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我是新來的PHP簡單的XML文件解析器 我有這樣PHP簡單的XML乘務長
<?xml version='1.0' encoding='UTF-8'?>
<feed xmlns='http://www.w3.org/2005/Atom' xmlns:openSearch='http://a9.com/-/spec/opensearchrss/1.0/' xmlns:gContact='http://schemas.google.com/contact/2008' xmlns:batch='http://schemas.google.com/gdata/batch' xmlns:gd='http://schemas.google.com/g/2005'>
<category scheme='http://schemas.google.com/g/2005#kind' term='http://schemas.google.com/contact/2008#contact'/>
<title type='text'>Aditya Technobd's Contacts</title>
<generator version='1.0' uri='http://www.google.com/m8/feeds'>Contacts</generator>
<entry>
<category scheme='http://schemas.google.com/g/2005#kind' term='http://schemas.google.com/contact/2008#contact'/>
<title type='text'>Ashfaq Ali</title>
<gd:email rel='http://schemas.google.com/g/2005#other' address='[email protected]' primary='true'/>
</entry>
<entry>
<category scheme='http://schemas.google.com/g/2005#kind' term='http://schemas.google.com/contact/2008#contact'/>
<title type='text'>Farhad Hossain</title>
<gd:email rel='http://schemas.google.com/g/2005#other' address='[email protected]' primary='true'/>
</entry>
</feed>
我需要的輸出看起來像這樣
名稱的XML文件:阿什法克阿里 電子郵件:阿什法克@ technobd.com
名稱:法哈德·侯賽因 電子郵件:[email protected]
我嘗試它吹代碼不能˚F IND任何線索
$contacts = file_get_contents("public/temporary/contacts_sample.xml");
$feed = simplexml_load_string($contacts);
foreach($feed->entry as $entry){
echo "Name: " .$entry->title;
echo "<br>";
}
任何人可以幫助我,我怎樣才能得到