我需要从RSS Feed(Workopolis)
上显示的描述中提取日期以下是Feed项的示例:
<item>
<guid isPermaLink="true">http://www.workopolis.com/jobsearch/job/15161566?uc=?RSS=Y</guid>
<link>http://www.workopolis.com/jobsearch/job/15161566?uc=?RSS=Y</link>
<title>Financial Services Representative</title>
<description><strong>Location:</strong> Barrie, Ontario; Owen Sound, Ontario; Port Elgin, Ontario<br><strong>Job Category:</strong> Sales and Business Development <br><strong>Job Industry:</strong> Financial Services and Banking<br><strong>Career Level:</strong> <br><strong>Position Type:</strong> Full Time<br><strong>Date Posted:</strong> 06/19/2014<br><strong>Company Name:</strong> CIBC<br><strong>Company URL:</strong> <a href="http:// ">http:// </a></description>
</item>
我已经做了大约2个小时的研究了。我对更高级的PHP编码不太满意,所以我真的可以使用一些帮助。日期在DD / MM / YY里面。
我想我的问题是:如何在DD / MM / YYYY中提取日期并将其转换为$ date变量?
谢谢!
答案 0 :(得分:1)
您可以将preg_match
与专为日期量身定制的正则表达式模式一起使用:
\d{2}\/\d{2}\/\d{4}
代码:
$str = '<item>
<guid isPermaLink="true">http://www.workopolis.com/jobsearch/job/15161566?uc=?RSS=Y</guid>
<link>http://www.workopolis.com/jobsearch/job/15161566?uc=?RSS=Y</link>
<title>Financial Services Representative</title>
<description><strong>Location:</strong> Barrie, Ontario; Owen Sound, Ontario; Port Elgin, Ontario<br><strong>Job Category:</strong> Sales and Business Development <br><strong>Job Industry:</strong> Financial Services and Banking<br><strong>Career Level:</strong> <br><strong>Position Type:</strong> Full Time<br><strong>Date Posted:</strong> 06/19/2014<br><strong>Company Name:</strong> CIBC<br><strong>Company URL:</strong> <a href="http:// ">http:// </a></description>
</item>';
preg_match("/\\d{2}\/\\d{2}\/\\d{4}/u", $str, $date);
echo $date[0];
输出:
06/19/2014