I'm making a people directory with the A - B - C etc. links, where clicking a letter will take you to the people whose last names begin with that letter. I have a People channel with a last_name field that I can use.

Can anyone recommend a good way to approach this? Thank you.

3 Answers 3


You might use "Low Alphabet"

See the tags from here: http://gotolow.com/addons/low-alphabet/docs/tags

For last_name, just pass parameter orderby="last_name" within {exp:low_alphabet:entries}


Disclosure: This is probably insane.

That said, here's what I did to "solve" the problem utilizing P&T's Iterate plugin and Stash:

    <ul class="item-list-nav">
{exp:iterate alpha='a|b|c|d|e|f|g|h|i|j|k|l|m|n|o|p|q|r|s|t|u|v|w|x|y|z' parse='inward'}
        <li class="ig-{alpha}"><a href="#ig-{alpha}">{alpha}</a></li>

{exp:iterate alpha='a|b|c|d|e|f|g|h|i|j|k|l|m|n|o|p|q|r|s|t|u|v|w|x|y|z' parse='inward'}
    <div class="item-list-group" id="ig-{alpha}"><a name="ig-{alpha}"></a>
        <span class="item-list-group-title">{alpha}</span>
    {exp:stash:get_list name='service_list' match='#^(?i){alpha}#' against='item_title'}
            <li><a href="{item_url}">{item_title}</a></li>

Insane? Maybe. But does it work? You bet.

  • Does this take into account letters that do not have an associated item? For example, will it output "Q" and then an empty list, because I don't have any last names that start with "Q"?
    – Bryan
    Jul 19, 2013 at 18:24
  • In my project an empty placeholder div (with the letter) was precisely the desired outcome. There are a few ways that you could avoid the empty lists--using something like {exp:stash:list_count}, for example. Jul 19, 2013 at 18:30

At first I was thinking of trying the search:field parameter, but instead used a href="#A" links at the top with named anchors with PHP on Output in my channel entries loop:

    <td id="<?
                    $lchar1 = substr("{last_name}", 0, 1);
                    if ($lchar1 != @$lchar2)
                        echo "$lchar1";
                    $lchar2 = $lchar1;

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