php output from mysql to html table

Question

currently I'm working on a site uptime search engine however I'm having a silly issue. I want to output more than one mysql row in a table however my code below causes a separate table to be created for each row thats found. Thank you in advance for your assistance

 $searchTerm = trim($_GET['searchterm']);

 //check whether the name parsed is empty
 if($searchTerm == "")
{
echo "Please enter something to search for...";
exit();
} 

//database connection info
$host = "localhost"; //server
$db = "DB NAME"; //database name
$user = "USER"; //dabases user name
$pwd = "PASSWORD"; //password


$link = mysqli_connect($host, $user, $pwd, $db);


 $query = "SELECT * FROM sites WHERE name OR des LIKE '%$searchTerm%'";

 $results = mysqli_query($link, $query);


 if(mysqli_num_rows($results) >= 1)
 {

while($row = mysqli_fetch_array($results))
{
echo '<table class="table table-striped table-bordered table-hover">'; 
echo"<TR><TD>Name</TD><TD>Description:</TD><TD>Status</TD></TR>"; 
echo "<tr><td>"; 
echo $row['name'];
echo "</td><td>";   
echo $row['des'];
echo "</td><td>";    
echo $row['status'];
echo "</TD></tr>";  
echo "</table>";    
}

    }
    else
echo "There was no matching record for the name " . $searchTerm;
?>

Solution

Replace the while inside the if with this:

echo '<table class="table table-striped table-bordered table-hover">'; 
echo "<tr><th>Name</th><th>Description:</th><th>Status</th></tr>"; 
while($row = mysqli_fetch_array($results))
{
  echo "<tr><td>"; 
  echo $row['name'];
  echo "</td><td>";   
  echo $row['des'];
  echo "</td><td>";    
  echo $row['status'];
  echo "</td></tr>";  
}
echo "</table>";    

Comments

[Ichigo Kurosaki]

You can put your table tag outside the loop and only tr in the loop

[Patrick Kostjens]

You are printing the headers between all the rows.