我的代码

Question

我正在创建一个导航栏，如果页面当前处于活动状态，则自动将类更改为“活动”（使用php if语句[使用当前URL匹配]）

我还希望能够根据用户是否登录来更改标题...现在我通常对此没有任何问题，因为变量中有if语句，我这样做不知道如何继续。

我的问题是，如果在变量建立中发表声明是不可能的......例如，这是我正在尝试做的事情，但它不起作用......有没有办法做到这一点，并且实际制作这行得通... 先谢谢你！

我的代码

--- PHP ---

在头部：

<?php
///// (GETS THE PARTS OF THE CURRENT URL)
error_reporting(0);
$directoryURIbody = $_SERVER['REQUEST_URI'];
$pathbody = parse_url($directoryURIbody, PHP_URL_PATH);
$componentsbody = explode('/', $pathbody);
$first_partsbody = $componentsbody[1];
$second_partsbody = $componentsbody[2];
$third_partsbody = $componentsbody[3];
$fourth_partsbody = $componentsbody[4];
$fifth_partsbody = $componentsbody[5];
?>

在正文中：

：

<?php

if (!isset($_SESSION['idx'])) { ///////////IF NOT LOGGED IN
  if (!isset($_COOKIE['idCookie'])) {//////IF NOT LOGGED IN
      $navbar = '
        <li class="<?php if ($first_partmainnav=="") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>">Home</a></li>
        <li class="<?php if ($first_partmainnav=="tutorials") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>tutorials">Tutorials</a></li>
        <li class="<?php if ($first_partmainnav=="resources") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>resources">Resources</a></li>
        <li class="<?php if ($first_partmainnav=="library") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>library">Library</a></li>
        <li class="<?php if ($first_partmainnav=="our-projects") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>our-projects">Our Projects</a></li>
        <li class="<?php if ($first_partmainnav=="community") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>community">Community</a></li>';
}
if (isset($_SESSION['idx'])) { ////////////IF LOGGED IN (WITHOUT COOKIES)

      $navbar = '
        <li class="<?php if ($first_partmainnav=="") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>">Home</a></li>
        <li class="<?php if ($first_partmainnav=="whatever") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>whatever">whatever</a></li>
        <li class="<?php if ($first_partmainnav=="justanother") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>justanother">Just Another</a></li>
        ';

} else if (isset($_COOKIE['idCookie'])) {//IF LOGGED IN (WITH COOKIES)

      $navbar = '
        <li class="<?php if ($first_partmainnav=="") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>">Home</a></li>
        <li class="<?php if ($first_partmainnav=="whatever") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>whatever">whatever</a></li>
        <li class="<?php if ($first_partmainnav=="justanother") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>justanother">Just Another</a></li>
        ';    

}
?>

<?php echo $navbar; ?>

Answer 1

您可以使用三元运算符并使用字符串连接而不是echoing：

<?php

if (!isset($_SESSION['idx'])) { ///////////IF NOT LOGGED IN
  if (!isset($_COOKIE['idCookie'])) {//////IF NOT LOGGED IN
  $navbar = '
    <li class="'. ($first_partmainnav=="" ? "active" : "noactive")  .'"><a href="<?php echo $dyn_wwwFULL; ?>">Home</a></li>

?>

其余的比喻

Answer 2

li标签中的变量没有任何问题，问题在于将HTML和PHP代码和标签混合在单引号中。除非你纠正，否则什么都不会奏效。以下是使用您自己的代码正确执行此操作的方法：

<?php
if ( !isset( $_SESSION[ 'idx' ] ) ) { ///////////IF NOT LOGGED IN
  if ( !isset( $_COOKIE[ 'idCookie' ] ) ) { //////IF NOT LOGGED IN
    ?>

  <li class="<?php if ( $first_partmainnav == "" ) { echo "active";   } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>">Home</a></li>
  <li class="<?php if ( $first_partmainnav == "tutorials" ) { echo "active"; } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>tutorials">Tutorials</a></li>

  <li class="<?php if ( $first_partmainnav == "resources" ) { echo "active"; } else { echo "noactive";  }?>"><a href="<?php echo $dyn_wwwFULL; ?>resources">Resources</a></li>
  <li class="<?php if ( $first_partmainnav == "library" ) { echo "active"; } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>library">Library</a></li>

