Question

我需要处理XML文档，我需要将一些字段复制到另一个XML文件中。我有这个领域：

<cast>
    <person name="Nanda Anand" character="" job="Director" id="589088" thumb="" department="Directing" url="http://www.themoviedb.org/person/589088" order="0" cast_id="1000"/>
    <person name="Lynn Collins" character="" job="Actor" id="21044" thumb="http://cf2.imgobject.com/t/p/w185/berS11tKvXqTFThUWAYrH279cvn.jpg" department="Actors" url="http://www.themoviedb.org/person/21044" order="0" cast_id="1001"/>
</cast>

我需要将所有元素复制到另一个XML但使用属性数据。我使用此代码，但它不起作用：

foreach ($movies->movies->movie->cast->children() as $person)
            {
            $person = $cast->appendChild(
                    $xmldoc->createElement(("person"), str_replace("&", "&amp;",$person)));
            }

Answer 1

如果您想操作XML文档（即将节点从一个文档添加到另一个文档），您应该使用DOMDocument，而不是SimpleXML。

这是使用DOMDocument从一个文档复制到另一个文档的一些代码。请注意，这是一个两步过程。首先将节点作为$ ndTemp导入文档。其次，将导入的$ ndTemp附加到指定的父级（我只使用根文档元素，但它可能是另一个节点）。

注意：如果你只是做一个简单的副本，你可能想考虑使用XSL，但这是另一篇文章......

输入XML（movie.xml）

<xml>
<movie name='first'>
    <cast>
        <person name="Nanda Anand" character="" job="Director" id="589088" thumb="" department="Directing" url="http://www.themoviedb.org/person/589088" order="0" cast_id="1000"/>
        <person name="Lynn Collins" character="" job="Actor" id="21044" thumb="http://cf2.imgobject.com/t/p/w185/berS11tKvXqTFThUWAYrH279cvn.jpg" department="Actors" url="http://www.themoviedb.org/person/21044" order="0" cast_id="1001"/>
    </cast>
</movie>
<movie name='Second'>
    <cast>
        <person name="Zaphod Beeblebrox" character="" job="Director" id="589088" thumb="" department="Directing" url="http://www.themoviedb.org/person/589088" order="0" cast_id="1000"/>
    </cast>
</movie>
</xml>

PHP

<?php
    $xml = new DOMDocument();
    $strFileName = "movie.xml";
    $xml->load($strFileName);

    $xmlCopy = new DOMDocument();
    $xmlCopy->loadXML( "<xml/>" );

    $xpath = new domxpath( $xml );
    $strXPath = "/xml/movie/cast/person";

    $elements = $xpath->query( $strXPath, $xml );
    foreach( $elements as $element ) {
        $ndTemp = $xmlCopy->importNode( $element, true );
        $xmlCopy->documentElement->appendChild( $ndTemp );
    }
    echo $xmlCopy->saveXML();

?>

输出

<?xml version="1.0"?>
<xml>
    <person name="Nanda Anand" character="" job="Director" id="589088" thumb="" department="Directing" url="http://www.themoviedb.org/person/589088" order="0" cast_id="1000" />
    <person name="Lynn Collins" character="" job="Actor" id="21044" thumb="http://cf2.imgobject.com/t/p/w185/berS11tKvXqTFThUWAYrH279cvn.jpg" department="Actors"
    url="http://www.themoviedb.org/person/21044" order="0" cast_id="1001" />
    <person name="Zaphod Beeblebrox" character="" job="Director" id="589088" thumb="" department="Directing" url="http://www.themoviedb.org/person/589088" order="0" cast_id="1000" />
</xml>

Answer 2

我认为这样可行，但未经过测试

foreach ($movies->movies->movie->cast->children() as $person)
{
   $cast->appendChild($person);
}

复制XML属性PHP

2 个答案: