CODEWITHSUNDEEP
phpsymfonyezpublishezplatform
user1037355
user1037355

Reputation:

ezplatform render links with url and object name from multi-relational content item in content type

does anyone know now to create a custom view type for ez platform? The default 3 have been exhausted and we need a new one for 'link' Alternatively, does anyone know how to use the render( controller( with a custom template as this would also resolve out block right now.

Basically, we have a multi-relational field in a content object used and we need to print links to all the related contentIds, path works great but we cannot find a way to extract the name of the content object for the link without doing some fairly funky tpl logic of passing in params.

EG: As a hack for now we can pass in "embed_type" as a custom param with the render(controller("ez_content:viewAction" to pull in an alternate view for the content object for a specific content type and view type.

{% if embed_type is defined %}
    {% include "embed/#{embed_type}.html.twig" %}
{% else %}
        <h1>{{ ez_field_value( content, 'name') }}</h1>
{% endif %}

However, this is very ugly and all we really want to do is use 1 template for all content types, so all we need to do is loop through the relational field and print links (as the only thing available in the content field: "destination ids"). I am sure there used to be this option in the docs but i cannot find it anymore eg:

{% set links =  ez_field_value( footer, "first_links_row" ).destinationContentIds%}
{% for id in links %}
   {{ render(controller("ez_content:viewAction", {"contentId": id, "template": "link.html.twig"})) }}
{% endfor %}

Where the link.html.twig would simple print the link:

<a href="{{ path( "ez_urlalias", {"contentId": id} ) }}">
   {{ ez_field_value( content, "name" ) }}
</a>

If using a custom tpl is not possible with the render (controller ( helper then a new custom view type would also fix this issue, but i cannot find documentation for either.

Upvotes: 0

Views: 891

Answers (1)

iherak
iherak

Reputation: 61

You can create a twig function that would do that. We have something like this:

Definition:

new Twig_SimpleFunction(
   'content_name',
    array($this, 'getContentName')
),

Implementation:

public function getContentName($content, $forcedLanguage = null)
{
    if (!$content instanceof Content && !$content instanceof ContentInfo) {
        $contentInfo = $this->repository->getContentService()->loadContentInfo($content);
    } elseif ($content instanceof Content) {
        $contentInfo = $content->contentInfo;
    } else {
        $contentInfo = $content;
    }

    return $this->translationHelper->getTranslatedContentNameByContentInfo($contentInfo, $forcedLanguage);
}

which enables you to provide either content id, content info or content itself, and it returns translated content name

Upvotes: 2

Related Questions