PHP：搜索字符串中的重合处

发布于 2024-09-13 09:15:07 字数 506 浏览 12 评论 0原文

我想过滤访问者从搜索引擎访问我的网站的 HTTP_REFERER。我想忽略为来自搜索引擎的访问者存储 HTTP_REFERER 信息。您能帮忙编写一下 PHP 脚本吗？

我有这个，但不正确的脚本：

<?
$exp_list = array('google', 'yahoo');

// exapmple of one HTTP_REFERER link from the Goggle search engine
$link = 'http://www.google.com/search?hl=ru&source=hp&q=bigazart&aq=f&aqi=&aql=&oq=&gs_rfai=';

for ($j = 0; $j < sizeof($exp_list); $j++){

if(!eregi($exp_list[$j], $link)){

// storing link to mysql...

break;

}

}
?>

原文

I would like to filter HTTP_REFERER where visitors come to my site from search engines. I want to ignore storing HTTP_REFERER information for the visitors came from the search engines. Could you please help with the PHP script?

I have this, but not correct script:

<?
$exp_list = array('google', 'yahoo');

// exapmple of one HTTP_REFERER link from the Goggle search engine
$link = 'http://www.google.com/search?hl=ru&source=hp&q=bigazart&aq=f&aqi=&aql=&oq=&gs_rfai=';

for ($j = 0; $j < sizeof($exp_list); $j++){

if(!eregi($exp_list[$j], $link)){

// storing link to mysql...

break;

}

}
?>

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甜宝宝 2024-09-20 09:15:07

尝试这样的事情：

if (isset($_SERVER['HTTP_REFERER'])) {
    $host = strtolower(parse_url($_SERVER['HTTP_REFERER'], PHP_URL_HOST));
    $exp_list = array('google', 'yahoo');
    $pattern = '/^(?:www\.)?(?:'.implode('|', array_map('preg_quote', $exp_list)).')\./'
    if (preg_match($pattern, $host)) {
        // match found
    }
}

重要的事情：

检查 $_SERVER['HTTP_REFERER'] 是否存在
使用 parse_url 从 URL 获取主机并仅在那里搜索
测试术语是否被点包围

但这仍然会错误地识别像 www.google.example 这样的主机。 com。因此您可能还想指定顶级/二级域名。

Try something like this:

if (isset($_SERVER['HTTP_REFERER'])) {
    $host = strtolower(parse_url($_SERVER['HTTP_REFERER'], PHP_URL_HOST));
    $exp_list = array('google', 'yahoo');
    $pattern = '/^(?:www\.)?(?:'.implode('|', array_map('preg_quote', $exp_list)).')\./'
    if (preg_match($pattern, $host)) {
        // match found
    }
}

The important things:

Check whether $_SERVER['HTTP_REFERER'] exists or not
Use parse_url to get the host from the URL to only search there
Test if the terms are surrounded by dots

But this will still falsely identify a host like www.google.example.com. So you might also want to specify the top/second level domain names.

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苦妄 2024-09-20 09:15:07

您应该能够自定义以下模式以匹配更多域。

<?php

$ignore_hosts = array(
    '/^www.google.com$/',
    '/^www.yahoo.com$/'
    );

$host = parse_url($_SERVER['HTTP_REFERRER'], PHP_URL_HOST);

$ignore = FALSE;
foreach ($ignore_hosts as $pattern) {
    if (preg_match($pattern, $host) == 0){
        $ignore = TRUE;
        break;
    }
}

if (! $ignore)
    echo "Here you should store the referrer.";

You should be able to customize the patterns below to match more domains.

<?php

$ignore_hosts = array(
    '/^www.google.com$/',
    '/^www.yahoo.com$/'
    );

$host = parse_url($_SERVER['HTTP_REFERRER'], PHP_URL_HOST);

$ignore = FALSE;
foreach ($ignore_hosts as $pattern) {
    if (preg_match($pattern, $host) == 0){
        $ignore = TRUE;
        break;
    }
}

if (! $ignore)
    echo "Here you should store the referrer.";

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