Question

当我加载它时会出现这个错误：注意：尝试在第29行的/public_html/php/application/views/admin/index.php中获取非对象的属性，并表示它不活动时很活跃。

请帮忙

    <div class="content">
    <?php 

    $con = mysql_connect("","","");
    if (!$con)
    {
      die('Could not connect: ' . mysql_error());
    }

    mysql_select_db("", $con);
    $result = mysql_query("SELECT * FROM users ORDER by user_id");

    echo "<link rel='stylesheet' href='<?php echo URL; ?>public/css/style.css' /><div class='CSSTableGenerator' >
      <table >
          <tr>
             <td>ID</td>
             <td >Username</td>
             <td>User Active</td>
          </tr>";

    while($row = mysql_fetch_array($result))
    {
      if ($row->user_active == "1") 
      {
        $mark = 'Yes';
      }
      else 
      {
        $mark = 'No';
      }

      echo "<tr>";
      echo "<td>" . $row['user_id'] . "</td>";
      echo "<td>" . $row['user_name'] . "</td>";
      echo "<td>" . $mark . "</td>";
      echo "</tr>";
   }
   echo "</table></div>";

   mysql_close($con);

   ?>

 </div>

谢谢josh_24_2

Answer 1

 while($row = mysql_fetch_array($result))
 {
  if ($row['user_active'] == "1") 
  {
    $mark = 'Yes';
  }
  else 
  {
    $mark = 'No';
  }
  echo "<tr>";
  echo "<td>{$row['user_id']}</td>";
  echo "<td>{$row['user_name']}</td>";
  echo "<td>{$mark}</td>";
  echo "</tr>";
  }
 echo "</table></div>";

Mysql数据库表错误

1 个答案: