Question

我拥有动态制作的巨大在线php文件集。它有链接，甚至一些带引号的无效链接（用首页制作）

index2.php?page=xd
index2.php?page=xj asdfa
index2.php?page=xj%20aas
index2.php?page=xj#jumpword
index2.php?page=gj#jumpword with spaces that arenot%20
index2.php?page=afdsdj#jumpword%20with
index2.php?page=xj#jumpword with "quotes" iknow


$input_lines=preg_replace("/(index2.php?page\=.*)(#[a-zA-Z0-9_ \\"]*)(\"\>)/U", "$0 --> $2", $input_lines);

我希望所有这些只是＃-part而没有index2.php？page = * part。

整个晚上我都无法工作。所以请帮忙。

Answer 1

在某些情况下，您可以使用parse_url从网址获取属性（例如：#之后的内容），如下所示：

$urls = array(
    'index2.php?page=xd',
    'index2.php?page=xj asdfa',
    'index2.php?page=xj%20aas',
    'index2.php?page=xj#jumpword',
    'index2.php?page=gj#jumpword with spaces that arenot%20',
    'index2.php?page=afdsdj#jumpword%20with',
    'index2.php?page=xj#jumpword with "quotes" iknow',
    );

foreach($urls as $url){
    echo 'For "' . $url . '": ';
    $parsed = parse_url($url);
    echo isset($parsed['fragment']) ? $parsed['fragment'] : 'DID NOT WORK';
    echo '<br>';
}

输出：

对于“index2.php？page = xd”：DID NOT WORK

对于“index2.php？page = xj asdfa”：DID NOT WORK

对于“index2.php？page = xj％20aas”：DID NOT WORK

对于“index2.php？page = xj #jackword”：jumpword

对于“index2.php？page = gj＃jumpword with spaces not not percent 20”：jumpword with spaces notnot％20

对于“index2.php？page = afdsdj #jackword％20with”：jumpword％20with

对于“index2.php？page = xj＃jumpword with”quotes“iknow”：jumpword with“quotes”iknow

如何preg_replace链接，以便保留哈希

1 个答案: