PHP를 사용하여 XML을 표시 할 수 없습니다.

Question

웹 페이지에 XML을 표시하려고했지만 아무 것도 표시되지 않습니다.

<item> 
<songid>US2HK18715</songid> 
<songtitle>Take Me To The River</songtitle> 
<artist>Al Green</artist> 
<genre>Soul</genre> 
<link>https://www.amazon.com/Take-Me-To-The-River/dp/B0023RP62U</link> 
<releaseyear>1974</releaseyear> 
</item> 

내 XML 항목 중 하나의 예

, 여기 내가 사용하려고했던 코드입니다 : 심지어 에코 안녕하세요 어떤 이유로

Answer 1

표시되지 않습니다

<?php 

$songs = simplexml_load_file('playlist.xml'); 

foreach($songs->item as $song){ 
    $id = $song->songid; 
    $title = $song->songtitle; 
    $artist = $song->artist; 
    $genre = $song->genre; 
    $link = $song->link; 
    $releaseYear = $song->releaseyear; 
    echo "<p><span>$id</span><span>$title</span></p>\n"; 
    echo "hello"; 


} 
?> 

을 아래 XML 파일 형식을 고려하십시오.

<?xml version="1.0" encoding="ISO-8859-1" ?> 
<rss version="0.91"> 
    <channel> 
    <title>Music Playlist Sharing</title> 
    <description>Share music playlists - import from other applications and export them</description> 
    <language>en-us</language> 
    <item> 
     <songid>US2HK18715</songid> 
     <songtitle>Take Me To The River</songtitle> 
     <artist>Al Green</artist> 
     <genre>Soul</genre> 
     <link>https://www.amazon.com/Take-Me-To-The-River/dp/B0023RP62U</link> 
     <releaseyear>1974</releaseyear> 
    </item> 
    <item> 
    </item> 
    </channel> 
</rss> 

아래 코드가 작동해야합니다.

foreach($songs->channel->item as $song){ 
    $id = $song->songid; 
    $title = $song->songtitle; 
    $artist = $song->artist; 
    $genre = $song->genre; 
    $link = $song->link; 
    $releaseYear = $song->releaseyear; 
    echo "<p><span>$id</span><span>$title</span></p>\n"; 
    echo "hello"; 
}