php - 从另一个域获取 XML 文件到 Mysql 数据库

Question

我正在尝试通过 php 脚本从要导入到 mysql 的 url 获取 xml 信息，但我遇到了一些麻烦，而且我的经验并未涵盖此领域。XML 的构成如下例所示:

 <?xml version="1.0" encoding="utf-8" ?>
<rss version="2.0" xmlns:g="http://base.google.se/ns/1.0">
<channel>
  <title></title>
  <description></description>
  <link></link>
  <item>
    <g:id></g:id>
    <title></title>
    <g:product></g:product>
  </item>
  <item>
    <g:id></g:id>
    <title></title>
    <g:product></g:product>
  </item>
and so on...

使用 php 脚本:

<?php

include '../connection-to-db.php';

$str_xml = file_get_contents('http://www.example.com/xmls/xmlfile.xml');
$library = new SimpleXMLElement($str_xml);

$arr = json_decode( json_encode($library) , true);
var_dump ($arr);

echo "Array got " .sizeof($arr['item']) . " items.<br> <br>";
if (sizeof($arr['item']) > 155555500) {

    mysql_query("TRUNCATE TABLE google_stat");

    $count = 0;
    foreach ($arr['item'] as $shelf)
    {
        $gId = mysql_real_escape_string($shelf['g:id']);
        $Title = mysql_real_escape_string($shelf['title']);
        $gProductType = mysql_real_escape_string($shelf['g:product']);

        mysql_query("INSERT INTO google_stat (gid, title, gcategory) 
                    VALUES ('$gID', '$Title', '$gCategory')")
                    or die(mysql_error());

        $count ++;
    }
    echo " Counted: " . $count . "inserts";
} else {
    echo "Non counted, no insert done";
}
?>

问题是，当 SimpleXMLElement 时，当我查看输出时，名称中带有 g: 的所有项目似乎都消失了，它甚至找不到任何项目。我什至尝试过使用具有相同 XML 树的本地文件，但甚至无法使其工作。我很感谢所提供的任何帮助，因为我越来越意识到我在这方面陷入了困境。

更新:

<?php

    include '../connection-to-db.php';

    $str_xml = file_get_contents('http://www.example.com/xmls/xmlfile.xml');
    $library = new SimpleXMLElement($str_xml);

    $arr = json_decode( json_encode($library) , true);


    echo "Array got " .sizeof($library->channel->item) . " items.<br> <br>";
    if (sizeof($library->channel->item) > 100) {

        mysql_query("TRUNCATE TABLE google_stat");

        $count = 0;
        foreach ($library->channel->item as $shelf)
        {
            $gId = (string) $shelf->children('g', TRUE)->id;
            $Title = (string) $shelf->title;
            $gProductType = $shelf->children('g', TRUE)->product;

             echo $gId."<br />";
             echo $Title."<br />";
             echo $gProductType."<br />";

            $count ++;
        }
        echo " Counted: " . $count . "inserts";
    } else {
        echo "Non counted, no insert done";
    }
    ?>

现在我得到了数组中的项目数，但 $gId、$Title 等不回显任何值。

Edit2:必须进行高数组检查，它有效。

Answer 1

它与您想要的 namespace 前缀有关。您可以像这样访问 g: 项目:

<?php

$str_xml = file_get_contents('test.xml');
$library = new SimpleXMLElement($str_xml);

$count = 0;
foreach ($library->channel->item as $shelf)
{

  $gId = (string) $shelf->title;
  $Title = (string) $shelf->children('g', TRUE)->id;
  $gProductType = (string) $shelf->children('g', TRUE)->product;

      echo $gId."<br />";
      echo $Title."<br />";
      echo $gProductType."<br />";

          $count ++;
}
echo " Counted: " . $count . " inserts";

?>

参见https://www.sitepoint.com/simplexml-and-namespaces/以供进一步引用。

xml.测试

<?xml version="1.0" encoding="utf-8" ?>
<rss version="2.0" xmlns:g="http://base.google.se/ns/1.0">
<channel>
 <title>product</title>
 <description>lots of products</description>
 <link>www.example.com</link>
 <item>
   <g:id>ID 1</g:id>
   <title>Title 1</title>
   <g:product>Product 1</g:product>
 </item>
 <item>
   <g:id>ID 2</g:id>
   <title>Title 2</title>
   <g:product>Product 2</g:product>
 </item>
 </channel>
</rss>

输出:

标题 1

ID 1

产品1

标题 2

ID 2

产品2

计数:2 个插入