用于阻止页面中任何代码的PHP代码

Question

我目前在我的页面上有以下代码.

<?php 
    if($result["r_approved"] == "APPROVED"){
echo "<!--";
} 
?>
<div class="main">
    <div class="main-sub">
<?php include('http://www.contractorsintelligence.com/contractors-license/includes-page-elements/navigation1.php'); ?>
<div id="mid-top"><img src="https://www.contractorsintelligence.com/images/shadowbg-top.png" width="990" height="20" alt="Top Spacer"/></div>
            <div id="mid_shdw">
    <?php
    if($result["r_approved"] == "APPROVED"){
echo "-->";
} 
?>

有了这个代码,我试图阻止/忽略的代码块用<!--和-->,但它并不想忽略PHP代码.我如何使用PHP来阻止代码的整个部分？如果您使用我当前的"if"语句变量,我将非常感激.

谢谢!

Answer 1

为什么不这样:

<?php
if($result["r_approved"] != "APPROVED"){
?>
<div class="main">
    <div class="main-sub">
<?php include('http://www.contractorsintelligence.com/contractors-license/includes-page-elements/navigation1.php'); ?>
<div id="mid-top"><img src="https://www.contractorsintelligence.com/images/shadowbg-top.png" width="990" height="20" alt="Top Spacer"/></div>
            <div id="mid_shdw">
<?php
}
?>