MYSQLi-命令不同步错误

这里有一些代码。我已经坚持了很长时间了，似乎无法摆脱这个错误。

    <?PHP
error_reporting(E_ALL);
ini_set('display_errors',1);
$mysqli = new mysqli('localhost', 'username', 'password', 'table');
$statsObjects = array();
$collatedObjects = array();
/* check connection */
if ($mysqli->connect_errno) {
    printf("Connect failed: %s'n", $mysqli->connect_error);
    exit();
}
Global $areRows;
$areRows = 2;


if( $result = $mysqli->query("SELECT * FROM stats WHERE collated = 0", MYSQLI_USE_RESULT) )
{
    while($row = $result->fetch_assoc())
    {
        array_push($statsObjects, 
            new Statistic(
                $row['ID'],
                $row['player_GUID'],
                $row['shots_fired'],
                $row['shots_hit'],
                $row['damage_done'],
                $row['damage_taken'],
                $row['friendly_damage_done'],
                $row['friendly_damage_taken'] 
                ));
    }
 $success = true;
} //end if
$result->free_result();
if($success)
{
    foreach($statsObjects as $stat)
    {
        $statsGuid = $stat->getGuid();
        $query = "SELECT COUNT(*) AS total FROM collatedStats WHERE player_GUID = '" . $statsGuid . "'";
        if( $result2 = $mysqli->query($query, MYSQLI_USE_RESULT) )
        {
            $value = $result2->fetch_assoc();
        $rows = $value['total'];
        if($rows > 0)
        {
            $areRows = 1;
        }
        else
        {
            $areRows = 0;
        }
        }
        else 
        {
            echo("Error <br/>");
            echo($mysqli->error);
        }
        if($areRows == 1)
        {
            echo("Found a row! <br/>");
        }
        elseif($areRows == 0)
        {
            Echo("No rows found. =) <br/>");
        }
    } //end foreach
}

//OBJECT
class Statistic
{
    var $GUID;
    var $shotsfired;
    var $shotshit;
    var $damagedone;
    var $damagetaken;
    var $friendlydamagedone;
    var $friendlydamagetaken;
    var $ID;
    function Statistic($ID, $GUID, $fired, $hit, $ddone, $dtaken, $fddone, $fdtaken)
    {
        $this->id = $ID;
        $this->GUID = $GUID;
        $this->shotsfired = $fired;
        $this->shotshit = $hit;
        $this->damagedone = $ddone;
        $this->damagetake = $dtaken;
        $this->friendlydamagedone = $fddone;
        $this->friendlydamagetaken = $fdtaken;
    }
    function getID()
    {
        return $this->ID;
    }
    function getGuid()
    {
        return $this->GUID;
    }
    function getShotsFired()
    {
        return $this->shotsfired;
    }
    function getShotsHit()
    {
        return $this->shotshit;
    }
    function getDamageDone()
    {
        return $this->damagedone;
    }
    function getDamageTaken()
    {
        return $this->damagetaken;
    }
    function getFriendlyDDone()
    {
        return $this->friendlydamagedone;
    }
    function getFriendlyDTaken()
    {
        return $this->friendlydamagetaken;
    }
    function getAccuracy()
    {
        if($shotsfired == 0)
        {
            $accuracy = 0;
        }
        else
        {
            $accuracydec = $shotshit / $shotsfired;
            $accuracy = $accuracydec * 100;
        }
        return $accuracy;
    }
}

基本上，每次我运行代码时，它都会出现错误"命令不同步；你现在不能运行这个命令"。我花了两天时间试图修复它——按照人们的指示，在运行下一个之前释放结果。我甚至在以前的代码中使用了一个准备好的语句，但它也不起作用——这是新编写的代码，试图让它起作用。我所做的所有阅读都表明，当你试图运行一个sql命令，而另一个命令仍在接收数据时，就会发生这个错误——然而，我已经调用了我的第一个查询，将其全部存储在一个数组中——然后我在数组中循环以获取下一批数据。。这给了我一个错误，这就是我感到困惑的地方。

任何帮助都将不胜感激！