1

I need to output a list of channel authors. There are about 40 authors but I want the 8 authors with the most entries. Not sure how to achieve this.

What I have so far:

    {# get all the itbuzz entries #}
    {% set allEntries = craft.entries.section('itbuzz').all() %}
    {# get all the authors #}
    {% set authors = craft.users.group('itBuzzAuthor').limit(8).all() %}

    <ul class="flex flex-row flex-wrap">
      {% for author in authors %}
        <li class="inline-block mb-4 mr-4 text-sm text-center">
          <img src="{{ craft.gravatar.url( author.email ) }}" alt="">
          <a href="#" class="text-blackBrand">{{ author.firstName }}<br>{{ author.lastName }}</a>
        </li>
      {% endfor %}
    </ul>

All this does is output the first 8 authors, it's not finding the 8 with the most entries.

2 Answers 2

2

You could do something like this:

{# Fetch entries + authors and group by author id #}
{% set entriesByAuthors = craft
    .entries()
    .section('itbuzz')
    .limit(null)
    .with('author')
    .all() | group(e => e.author.id) %}

{# Build an array with authorId and entry count for each that we can sort them #}
{% set leaderBoard = [] %}
{% for author, entries in entriesByAuthors  %}
    {% set leaderBoard = leaderBoard | merge([{'author' : author, 'count' : entries | length}]) %}
{% endfor %}

{# Build an array of user ids only sorted by entry count and extract the top 8 #}
{% set authorIds = [] %}
{% for result in leaderBoard | multisort(e => e.count, direction=SORT_DESC) | slice(0,8) %}
    {% set authorIds = authorIds | merge([result.author]) %}
{% endfor %}

{# Fetch users based on our sorted and sliced array of ids #}
{% for author in craft.users().id(authorIds).all() %}
    {{ author.firstName }} - {{ author.lastName }}<br>
{% endfor %}
1
  • Oli, this works perfectly, thank you. I"m not sure what's going on in with .with(author) and the |group filter. Could you explain that a little for me. Thank you.
    – CreateSean
    Mar 4, 2020 at 14:36
1

Maybe a good excuse to use the spaceship operator (<=>)... Combined with the group filter, this seems to work for me:

{% set groupedEntries = craft.entries.section('itbuzz').all()|group('author') %}

{% for author, entries in groupedEntries|sort((a, b) => a|length <=> b|length)|reverse|slice(0,8) %}
    <h3>{{ author }}</h3>
    {# optionally output their entries too... #}
    <ul>
        {% for entry in entries %}
            <li>{{ entry.title }}</li>
        {% endfor %}
    </ul>
{% endfor %}
3
  • Thought of that but the simple fact of grouping means you loose your user object and Sean seems to need to access author.stuff ¯_(ツ)_/¯
    – Oli
    Mar 4, 2020 at 10:05
  • Ah ha good point yerp. Mar 4, 2020 at 10:06
  • That sort filter - wow, I would never have thought of that.
    – CreateSean
    Mar 4, 2020 at 14:38

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