用两个变量连接URL

我正在做的是从数据库中获取纬度和经度，并以$row['latitude']格式显示。现在我要做的是，无论用户点击哪个超链接，它都会把用户带到maps.google.com，并根据特定的纬度和经度显示位置

请帮忙!这是我的PHP代码。

    <?php
    $location=$_POST["location"]; 
    $myArray = json_decode($location,true);
    $jsonArray = $myArray[0];
    $longitude = $jsonArray['longitude']; $adress = $jsonArray['adress'];
    $con=mysqli_connect("","","","");
   // Check connection if (mysqli_connect_errno())   {   echo "Failed to
   connect to MySQL: " . mysqli_connect_error();   }
   $sql = "INSERT INTO chokh_db. gpslocations (     latitude,   longitude ,
   datetime     ) VALUES ( '$latitude' , '$longitude' ,  NOW() );";

 if (!mysqli_query($con,$sql)) {
   die('Error: ' . mysqli_error($con)); } $result = mysqli_query($con,"SELECT * FROM      _db. gpslocations where
      latitude <> 0");
        while($row = mysqli_fetch_array($result))   {       
            $lat = $row['latitude'];    
            $long = $row['longitude']   
            echo $row['datetime'] . 
                "&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;" .          
                $row['latitude'] .
                "&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;" .
                $row['longitude'] ;      
            $url = "http://maps.google.com/?q='.$lat,$long.' "; 
            echo "<a href='$url'>View</a>";    echo "<br>";   
        }
        mysqli_close($con);
    ?>