php - Parse error: syntax error, unexpected T_IF -

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i have become stuck problem week or now, , can't seem wrap head around it. put hands , admit not greatest php any highly appreciated.. trying achieve drop down menu has 3 categorys 'all', 'fencing' , 'driveways'. when selected want specific images displayed according category, , when 'all' selected display every image in table.. have mysql table set 'image' 3 columns; 'id', 'img' , 'category'.

edit: warning: mysqli_fetch_array() expects parameter 1 mysqli_result, boolean given in /home1/wlarter/public_html/portfolio.php on line 80

while($row = mysqli_fetch_array($result))

<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 }  $queries=$query; $result=@mysqli_query($db,"$query");  while($row = mysqli_fetch_array($result)) { ?>  <div class="box-portfolio"> <?php echo $row['img']; ?> </div>   <?php } mysqli_close($db); ?>

replace php code this, had syntax errors

<?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'];  $queries = "select * image"; //code updated if ($option != 'none') {      $queries = "select * image category=".$option; //code updated }  $queries=$query; $result=@mysqli_query($db,"$query");  while($row = mysqli_fetch_array($result)) { ?>     <div class="box-portfolio"> <?php echo $row['img']; ?> </div> <?php }                       //code updated mysqli_close($con); ?>

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