我有一个SOAP响应,我想保存到XML文件。当响应写入文件时,SOAP信封会随之写入,因为错误导致XML文件无效:
XML declaration allowed only at the start of the document in ...
在这种情况下,XML被声明两次:
<?xml version="1.0" encoding="ISO-8859-1"?>
<SOAP-ENV:Envelope SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/">
<SOAP-ENV:Body><ns1:NDFDgenResponse xmlns:ns1="http://graphical.weather.gov/xml/DWMLgen/wsdl/ndfdXML.wsdl">
<dwmlOut xsi:type="xsd:string">
<?xml version="1.0"?>
...
有没有一种很好的方法来删除这个SOAP信封,只是保存它之间的内容?
以下是我如何编写文件的响应:
$toWrite = htmlspecialchars_decode($client->__getLastResponse());
$fp = fopen('weather.xml', 'w');
fwrite($fp, $toWrite);
fclose($fp);
答案 0 :(得分:0)
问题是htmlspecialchars_decode()
。信封文档包含其他XML文档作为文本节点。如果解码XML文档中的实体,则会将其销毁。切勿在XML文档上使用htmlspecialchars_decode()
。
将(信封)XML加载到DOM中并从中读取所需的值。
$xml = <<<'XML'
<?xml version="1.0" encoding="ISO-8859-1"?>
<SOAP-ENV:Envelope
SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"
xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/">
<SOAP-ENV:Body>
<ns1:NDFDgenResponse
xmlns:ns1="http://graphical.weather.gov/xml/DWMLgen/wsdl/ndfdXML.wsdl">
<dwmlOut xsi:type="xsd:string">
<?xml version="1.0"?>
<weather>XML</weather>
</dwmlOut>
</ns1:NDFDgenResponse>
</SOAP-ENV:Body>
</SOAP-ENV:Envelope>
XML;
$dom = new DOMDocument();
$dom->loadXml($xml);
$xpath = new DOMXPath($dom);
$xpath->registerNamespace('soap', 'http://schemas.xmlsoap.org/soap/envelope/');
$xpath->registerNamespace('ndfd', 'http://graphical.weather.gov/xml/DWMLgen/wsdl/ndfdXML.wsdl');
$innerXml = $xpath->evaluate(
'string(/soap:Envelope/soap:Body/ndfd:NDFDgenResponse/dwmlOut)'
);
echo $innerXml;
输出:
<?xml version="1.0"?>
<weather>XML</weather>