Question

我编写了这段代码 - 一个Google搜索结果解析器 - 但它只获得了标题：

$url = "http://www.google.com/search?client=opera&q=example&sourceid=opera&ie=UTF-8&oe=UTF-8";
$ch = curl_init();
$timeout = 5;
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch,CURLOPT_CONNECTTIMEOUT, $timeout);
$html = curl_exec($ch);
curl_close($ch);

$dom = new DOMDocument();
@$dom->loadHTML($html);

foreach ($dom->getElementsByTagName('h3') as $link) {
    echo $link->nodeValue;
    echo "<br />";
}

我如何获取网址和说明？

Answer 1

您可以获取链接名称和网址，如此

示例：

<?php
$url = "http://www.google.com/search?client=opera&q=example&sourceid=opera&ie=UTF-8&oe=UTF-8";
$ch = curl_init();
$timeout = 5;
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
curl_setopt($ch,CURLOPT_CONNECTTIMEOUT, $timeout);
$html = curl_exec($ch);


curl_close($ch);
$dom = new DOMDocument();
@$dom->loadHTML($html);

foreach ($dom->getElementsByTagName('h3') as $link) {
    echo $link->nodeValue."\n";
    echo str_replace('/url?q=', '',$link->firstChild->getAttribute('href'))."\n";

    echo "<br />";
}

除标题外，还可获取Google搜索结果的网址和说明

1 个答案: