警告：mysqli_fetch_array（）期望参数1为mysqli_result，[closed]中给出的字符串

Question

<?php
 $bla=$_GET['category'];
   $bla1= " SELECT * FROM project where id=$bla ";
      $result1= mysqli_query($db, $bla1);
                if(!$result1) {
                die("hehehe");
                             } 
   while($result2 = mysqli_fetch_array('result1')){ 
   echo $result2['project_name'] ;
        }
?>

whats wrong in this? I am getting this warning

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in ...

Answer 1

Use

mysqli_fetch_array($result1)

instead of

mysqli_fetch_array('result1')