php - Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given -

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wtf? - becoming nightmare, seem error after error.. im willing pay time me achieve seems unachievable, please.. alls want dropdown menu selects data table according category , displays it.. help! :(

<form action="portfolio.php" method="post">  <select onload="displayproject(this.value);" onchange="displayproject(this.value);">   <option value='none'>all</option>   <option value='1'>fencing</option>   <option value='2'>driveway</option>  </select> </form>  <?php   $db = new mysqli('localhost', 'wlarter_user', 'pw', 'wlarter_portfolio');  if($db->connect_errno > 0){     die('unable connect database [' . $db->connect_error . ']'); }  $option= $_post['option'];  $queries = "select * image";  if ($option != 'none'){  $queries = "select * image category=".$option;  }  $result=@mysqli_query($db,"$queries");  while($row = mysqli_fetch_array($result)) { ?>  <div class="box-portfolio"> <?php echo $row['img']; ?> </div>   <?php }                       mysqli_close($db); ?>

simply turn around, want:

$queries=$query;

into

$query = $queries;

full code:

<form action="portfolio.php" method="post">     <select onload="displayproject(this.value);" onchange="displayproject(this.value);">         <option value='none'>all</option>         <option value='1'>fencing</option>         <option value='2'>driveway</option>     </select> </form>  <?php  $db = new mysqli('localhost', 'wlarter_user', 'pw', 'wlarter_portfolio');  if($db->connect_errno > 0){     die('unable connect database [' . $db->connect_error . ']'); }  $option= $_post['option'];  $queries = "select * image"; if ($option != 'none'){     $queries = "select * image category=".$option; }  $query=$queries; $result=mysqli_query($db,"$query");  while($row = mysqli_fetch_array($result)) {     ?>      <div class="box-portfolio"> <?php echo $row['img']; ?> </div>   <?php } mysqli_close($db); ?>

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