嗨，我正在尝试解析谷歌结果页面以提取所有 URL

Question

我已尝试使用以下代码从 google 搜索结果中获取所有 URL 的列表。但我能够找到带有描述的 URL。我只想获取 URL 作为列表。

<?php
include('simple_html_dom.php');
function file_get_contents_curl($url)
{
  $ch = curl_init();
  curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); //Set curl to return the data instead of printing it to the browser.
  curl_setopt($ch, CURLOPT_URL, $url);
  $data = curl_exec($ch);
  curl_close($ch);
  return $data;
}
$query      = "what is a car";
$url        = 'http://www.google.com/search?q=' . urlencode($query) . '';
$scrape     = file_get_contents_curl($url);
//gettype($scrape); 
//echo $scrape;
$domResults = new simple_html_dom();
$domResults->load($scrape);
foreach ($domResults->find('a[href^=/url?q]') as $element) {
  echo $element . '<br><br>';
}
?>

Answer 1

更改最后一个循环：

$result = "";
foreach($domResults->find('a[href^=/url?q]') as $element) {
    $result .=  $element->href."\n";
    //echo $element.'<br><br>';
}
file_put_contents("myresult.txt", $result, FILE_APPEND);

但你只能用这个来做同样的事情:)（加上你最后的 2 个“需求”）

<?php
include ('simple_html_dom.php');
$query = "what is a car";
$html = file_get_html('http://www.google.com/search?q='.urlencode($query));
$result = "";
foreach($html->find('a[href^=/url?q]') as $element) {
   $url = $element->href;
   $url = str_replace("/url?q=","",$url);
   $result .=  $url."\n";
   echo "<a href='".$url."'>".$url."</a><br>";

}
file_put_contents("myresult.txt", $result, FILE_APPEND);
?>