PHP Function and array Syntax Issue

Question

I'm really struggling with something and wondered if anyone could spare a few moments to have a look at this code block.

The original line looked like this: $home_collectionsx=get_home_page_promoted_collections();

This brought back all the promoted to homepage items and displayed them on the homepage. I however simply want to pull 1 item in using the same code and an array function, the id is 5 for this purpose so I thought adding =array(5) or (array (5)) would work - but it doesn't.

I'm hoping it's something simple, or something that I have missed or not written correctly.

<?php 
if(!hook("EditorsPick")):
/* ------------ Collections promoted to the home page ------------------- */
$home_collectionsx=get_home_page_promoted_collections (array(5));
foreach ($home_collectionsx as $home_collectionx)
{
?>
<div class="EditorsPick">
<div class="HomePanel"><div class="HomePanelINtopEditors">
<div class="HomePanelINtopHeader">Editors Pick</div>
<div class="HomePanelINtopText">This is the editors pick of Asset Space...</div>
<div class="EditorsPicImage"><div style="padding-top:<?php echo floor((155-$home_collectionx["thumb_height"])/2) ?>px; margin-top: -24px; margin-bottom: -15px;">
<a href="<?php echo $baseurl_short?>pages/search.php?search=!collection<?php echo   $home_collectionx["ref"] ?>" onClick="return CentralSpaceLoad(this,true);"><img     class="ImageBorder" src="<?php echo   get_resource_path($home_collectionx["home_page_image"],false,"thm",false) ?>" width="<?php echo $home_collectionx["thumb_width"] ?>" height="<?php echo $home_collectionx["thumb_height"] ?>" /></div>
</div></div>
</div>
</div>
</div>
<?php
}
endif; # end hook homefeaturedcol
?>

This is the function to the DB itself that the above code is connecting to…

function get_home_page_promoted_collections()
{
return sql_query("select   collection.ref,collection.home_page_publish,collection.home_page_text,collection.home_page_image,resource.thumb_height,resource.thumb_width from collection left outer join resource on collection.home_page_image=resource.ref where collection.public=1 and collection.home_page_publish=1 order by collection.ref desc");
}

Any help would be hugely appreciated :-)

Many many thanks Rich

array() is not a function (even though the syntax looks like one). And you cannot magically pass arguments to that function and expect them to do something -- it was not designed for that. — Jon
– Jon, Commented Jul 26, 2013 at 8:45

MDEV · Accepted Answer · 2013-07-26 09:58:59Z

1

That function doesn't take a parameter: get_home_page_promoted_collections()

You want something like:

$home_collectionsx=get_home_page_promoted_collections(5);

And:

function get_home_page_promoted_collections($id=null)
{
    $filterClause = '';
    if(!is_null($id))
    {
        //to only return this id
        $filterClause = ' AND collection.ref = '.intval($id);
        //to get all but that id
        $filterClause = ' AND collection.ref != '.intval($id);
    }
    return sql_query("SELECT collection.ref,collection.home_page_publish,collection.home_page_text,collection.home_page_image,resource.thumb_height,resource.thumb_width FROM collection LEFT OUTER JOIN resource on collection.home_page_image=resource.ref WHERE collection.public=1 AND collection.home_page_publish=1".$filterClause." ORDER BY collection.ref DESC");
}

edited Jul 26, 2013 at 9:58

answered Jul 26, 2013 at 8:43

MDEV

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14 Comments

griffin Over a year ago

Unfortunately that (not taking arguments) in itself is not a problem in PHP as per default all functions support variable arguments - see php.net/manual/en/function.func-get-args.php

MDEV Over a year ago

Yes, but if the function doesn't look for any arguments, it's not exactly going to do anything no matter what/how many arguments you pass it

richyp147 Over a year ago

Thanks Smokey, but I'm getting this error now using your code: (database) line N/A: Unknown column 'collection.id' in 'where clause'<br/><br/>select collection.ref,collection.home_page_publish,collection.home_page_text,collection.home_page_image,resource.thumb_height,resource.thumb_width from collection left outer join resource on collection.home_page_image=resource.ref where collection.public=1 and collection.home_page_publish=1 AND collection.id = 5 order by collection.ref desc

MDEV Over a year ago

@user2621921 You mentioned the use of an id - if it's not in the collection table, change collection.id to resource.id or whatever table your ID of 5 refers to

richyp147 Over a year ago

Ok brill this worked collection.ref :-) However now I want to do the opposite in this function and eclude the id of 5: function get_home_page_promoted_collections() { return sql_query("select collection.ref,collection.home_page_publish,collection.home_page_text,collection.home_page_image,resource.thumb_height,resource.thumb_width from collection left outer join resource on collection.home_page_image=resource.ref where collection.public=1 and collection.home_page_publish=1 order by collection.ref desc"); }

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