Question

所以我有一个数据库，我希望以表格的形式输出数据。

<table>
  <tr>
    <th class="tg-031e">Username</th>
    <th class="tg-031e">Password</th>
  </tr>
..
</table>

现在我得到这样的数据：

 <?php include 'connect.php';
 echo "<tr><td class='tg-031e'>";
 $SQL = "SELECT `Username` FROM `Users` WHERE ID=1";
 $exec = mysql_query($SQL, $connection);
 while($row = mysql_fetch_array($exec)){
    echo $row['Username'] . "</td>";
 } ?>

然而，在这种情况下，我需要独立地回显每列。当MySQL列中出现新数据时，如何使PHP动态创建包含数据库信息的HTML表行？

Answer 1

<?php include 'connect.php';

$SQL = "SELECT `Username`, Password FROM `Users`";
$exec = mysql_query($SQL, $connection);
echo "<table><tr><td class="tg-031e">Username</td><td class="tg-031e">Password</td></tr>";
while($row = mysql_fetch_array($exec))
{
    echo "<tr><td class='tg-031e'>";
    echo $row['Username'] . "</td>";
    echo "<td class='tg-031e'>".$row['Password'] . "</td></tr>";
 }
echo "</table>";

?>

Answer 2

像这样修改你的PHP代码：

 <?php include 'connect.php';

   $SQL = "SELECT `Username` FROM `Users` WHERE ID=1";
   $exec = mysql_query($SQL, $connection);

   echo "<table>";
   echo "<tr>";
   echo "<th class="tg-031e">Username</th>";
   echo "<th class="tg-031e">Password</th>";
   echo "</tr>";    

   while($row = mysql_fetch_array($exec)){
      //add as many fields in record, i.e: username, password... etc.
      echo "<tr>";
      echo "<td class='tg-031e'>" . $row['Username'] . "</td>";
      echo "<td class='tg-031e'>" . $row['Password'] . "</td>";
      echo "</tr>"; 
   }

   echo "</table>";  

 ?>

将MySQL数据提取为表格内容

2 个答案: