Question

我想进行一些编辑，但是当用户在网站上时，我需要同时进行编辑。所以我有这个，但是网站源代码中评论了php：

 <?php

if($_SERVER['REMOTE_ADDR'] == "86.115.14.77"){ 
                echo "<li class='dropdown hv'>
                    <a style='cursor:pointer' class='dropdown-toggle ' data-toggle='dropdown'>
                        <i class='fa fa-user' aria-hidden='true'></i> Contul meu
                        <ul class='dropdown-menu'>
                            <li>
                                <?php include('form_logare.php'); ?>
                            </li>
                        </ul>
                    </a>
                </li>";
 } else{ 
            echo   "<li class='dropdown hv'>
                    <a style='cursor:pointer' class='dropdown-toggle ' data-toggle='dropdown'>
                        <i class='fa fa-user' aria-hidden='true'></i> Contul meu
                        <ul class='dropdown-menu'>
                            <li>
                                <?php include('form_logare.php'); ?>
                            </li>
                        </ul>
                    </a>
                </li>";
}
?>

因此<?php include('form_logare.php');?>在正面被评论。

<<在网站的来源中成为注释。

Answer 1

如果要在回显中包括php文件内容，则应将其连接起来：

<?php

if($_SERVER['REMOTE_ADDR'] == "86.115.14.77"){ 
                echo "<li class='dropdown hv'>
                    <a style='cursor:pointer' class='dropdown-toggle ' data-toggle='dropdown'>
                        <i class='fa fa-user' aria-hidden='true'></i> Contul meu
                        <ul class='dropdown-menu'>
                            <li>" . include('form_logare.php') . "</li>
                        </ul>
                    </a>
                </li>";
 } else{ 
            echo   "<li class='dropdown hv'>
                    <a style='cursor:pointer' class='dropdown-toggle ' data-toggle='dropdown'>
                        <i class='fa fa-user' aria-hidden='true'></i> Contul meu
                        <ul class='dropdown-menu'>
                            <li>". include('form_logare.php') . "</li>
                        </ul>
                    </a>
                </li>";
}
?>

Answer 2

include函数将按原样呈现，因为它位于引号内。您需要像这样将其连接起来。

  "<li>" . include('form_logare.php') . "</li>";

如何回显html并包含php

2 个答案: