PHP：删除除一个查询字符串变量之外的所有变量

发布于 2025-01-06 11:55:06 字数 374 浏览 3 评论 0原文

假设我有以下内容，即当前 url：

http://myurl.com/locations/?s=bricks&style=funky-quirky+rustic&feature=floors-concrete+kitchen+bathroom

How will I remove allparameters except S using PHP?换句话说，只需离开： http://myurl.com/locations/?s=bricks 其他参数将始终是 Style、Type、Feature 和 Area，其中部分或全部可能出席。

谢谢！

原文

Suppose I have the following, which is the current url:

http://myurl.com/locations/?s=bricks&style=funky-quirky+rustic&feature=floors-concrete+kitchen+bathroom

How would I remove all parameters except S using PHP? In other words, just leaving: http://myurl.com/locations/?s=bricks The other parameters will always be Style, Type, Feature, and Area, and some or all of them may be present.

Thanks!

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醉生梦死 2025-01-13 11:55:06

if ( isset($_GET['s']) && count($_GET) > 1 ) {
    header('Location: ?s='. rawurlencode($_GET['s']));
    exit;
}

编辑是为了安抚乔恩。

if ( isset($_GET['s']) && count($_GET) > 1 ) {
    header('Location: ?s='. rawurlencode($_GET['s']));
    exit;
}

Edited to somewhat appease Jon.

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樱娆 2025-01-13 11:55:06

错误检查的方式不多，但如果保证 s 存在，这将起作用：

list($base, $query) = explode('?', $url);
parse_str($query, $vars);
$result = $base.'?s='.rawurlencode($vars['s']);

无论 s 出现在查询字符串中的哪个位置，它都将起作用。我正在使用 explode 进行最小的初始“解析”步骤，但如果感觉不合适，您可以随时使用大枪并使用 parse_url 相反，代价是更加冗长。

Not much in the way of error checking, but if s is guaranteed to exist this will work:

list($base, $query) = explode('?', $url);
parse_str($query, $vars);
$result = $base.'?s='.rawurlencode($vars['s']);

It will also work no matter in what position s appears in the query string. I 'm doing a minimal initial "parsing" step with explode, but if that doesn't feel right you can always bring in the big guns and go with parse_url instead at the expense of being much more verbose.

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走野 2025-01-13 11:55:06

您可以通过几种不同的方式来实现这一点。

我将向您展示我在处理 url 时遇到的最佳标准：

$originUrl = parse_url('http://myurl.com/locations/?s=bricks&style=funky-quirky+rustic&feature=floors-concrete+kitchen+bathroom'); 
parse_str($originUrl['query'], $queryString);
echo $newUrl = $originUrl['scheme'].'://'.$originUrl['host'].$originUrl['path'].'?s='.$queryString['s'];

祝您好运！

You can achieve this in several different manners.

I will show you the best standard I ever faced to work with urls:

$originUrl = parse_url('http://myurl.com/locations/?s=bricks&style=funky-quirky+rustic&feature=floors-concrete+kitchen+bathroom'); 
parse_str($originUrl['query'], $queryString);
echo $newUrl = $originUrl['scheme'].'://'.$originUrl['host'].$originUrl['path'].'?s='.$queryString['s'];

Good Luck!

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故人爱我别走 2025-01-13 11:55:06

parse_str(parse_url($url, PHP_URL_QUERY), $query);
$newurl = 'http://myurl.com/locations/?s=' . $query['s'];

parse_str(parse_url($url, PHP_URL_QUERY), $query);
$newurl = 'http://myurl.com/locations/?s=' . $query['s'];

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反话 2025-01-13 11:55:06

如果您知道 s 始终是第一个参数，那么您可以简单地执行以下操作：

$url = "http://myurl.com/locations/?s=bricks&style=funky-quirky+rustic&feature=floors-concrete+kitchen+bathroom";
$split = explode('&',$url);
echo $split[0];

但是，如果 s 在参数中的位置未知，那么您可以这样做:

$url = "http://myurl.com/locations/?s=bricks&style=funky-quirky+rustic&feature=floors-concrete+kitchen+bathroom";
$split = explode('?',$url);
parse_str($split[1],$params);
echo $split[0].'?s='.$params['s'];

If you know that s will always be the first parameter, then you can simply do:

$url = "http://myurl.com/locations/?s=bricks&style=funky-quirky+rustic&feature=floors-concrete+kitchen+bathroom";
$split = explode('&',$url);
echo $split[0];

However, if the position of s in the parameters is unkown, then you can do this:

$url = "http://myurl.com/locations/?s=bricks&style=funky-quirky+rustic&feature=floors-concrete+kitchen+bathroom";
$split = explode('?',$url);
parse_str($split[1],$params);
echo $split[0].'?s='.$params['s'];

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