php输出从mysql到html表

目前我正在开发一个网站正常运行时搜索引擎但是我有一个愚蠢的问题.我想在表中输出多个mysql行,但是下面的代码会导致为找到的每一行创建一个单独的表.提前感谢您的协助 $searchTerm = trim($_GET[searchterm...

目前我正在开发一个网站正常运行时搜索引擎但是我有一个愚蠢的问题.我想在表中输出多个mysql行,但是下面的代码会导致为找到的每一行创建一个单独的表.提前感谢您的协助
        

 $searchTerm = trim($_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;
?>

解决方法:

用if替换if中的while：

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>";