我有我想要提取一些信息,这个PHP代码,但我停下来HREF步:如何让只有一些href属性
$site = "http://www.sports-reference.com/olympics/countries";
$site_html = file_get_html($site);
$country_dirty = $site_html->getElementById('div_countries');
foreach($country_dirty->find('img') as $link){
$country = $link->alt;
$link_country = "$site/$country";
$link_country_html = file_get_html($link_country);
$link_season = $link_country_html->getElementById('div_medals');
foreach($link_season->find('a') as $season){
echo $link_year_season = $season->href . "\n";
//echo $link_season = strstr ($link_year_season,'summer') . "\n";
}
}
变量$ link_year_season让我以下的输出:
/olympics/countries/AFG/summer/2012/
/olympics/athletes/ba/nesar-ahmad-bahawi-1.html
/olympics/athletes/ni/rohullah-nikpai-1.html
/olympics/countries/AFG/summer/2008/
/olympics/athletes/ba/nesar-ahmad-bahawi-1.html
/olympics/athletes/ni/rohullah-nikpai-1.html
/olympics/countries/AFG/summer/2004/
/olympics/countries/AFG/summer/1996/
/olympics/countries/AFG/summer/1988/
/olympics/countries/AFG/summer/1980/
/olympics/countries/AFG/summer/1972/
.....
我想知道是否有可能获得仅此输出:
/olympics/countries/AFG/summer/2012/
/olympics/countries/AFG/summer/2008/
/olympics/countries/AFG/summer/2004/
/olympics/countries/AFG/summer/1996/
/olympics/countries/AFG/summer/1988/
/olympics/countries/AFG/summer/1980/
/olympics/countries/AFG/summer/1972/
这样做的一个快速方法是在输出中应用'preg_match'或'strpos'或类似的东西,您已经得到了。 – Maximus2012
下面的答案是否可以解决您的问题? http://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – chris85