网络抓取删除链接没有ID /类附加为PHP

Question

您好我使用网络抓取网站，但它的内容太多，我不需要的信息。 这里是我的代码：

<?php 
require('phpQuery.php'); 
$url = 'http://www.nasdaq.com/screening/companies-by-name.aspx?letter=A'; 
$html = file_get_contents($url); 
$pq = phpQuery::newDocumentHTML($html); 
echo $pq['#CompanylistResults']; 
?> 

，其结果是：

<table id="CompanylistResults"> 
<tbody> 
<tr> 
<tr> 
<td> 
<a target="_blank" rel="nofollow" href="http://www.1800flowers.com">1-800 FLOWERS.COM, Inc.</a> 
</td> 
<td> 
<td style="">$100.55M</td> 
<td style="display:none"></td> 
<td>United States</td> 
<td>1999</td> 
<td style="width:105px">Other Specialty Stores</td> 

我需要的是 “1-800 FLOWERS.COM公司”和“$ 100.55M”中的文字来说，该怎么做呢？

Answer 1

尝试使用此代码：

//the url you need to scrape 
$uri = ('http://www.nasdaq.com/screening/companies-by-name.aspx?letter=A'); 
//extracts HTML from the url 
$get = file_get_contents($uri); 

//Finding what you want removed 
$pos1 = strpos($get, "<a target=\"_blank\" rel=\"nofollow\" href=\"http://www.1800flowers.com\">"); 
$pos2 = strpos($get, "</a>", $pos1); 

$pos3 = strpos($get, "<td style=\"\">"); 
$pos4 = strpos($get, "</td>", $pos3); 

//Removing the parts that are not needed 
$text = substr($get,$pos1,$pos2-$pos1); 
$test3 = substr($get,$pos3,$pos4-$pos3); 

//Removing tags from is left after the above code, you should now have only the values that you are looking for 
$text1 = strip_tags($text); 
$text2 = strip tags($text3);