用PHP解析SOAP響應

Question

因此，我將以下SOAP響應作為字符串接收：

<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <soap:Body>
        <GetListItemsResponse xmlns="http://schemas.microsoft.com/sharepoint/soap/">
            <GetListItemsResult>
                <listitems xmlns:s="uuid:BDC6E3F0-6DA3-11d1-A2A3-00AA00C14882" xmlns:dt="uuid:C2F41010-65B3-11d1-A29F-00AA00C14882" xmlns:rs="urn:schemas-microsoft-com:rowset" xmlns:z="#RowsetSchema">
                    <rs:data ItemCount="3">
                       <z:row ows_DocIcon="jpg" ows_LinkFilename="18380014229851.jpg" ows_Modified="2016-10-08 17:27:40" ows_Editor="179440;#asdf" ows_Last_x0020_Modified="2;#2016-10-08 17:29:29" ows_ID="2" ows_Created_x0020_Date="2;#2016-10-08 17:27:40" ows_FileLeafRef="2;#18380014229851.jpg" />
                       <z:row ows_DocIcon="jpg" ows_LinkFilename="18380014229851_2.jpg" ows_Modified="2016-10-08 17:27:40" ows_Editor="179440;#asfd" ows_Last_x0020_Modified="3;#2016-10-08 17:29:29" ows_ID="3" ows_Created_x0020_Date="3;#2016-10-08 17:27:41" ows_FileLeafRef="3;#18380014229851_2.jpg" />
                       <z:row ows_DocIcon="jpg" ows_LinkFilename="18380014229851_3.jpg" ows_Modified="2016-10-08 17:27:40" ows_Editor="179440;#asdf" ows_Last_x0020_Modified="4;#2016-10-08 17:30:03" ows_ID="4" ows_Created_x0020_Date="4;#2016-10-08 17:27:41" ows_FileLeafRef="4;#18380014229851_3.jpg" />
                    </rs:data>
                </listitems>
            </GetListItemsResult>
        </GetListItemsResponse>
    </soap:Body>
</soap:Envelope>

我試圖獲取每個“ z：row”條目，但是由於名稱空間而苦苦掙扎（經過一些谷歌搜索之后，我才知道它們被稱為）。

這是我正在使用的代碼：

$xml = simplexml_load_string($sp->soapClient->__last_response);

foreach($xml->GetListItemsResult as $item)
{
    $ns_li = $item->children('uuid:BDC6E3F0-6DA3-11d1-A2A3-00AA00C14882');
    $ns_rs = $ns_li->children('urn:schemas-microsoft-com:rowset');
    $ns_z  = $nr_rs->children('#RowsetSchema');
    echo $ns_z->row;
}

現在，我沒有從echo語句得到任何輸出。 我究竟做錯了什么？

Answer 1

您可以使用XPath表達式使用各種條件來獲取XML文檔的特定部分。 例如，要在XML文檔中的任何位置獲取z:row元素，只需在注冊前綴z之后執行//z:row ：

$xml->registerXPathNamespace("z", "#RowsetSchema");
foreach($xml->xpath('//z:row') as $item)
{
    echo $item["ows_LinkFilename"] ."\n";
}

eval.in demo

輸出：

18380014229851.jpg
18380014229851_2.jpg
18380014229851_3.jpg