XF 2.2 How do I format this link in the template?

[AndrewSimm]

The values come from $sport, $team, and $year. I can't figure out how to format a link that includes 3 variables. $sport and $team are pulled in with the template and $year is looped through from $years. It is worth noting that the route works fine when I enter the information into the address bar.

                    <xf:foreach loop="$years" value="$year">
                        <a href="{{ link('players/commits', $sport) }}" class="menu-linkRow u-indentDepth0 js-offCanvasCopy">{$year.name}</a>
                    </xf:foreach>

 

[Jeremy P]

The link builder expects an associative array (or something that can be treated like an associative array, like an entity) with the corresponding route parameters as key/value pairs. You would want to rename the name parameter so it is unique for each fragment (sport_name, team_name, etc.). Then you could do something like this:

{{ link('players/commits', {
    'sport_id': $sport.sport_id,
    'sport_name': $sport.name,
    'team_id': $team.team_id,
    'team_name': $team.name,
    'year_id': $year.year_id, 
    'year_name': $year.name
}) }}

Of course that's rather tedious to do more than once, so you could create a method somewhere that builds the array and then call it in your template or pass that through the view parameters.