0

我有一个这样的 XML 文件

 <report>
    <table>
      <columns>
         <column name="dateTime"/>
         <column name="userLogin"/>
         <column name="campaignName"/>
         <column name="adGroupName"/>
         <column name="changes"/>
      </columns>
      <row dateTime="Aug 13, 2013 11:56:34 PM" userLogin="hello123" campaignName="Search Exact Parts - 2012 USA" adGroupName="12 volt razor battery" changes="Text ad changed  Status changed from Enabled to Paused"/>
      <row dateTime="Aug 13, 2013 11:56:34 PM" userLogin="hello123" campaignName="Search Exact Parts - 2012 USA" adGroupName="Razor Quad Battery" changes="Text ad changed Status changed from Enabled to Paused"/>
      <row dateTime="Aug 13, 2013 11:56:34 PM" userLogin="hello123" campaignName="Search Exact Parts - 2012 USA" adGroupName="Razor Quad Tires" changes="Text ad changed Status changed from Enabled to Paused"/>
      <row dateTime="Aug 13, 2013 11:56:34 PM" userLogin="hello123" campaignName="Search Exact Parts - 2012 USA" adGroupName="Razor Replacement Battery" changes="Text ad changed Status changed from Enabled to Paused"/>
      <row dateTime="Aug 13, 2013 11:56:34 PM" userLogin="hello123" campaignName="Search Exact Parts - 2012 USA" adGroupName="Razor.com" changes="Text ad changed Status changed from Enabled to Paused"/>
    </table>
  </report>

我的问题是我正在尝试读取这个 XML 文件。我写的代码是，

foreach($xml->children() as $secgen=>$value){
        foreach($value->columns->children() as $thrgen){
            $feilds[] = $thrgen['name'];
        }
    $data[] = $value->row['dateTime'];
}

此代码仅读取第一行值，即“2013 年 8 月 13 日晚上 11:56:34”

即使我尝试了我所知道的所有可能的方式。但我无法以正确的方式检索方式。如果有人能解决这个问题，我将非常感激。

4

1 回答 1

0

用于xpath选择所有节点的datetime属性：<row>

$results = $xml->xpath("//row/@dateTime");

全部查看：

foreach ($results as $result) echo "$result <br />";
于 2013-08-16T08:19:05.760 回答