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Home/ Questions/Q 6926873
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T10:58:28+00:00 2026-05-27T10:58:28+00:00

Another problem I’ve come across is this one. I’m trying to call this function

  • 0

Another problem I’ve come across is this one. I’m trying to call this function from another function but the SQL query is giving me this error:

Catchable fatal error: Object of class mysqli could not be converted to string in …..

I’m clueless as to why this keeps occurring. Could someone please explain to me why this fatal error wont go away? I’m using procedural statements.

    function answered($you, $dbc)
    {
        $SQLanswered = "SELECT readcount FROM Faq WHERE sender = '$you' ";
        $Ranswered = mysqli_query ($dbc. $SQLanswered)
            or trigger_error("Query: $SQLanswered\n<br />MySQL Error: " . mysqli_error($dbc));

        while($row =  mysqli_fetch_assoc($Ranswered))
        {
             $answered[] = $row['readcount'];
        }

        mysqli_free_result($Ranswered);
        for($i = 0; $i < sizeof($answered); $i++)
        {
             $num += $answered[$i];
        }
        if($num > 0)
        {
              echo "   <a href='extrainfo.php'>($num answered)</a>";    
        }
    }
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  1. Editorial Team
    Editorial Team
    2026-05-27T10:58:29+00:00Added an answer on May 27, 2026 at 10:58 am
     mysqli_query ($dbc. $SQLanswered)
    

    Shouldn’t that be

      mysqli_query ($dbc, $SQLanswered)
    

    ?

    Looks like you’re accidentally doing string concatenation.

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