否则如果不在PHP工作

Kar*_* RK 0 php mysql if-statement while-loop

这是我的编码

          <?php
                try
                {
                    $stmt = $conn->prepare("SELECT * FROM registration");
                    $stmt->errorInfo();
                    $stmt->execute();

                    while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
                       {
                        echo "<tr>";
                        echo "<td>". $row['Autoid'] ."</td>";
                        echo "<td>". $row['Name'] ."</td>";
                        if($_SESSION['sesuname'])
                         {
                            echo "<td>Its You ". $_SESSION['sesuname'] ."</td>";
                         } 
                        else if(!$_SESSION['sesuname'])
                         {
                             echo '<td>Follow</td>';
                         }
                        echo "</tr>";
                       }
                } catch (PDOException $e) {
                       'Database Error : ' .$e->getMessage();
                }
            ?>
Run Code Online (Sandbox Code Playgroud)

我正在尝试显示Follow选项并在最终中登录用户名td.但是,此编码始终显示所有行中的登录用户名.我不知道我哪弄错了?

Rik*_*esh 5

我想你的病情应该是,

if($_SESSION['sesuname'] == $row['Name'])
{
   echo "<td>Its You ". $_SESSION['sesuname'] ."</td>";
} 
else
{
   echo '<td>Follow</td>';
}
Run Code Online (Sandbox Code Playgroud)