将PHP变量传递给HTML img标签中的src

Question

<?php
    $avatar_location = $_SESSION['p_avatar_location'];
    $avatar_location = "avatar/".$avatar_location;
    $avatar_location = (string)$avatar_location;
    echo "avatar location: " $avatar_location;
    ?>

<img src=$avatar_location alt="Avatar" class="avatarlogin">

我想将$avatar_location作为src传递，但显示给我的是错误图像，但是它会打印正确的图像位置：

  

头像位置：avatar / avatar5.png

我已经尝试过<img src="avatar/avatar5.png" alt="Avatar" class="avatarlogin">，并且可以正常工作。如何传递$avatar_location变量以显示图像？

Answer 1

使用header.liquid标签

<?php ?>

Answer 2

您必须在<?php ?>内打印img标签

<?php
echo "<img src='{$avatar_location}' alt='Avatar' class='avatarlogin>";
?>