我有这个有效的php xml代码:
<?php
$str=<<<XML
<bookstore>
<book category="COOKING">
<title lang="en">Everyday Italian</title>
<author>Giada De Laurentiis</author>
<year>2005</year>
<price>30.00</price>
</book>
<book category="CHILDREN">
<title lang="en">Harry Potter</title>
<author>J K. Rowling</author>
<year>2005</year>
<price>29.99</price>
</book>
<book category="WEB">
<title lang="en-us">XQuery Kick Start</title>
<author>James McGovern</author>
<year>2003</year>
<price>49.99</price>
</book>
<book category="WEB">
<title lang="en-us">Learning XML</title>
<author>Erik T. Ray</author>
<year>2003</year>
<price>39.95</price>
</book>
</bookstore>
XML;
$data = new SimpleXMLElement($str);
?>
<table border>
<tr>
<th>Title</th>
<th>Author</th>
<th>Year</th>
<th>Price</th>
</tr>
<?php
foreach ($data->book as $book) {
if ($book->price == 39.95) {
?>
<tr>
<td><?php echo $book->title; ?></td>
<td><?php echo $book->author; ?></td>
<td><?php echo $book->year; ?></td>
<td><?php echo $book->price; ?></td>
</tr>
<?php
}
}
?>
</table>
您注意到 XML 位于代码中。我现在想要的只是消除该XML代码并替换或只是从w3schools http://www.w3schools.com/php/books.xml 获取它。先生,你能帮我吗,因为我被这个困了一个星期?
如果要
从外部源加载它,只需使用该simplexml_load_file()
:
$data = simplexml_load_file('http://www.w3schools.com/php/books.xml');
输出的外观
检查以下内容
$xml=simplexml_load_file('http://www.w3schools.com/php/books.xml');
var_dump($xml);
试试这个:
$content = file_get_contents('http://www.w3schools.com/php/books.xml');回声$content;
简单!