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Home/ Questions/Q 9154855
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T12:29:48+00:00 2026-06-17T12:29:48+00:00

static $searchQuery=nothing; if ($searchQuery==nothing) { if ($referral==) { $referral=referrer(); } $searchQuery=getSearchQuery($referral); } return $searchQuery;

  • 0
static $searchQuery="nothing";

if ($searchQuery=="nothing")
{
    if ($referral=="")
    {
        $referral=referrer();
    }
    $searchQuery=getSearchQuery($referral);
}
return $searchQuery;

Basically I need to initialize $searchQuery with something.

I can initialize it with “” but the value of $searchQuery is often legitimately “”.

So how should I initialize it? “”? Nil? Null? array()? What?

I can also use

static $result;
if  (isset($result))
{
    return $result;
}

but I got warning because $result is not defined.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T12:29:49+00:00Added an answer on June 17, 2026 at 12:29 pm

    It’s up to you, but you need to be careful with PHP’s loose typing. I prefer NULL, and I would write that code as:

    static $searchQuery = NULL;
    
    if( is_null($searchQuery) ) {
        if( emtpy($referral) ) {
            $referral=referrer();
        }
        $searchQuery=getSearchQuery($referral);
    }
    return $searchQuery;
    
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