(PHP) 如何解析Google搜索结果中的URL？

发布于 2024-09-30 07:02:28 字数 1024 浏览 10 评论 0原文

如何获取google搜索结果的url？

（我使用 Zend_Gdata_Gbase 来获取搜索谷歌结果，而不是 DomDocument/htmlsimpleparser ，因为在我看来，Zend_Gdata_Gbase 专门用于解析谷歌结果。如果我的选择错误，请写下来。）

我获取“yahoo”或其他查询搜索字符串的谷歌搜索结果的函数： （该函数获取的 feed 应该包含单词“yahoo”的搜索结果，但是当我使用 prin_t($feed) 时，我看不到每个结果的 url）

<?php    
function queryGoogleSearch($queryString='yahoo'){
            $service = new Zend_Gdata_Gbase();
            $query = $service->newSnippetQuery();
            $query->setBq('['.$queryString.']');
            $query->setOrderBy('modification_time');
            $query->setSortOrder('descending');
            $query->setMaxResults('4');
            $feed = $service->getGbaseSnippetFeed($query);
            return $feed; 
    }
    print_r(queryGoogleSearch());
?>

我得到 4 个第一个 url 结果（当我在 google 中手动搜索时）：

www.yahoo.com、mail.yahoo.com、search.yahoo.com、maps.yahoo.com

但是当我打印 $feed 变量时我找不到它们。

请问我应该更改或添加 inqueryGoogleSearch() 函数吗？（或其他更好的代码）

谢谢

原文

How get google search results url?

(I use Zend_Gdata_Gbase for get search google results and not DomDocument/htmlsimpleparser because its looks to me that Zend_Gdata_Gbase done specially for parsing google results.
if I wrong in my selection, please write.)

My function to get google search results for 'yahoo' or other query search string:
(the function get a feed that should have search result for word 'yahoo', but when i use prin_t($feed) I don't see url for each result)

<?php    
function queryGoogleSearch($queryString='yahoo'){
            $service = new Zend_Gdata_Gbase();
            $query = $service->newSnippetQuery();
            $query->setBq('['.$queryString.']');
            $query->setOrderBy('modification_time');
            $query->setSortOrder('descending');
            $query->setMaxResults('4');
            $feed = $service->getGbaseSnippetFeed($query);
            return $feed; 
    }
    print_r(queryGoogleSearch());
?>

I get 4 first url results (when I search manually in google):

www.yahoo.com, mail.yahoo.com, search.yahoo.com, maps.yahoo.com

But I can't find them when I print $feed variable.

Please what should i change or add inqueryGoogleSearch() function? (Or other better code)

Thanks

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