提取MySQL数据作为表内容 - Extracting MySQL data as table content

Extracting MySQL data as table content

本文关键字：MySQL 数据提取 | 更新日期: 2023-09-27

所以我有一个数据库，我想从中以表的形式输出数据。

<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>";
 } ?>

但是，在这种情况下，我需要独立地回显每个列。当MySQL列中出现新数据时，我如何使PHP动态地使用来自数据库的信息创建HTML表行?

像这样修改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>";  
 ?>

<?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>";
?>