Question

我正在尝试使用带有以下XML的API：

<movies>
   <movie>
      <images>
         <image type="poster" url="http://cf1.imgobject.com/posters/b7a/4bc91de5017a3c57fe00bb7a/i-am-legend-original.jpg" size="original" width="675" height="1000" id="4bc91de5017a3c57fe00bb7a"/>
         <image type="poster" url="http://cf1.imgobject.com/posters/b7a/4bc91de5017a3c57fe00bb7a/i-am-legend-mid.jpg" size="mid" width="500" height="741" id="4bc91de5017a3c57fe00bb7a"/>
         <image type="poster" url="http://cf1.imgobject.com/posters/b7a/4bc91de5017a3c57fe00bb7a/i-am-legend-cover.jpg" size="cover" width="185" height="274" id="4bc91de5017a3c57fe00bb7a"/>
      </images>
   </movie>
</movies>

有人可以给我一个PHP代码示例，我应该使用它来获取图片网址大小=“封面”吗？

感谢。

Answer 1

SimpleXML可以做到这一点，因为缺少一个更好的词，简单地说：

$xml = new SimpleXMLElement($str);
$xpath = $xml->xpath("/movies/movie/images/image[@size = 'cover']");

echo $xpath[0]['url'];

Answer 2

使用XML Parser，DOMDocument，SimpleXML等加载xml。

http://se.php.net/manual/en/refs.xml.php

然后您可以使用XPath来选择图像。 http://www.w3schools.com/xpath/xpath_syntax.asp

使用属性大小=封面

获取影片的XPath

/movies/movie/images/image[@size=cover]

看起来像一个很好的教程：http://ditio.net/2008/12/01/php-xpath-tutorial-advanced-xml-part-1/

Answer 3

<?php
$string = <<<XML
<?xml version='1.0'?>
<movies>
  <movie>
     <images>
        <image type="poster" url="http://cf1.imgobject.com/posters/b7a/4bc91de5017a3c57fe00bb7a/i-am-legend-original.jpg" size="original" width="675" height="1000" id="4bc91de5017a3c57fe00bb7a"/>
        <image type="poster" url="http://cf1.imgobject.com/posters/b7a/4bc91de5017a3c57fe00bb7a/i-am-legend-mid.jpg" size="mid" width="500" height="741" id="4bc91de5017a3c57fe00bb7a"/>
        <image type="poster" url="http://cf1.imgobject.com/posters/b7a/4bc91de5017a3c57fe00bb7a/i-am-legend-cover.jpg" size="cover" width="185" height="274" id="4bc91de5017a3c57fe00bb7a"/>
     </images>
  </movie>
</movies>
XML;

$xml = simplexml_load_string($string);

foreach($xml->movie->images->image as $image) {

    if(strcmp($image['size'],"cover") == 0)
        echo $image['url'];
}

?>

Answer 4

$ xml = simplexml_load_string（$ string2）; foreach（$ xml-＆gt; movie-＆gt; images-＆gt; image as $ image）{

if(strcmp($image['size'],"cover"))
   // echo $image['url'];
    ?>

＆＃34;宽度=＆＃34; 200像素＆＃34;高度=＆＃34; 100像素＆＃34;＆GT;

从XML获取图像并使用PHP显示？

4 个答案: