories Odoo
我想在网站文章列表中显示类别树
我尝试使用此源代码,但它只显示了最后的类别
<div class="row">
<div class="col-sm-12">
<ol class="breadcrumb">
<li t-if="category"><a t-att-href="keep('/shop',category=0)" t-field="category.name"/></li>
<li t-if="category">
<a t-att-href="keep('/shop/category/' + slug(category), category=0)"
t-field="category.name"/>
</li>
</ol>
<ol>
<li t-if="category.child_id" t-attf-class="text-primary fa #{'fa-chevron-down' if category.id in parent_category_ids else 'fa-chevron-right'}">
<a t-att-href="keep('/shop/category/' + slug(category), category=0)" t-field="category.name"></a>
</li>
</ol>
</div>
</div>
我想要像图片一样的结果
我在这个模块中找到了解决方案
https://www.odoo.com/apps/modules/10.0/website_breadcrumb/
您可以在网站模块 xml 文件中添加此模板
这是真实代码 XML
<template id="breadcrumb" name="Breadcrumb snippet">
<!-- Know if current page is in a menu item -->
<t t-set="bc_item"
t-value="request.env['website.menu']
.search([('url', '=', request.httprequest.path)], limit=1)"/>
<!-- If so, create the breadcrumbs -->
<ol t-if="bc_item" class="breadcrumb">
<!-- Parents -->
<t t-foreach="bc_item.get_parents(True)" t-as="step">
<li t-if="step.url or step_first">
<a t-att-href="step.url or (step_first and '/')">
<t t-if="step_first and not step.url">
Home
</t>
<t t-if="not step_first and step.url">
<t t-esc="step.name_get()[0][1]"/>
</t>
</a>
</li>
</t>
<!-- Current -->
<li class="active" t-esc="bc_item.name_get()[0][1]"/>
</ol>
</template>
谢谢
我想在网站文章列表中显示类别树 我尝试使用此源代码,但它只显示了最后的类别
<div class="row">
<div class="col-sm-12">
<ol class="breadcrumb">
<li t-if="category"><a t-att-href="keep('/shop',category=0)" t-field="category.name"/></li>
<li t-if="category">
<a t-att-href="keep('/shop/category/' + slug(category), category=0)"
t-field="category.name"/>
</li>
</ol>
<ol>
<li t-if="category.child_id" t-attf-class="text-primary fa #{'fa-chevron-down' if category.id in parent_category_ids else 'fa-chevron-right'}">
<a t-att-href="keep('/shop/category/' + slug(category), category=0)" t-field="category.name"></a>
</li>
</ol>
</div>
</div>
我想要像图片一样的结果
我在这个模块中找到了解决方案
https://www.odoo.com/apps/modules/10.0/website_breadcrumb/
您可以在网站模块 xml 文件中添加此模板
这是真实代码 XML
<template id="breadcrumb" name="Breadcrumb snippet">
<!-- Know if current page is in a menu item -->
<t t-set="bc_item"
t-value="request.env['website.menu']
.search([('url', '=', request.httprequest.path)], limit=1)"/>
<!-- If so, create the breadcrumbs -->
<ol t-if="bc_item" class="breadcrumb">
<!-- Parents -->
<t t-foreach="bc_item.get_parents(True)" t-as="step">
<li t-if="step.url or step_first">
<a t-att-href="step.url or (step_first and '/')">
<t t-if="step_first and not step.url">
Home
</t>
<t t-if="not step_first and step.url">
<t t-esc="step.name_get()[0][1]"/>
</t>
</a>
</li>
</t>
<!-- Current -->
<li class="active" t-esc="bc_item.name_get()[0][1]"/>
</ol>
</template>
谢谢