  <li class="<?php if ( $first_partmainnav == "our-projects" ) { echo "active"; } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>our-projects">Our Projects</a></li>
  <li class="<?php if ( $first_partmainnav == "community" ) { echo "active"; } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>community">Community</a></li>

  <?php }
  if ( isset( $_SESSION[ 'idx' ] ) ) { ////////////IF LOGGED IN (WITHOUT COOKIES)
  ?>

  <li class="<?php if ( $first_partmainnav == "" ) { echo "active"; } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>">Home</a></li>
  <li class="<?php if ( $first_partmainnav == "whatever" ) { echo "active"; } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>whatever">whatever</a></li>

  <li class="<?php if ( $first_partmainnav == "justanother" ) { echo "active"; } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>justanother">Just Another</a></li>

  <?php }  else {
    if ( isset( $_COOKIE[ 'idCookie' ] ) ) { //IF LOGGED IN (WITH COOKIES)
  ?>

    <li class="<?php if ( $first_partmainnav == "" ) { echo "active"; } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>">Home</a></li>
    <li class="<?php if ( $first_partmainnav == "whatever" ) { echo "active"; } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>whatever">whatever</a></li>
    <li class="<?php if ( $first_partmainnav == "justanother" ) { echo "active"; } else { echo "noactive"; }?>"><a href="<?php echo $dyn_wwwFULL; ?>justanother">Just Another</a></li>
   <?php 
    }
  }
}
?>

Answer 3

虽然我个人会改变这样做的整个方法（有一个与你的列表相对应的数组并相应地回显），你的方法可以通过将变量与代码分开来完成，例如：

$array['tutorials'] = ($first_partmainnav == "tutorials") ? 'active' : 'noactive';
$array['resources'] = ($first_partmainnav == "resources") ? 'active' : 'noactive';
// etc...

然后将它与您的输出连接。

通常你应该使用字符串连接：

错：

echo '<a href="<?php echo $dyn_wwwFULL; ?>tutorials ...';

右：

echo '<a href="'.$dyn_wwwFULL.'tutorials ...';

Answer 4

我认为你正在寻找这样的东西：

$first_partmainnav = 'tutorials';
echo 'some text ' . ($first_partmainnav == 'tutorials' ? 'active' : '') . ' some more text';

Answer 5

有一段时间我用PHP编写代码，但它看起来像你

以下代码如何工作？

<?php

if (!isset($_SESSION['idx'])) { ///////////IF NOT LOGGED IN
if (!isset($_COOKIE['idCookie'])) {//////IF NOT LOGGED IN
$navbar = '
<li class="' if($first_partmainnav=="") {echo "active"; } else  {echo "noactive";} '"><a href="' echo $dyn_wwwFULL; '">Home</a></li>'
....
?>

相信你可以自己填写其余部分。

当我查看＆lt;第一行的语法高亮显示时李＆GT;你注意到有些奇怪的事情发生了。

Answer 6

真正的问题是您不能在引用的PHP代码中包含PHP标记。让我们在问题中检查第一个li块的最后一行：

    <?php
    //NOTE: Had to add a space to separate php tags in comments for proper code highlighting, like this: < ? php --- ? >

    $navbar = '
            <li class="<?php if ($first_partmainnav=="community") {echo "active"; } else  {echo "noactive";}?>"><a href="<?php echo $dyn_wwwFULL; ?>community">Community</a></li>'; ?>

    <?php
    /* Although the whole $navbar value is quoted (Single quotes), there is php code surrounded by php tags in 2 places: 
    < ? php if ($first_partmainnav=="community") {echo "active"; } else  {echo "noactive";} ? > Here
    < ? php echo $dyn_wwwFULL; ? > And here
    Same happens with the rest of the lines, so it is impossible for this code to work.
    */
    ?>

所选答案也是错误的。我们来看看：

    <?php
      $navbar = '<li class="'. ($first_partmainnav=="" ? "active" : "noactive")  .'"><a href="<?php echo $dyn_wwwFULL; ?>">Home</a></li>
      '; // Add the single quote an semicolon missing.

    /* Same problem: $navbar value is quoted but there is php code surrounded by php tags:
    <a href="< ? php echo $dyn_wwwFULL; ? > Here.                                       
    This code can't work either.
    */
    ?>

PHP“if”里面的变量echo

我的代码

--- PHP ---

在头部：

在正文中：

6 个答案: