4

我有一个要求，我必须在谷歌上反向查找图像并提取打印在“此图像的最佳猜测：”标题上的名称。不，我对网上现有的 curl 代码进行了一些修改，并走到了这一步：

<?php

function fetch_google($terms="sample search",$numpages=1,$user_agent='Mozilla/5.0 (Windows NT 6.1; rv:8.0) Gecko/20100101 Firefox/8.0')  
{
    $searched="";
    for($i=0;$i<=$numpages;$i++)
    {
        $ch = curl_init();
        $url="http://www.google.com/searchbyimage?hl=en&image_url=".urlencode($terms);
        curl_setopt ($ch, CURLOPT_URL, $url);
        curl_setopt ($ch, CURLOPT_USERAGENT, $user_agent);
        curl_setopt ($ch, CURLOPT_HEADER, 0);
        curl_setopt ($ch, CURLOPT_FOLLOWLOCATION, 1);
        curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
        curl_setopt ($ch, CURLOPT_REFERER, 'http://www.google.com/');
        curl_setopt ($ch,CURLOPT_CONNECTTIMEOUT,120);
        curl_setopt ($ch,CURLOPT_TIMEOUT,120);
        curl_setopt ($ch,CURLOPT_MAXREDIRS,10);
        curl_setopt ($ch,CURLOPT_COOKIEFILE,"cookie.txt");
        curl_setopt ($ch,CURLOPT_COOKIEJAR,"cookie.txt");
        $searched=$searched.curl_exec ($ch);
        curl_close ($ch);
    }

    $xml = new DOMDocument();
    @$xml->loadHTML($searched);
    foreach($xml->getElementsByTagName('div') as $div)
    {
        if(strpos($div->nodeValue,"Best guess for this image:"))
            return $div->nodeValue;
    } 
}

$content = fetch_google("http://media.il.edmunds-media.com/aston-martin/as/03/de/aston-martin_front_03-de-as_1_276.jpg",1);
echo $content."<br>";

?>

但它给了我很多文字，我无法获得确切的 div。因为“a”没有类属性，所以我必须这样做。

请帮忙！

4

2 回答 2

3

您可以改用 preg_match 。

当您从 CURL 获取 HTML 时，您可以使用 Regex 来匹配文本：

function fetch_google($terms="sample search",$numpages=1,$user_agent='Mozilla/5.0 (Windows NT 6.1; rv:8.0) Gecko/20100101 Firefox/8.0')  
{
    $searched="";
    for($i=0;$i<=$numpages;$i++)
    {
        $ch = curl_init();
        $url="http://www.google.com/searchbyimage?hl=en&image_url=".urlencode($terms);
        curl_setopt ($ch, CURLOPT_URL, $url);
        curl_setopt ($ch, CURLOPT_USERAGENT, $user_agent);
        curl_setopt ($ch, CURLOPT_HEADER, 0);
        curl_setopt ($ch, CURLOPT_FOLLOWLOCATION, 1);
        curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
        curl_setopt ($ch, CURLOPT_REFERER, 'http://www.google.com/');
        curl_setopt ($ch,CURLOPT_CONNECTTIMEOUT,120);
        curl_setopt ($ch,CURLOPT_TIMEOUT,120);
        curl_setopt ($ch,CURLOPT_MAXREDIRS,10);
        curl_setopt ($ch,CURLOPT_COOKIEFILE,"cookie.txt");
        curl_setopt ($ch,CURLOPT_COOKIEJAR,"cookie.txt");
        $searched=$searched.curl_exec ($ch);
        curl_close ($ch);
    }

    $matches = array();
    preg_match('/Best guess for this image:[^<]+<a[^>]+>([^<]+)/', $searched, $matches);
    return (count($matches) > 1 ? $matches[1] : false);
}
于 2012-05-17T11:49:01.800 回答
2

如果要使用 DOMDocument ，可以通过以下修改获取值。

    <?php

function fetch_google($terms="sample search",$numpages=1,$user_agent='Mozilla/5.0 (Windows NT 6.1; rv:8.0) Gecko/20100101 Firefox/8.0')  
{
    $searched="";
    for($i=0;$i<=$numpages;$i++)
    {
        $ch = curl_init();
        $url="http://www.google.com/searchbyimage?hl=en&image_url=".urlencode($terms);
        curl_setopt ($ch, CURLOPT_URL, $url);
        curl_setopt ($ch, CURLOPT_USERAGENT, $user_agent);
        curl_setopt ($ch, CURLOPT_HEADER, 0);
        curl_setopt ($ch, CURLOPT_FOLLOWLOCATION, 1);
        curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
        curl_setopt ($ch, CURLOPT_REFERER, 'http://www.google.com/');
        curl_setopt ($ch,CURLOPT_CONNECTTIMEOUT,120);
        curl_setopt ($ch,CURLOPT_TIMEOUT,120);
        curl_setopt ($ch,CURLOPT_MAXREDIRS,10);
        curl_setopt ($ch,CURLOPT_COOKIEFILE,"cookie.txt");
        curl_setopt ($ch,CURLOPT_COOKIEJAR,"cookie.txt");
        $searched=$searched.curl_exec ($ch);
        curl_close ($ch);
    }

    $xml = new DOMDocument();
    @$xml->loadHTML($searched);
    if(true == ($topblock = $xml->getElementByID('topstuff')))
    {

        foreach($topblock->getElementsByTagName('div') as $div){

            if(strstr(strtolower($div->nodeValue), "guess")){
                foreach($div->getElementsByTagName('a') as $a){
                    $last = $a->nodeValue;
                }
            }
        }
    }

    return $last; 
}

$content = fetch_google($_GET['img'],1);
echo $content."<br>";

?>
于 2013-03-06T05:42:45.890 回答