Checking if a page is not a draft and include its content

Kirby 3 Solved ✅
1

Hello,

I’m migrating from V2 to V3 and try to display the content of a project page on the homepage if the page is listed.

In V2 I was doing:
<?php if(page('projets')->isVisible() && page('projets')->hasVisibleChildren()): ?>

in V3 I tried:
<?php if(!page('projets')->isDraft() && page('projets')->hasListedChildren()): ?>

or

<?php if(page('projets')->exists() && page('projets')->isListed() && page('projets')->hasListedChildren()): ?>

But it throw the error Call to a member function exists() on null

I think it’s normal as the page does not really exists (just in draft state) but I don’t know if there is a solution to display this page only if it’s not a draft. I tried with find() but don’t succeed.

How can I “select” draft pages so my client can load/unload content of certain sections of his website?

Thank you.

2

The page() helper only fetches published pages, so checking if it is a draft is not necessary. However, before you check if the page has children, you should make sure you have a page. So your code should look like this:

<?php if (($page = page('projects')) and $page->hasListedChildren()): ?>
<!-- do something -->
<?php endif ?>
4

Thank you very much!
(doesn’t work as first as I past the code with $page = page('projects') in english instead of $page = page('projets') in french :slight_smile:)

5

Oh, sorry, missed the French spelling… But at least it didn’t throw an error, that’s why we introduce the check if the page exists.

6

No problem. And to achieve exactly the same behavior than in V2, the code is:

<?php if (($page = page('projets')) and $page->isListed() and $page->hasListedChildren()): ?>
<!-- do something -->
<?php endif ?>