PHP 获取变量爆炸

Question

我正在使用此代码展开并显示 GET 变量。但我想删除 link:

中的当前查询

我的爆破代码：

$k = $_GET['sef'];
$s_explode = explode("-",$k);

foreach($s_explode as $q) {
  if($q==$s_explode[0]) {
echo '<a class="active" href="/category/'.$q.'">'.$s_explode[0].' <span class="dismiss">&times;</span></a>';
  } else {
echo '<a href="/category/'.$q.'">'.$q.' <span class="dismiss">&times;</span></a>';
  }
}

如果我使用 GET

website.com/?sef=游戏书视频

打印为：

<a class="active" href="/category/game">game</a>
<a href="/category/book">book</a>
<a href="/category/video">video</a>

我想如果我使用 GET

website.com/?sef=游戏书视频

<a class="active" href="/category/book-video">game</a>
<a href="/category/game-video">book</a>
<a href="/category/book-game">video</a>

我希望我能为我糟糕的英语解释好抱歉。

Answer 1

你的代码应该是这样的：

$k         = $_GET['sef'];
$s_explode = explode("-", $k);
//game-book-video

foreach($s_explode as $i => $q) {
    $parts = $s_explode;

    if(($key = array_search($q, $parts)) !== false) {
        unset($parts[$key]);
    }

    $class = ($i == 0 ? "class='active'" : '');

    echo '<a ' . $class . ' href="/category/'.implode('-', $parts).'">'.$q.' <span class="dismiss">&times;</span></a>';
}

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