【发布时间】:2011-02-20 06:29:03
【问题描述】:
您好,我正在尝试在我的网站上实现屏幕抓取场景,并且到目前为止设置了以下内容。我最终要做的是替换 $results 变量中具有“ResultsDetails.aspx?”的所有链接。到“results-scrape-details/”然后再次输出。谁能指出我正确的方向?
<?php
$url = "http://mysite:90/Testing/label/stuff/ResultsIndex.aspx";
$raw = file_get_contents($url);
$newlines = array("\t","\n","\r","\x20\x20","\0","\x0B");
$content = str_replace($newlines, "", html_entity_decode($raw));
$start = strpos($content,"<div id='pageBack'");
$end = strpos($content,'</body>',$start) + 6;
$results = substr($content,$start,$end-$start);
$pattern = 'ResultsDetails.aspx?';
$replacement = 'results-scrape-details/';
preg_replace($pattern, $replacement, $results);
echo $results;
【问题讨论】:
标签: php screen screen-scraping