有可能显示这样的表吗？

ok is it possible to show something like this

$con=mysqli_connect("localhost","root","","login");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
  ////////////////////////////////user1
  if($session->username=='user1')
{
echo "<table border='1'>
<tr>
<th><b>column1</b></th>
<th><b>column2</b></th>
<th><b>column3</th>
/tr>";

$result = mysqli_query($con,"SELECT * FROM test WHERE user1 == 0 ");
//if($result =='0')
    //{
        //echo"Nothing to show";
    //}
//else

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  if ($row['user1'] == '1'){echo "<td> confirmed </td>";} elseif ($row['user1'] == '2') {echo "<td> not confirmed </td>";} else{echo "<td>waiting</td>";}
  if ($row['user2'] == '1'){echo "<td>confirmed </td>";} elseif ($row['user2'] == '2') {echo "<td> not confirmed </td>";} else{echo "<td>waiting</td>";}
  if ($row['user3'] == '1'){echo "<td> confirmed </td>";} elseif ($row['user3'] == '2') {echo "<td> not confirmed </td>";} else{echo "<td>waiting</td>";}
  echo "</tr>";
  }
echo "</table>";
}

just help me on this i got error Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in

in mysql the comparison operator is = and not, like in PHP, ==.

So write:

$result = mysqli_query($con,"SELECT * FROM test WHERE user1 = 0 ");
// Only one `=` sign here ----------------------------------^

If you experience further problems of this type, have first a look at the value of $con->error after the execution of your query.