python - 如何使用 python 在 SOAPpy 中设置 XML 属性？

Question

我正在形成一个 SOAPpy 请求，但我不知道如何在标记中设置属性。这是我的代码:

url = wsdlfile = 'https://stats2.overture.com/ExternalSOAP/statsPMCAPI_1_0.wsdl'
n = 'urn:yahoo:overture:stats:3.0'
server = WSDL.Proxy(wsdlfile)
server.soapproxy.config.dumpSOAPOut = 1
server.soapproxy.config.dumpSOAPIn = 1
result = server.getAvailablePmcReports(ReportAuth = {'username': username, 'cookie': YBY}, ReportRequest= '')
print(result)

输出这个:

<?xml version="1.0" encoding="UTF-8"?>
<SOAP-ENV:Envelope SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/" xmlns:xsi="http://www.w3.org/1999/XMLSchema-instance" xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/1999/XMLSchema">
<SOAP-ENV:Body>
<ns1:getAvailablePmcReports xmlns:ns1="urn:yahoo:overture:stats:3.0" SOAP-ENC:root="1">
<ReportRequest xsi:type="xsd:string"></ReportRequest>
<ReportAuth>
<username xsi:type="xsd:string">myuser</username>
<cookie xsi:type="xsd:string">cookie here...</cookie>
</ReportAuth>
</ns1:getAvailablePmcReports>
</SOAP-ENV:Body>
</SOAP-ENV:Envelope>

但我想要的是:

<ReportRequest startRow="0" shownRows="200">

如何添加属性？谢谢。

Answer 1

您可以通过为 ReportRequest 关键字参数传递一个类型化的值来做到这一点。例如，如果我将 getAvailablePmcReports 行更改为:

from SOAPpy import Types
result = server.getAvailablePmcReports(
    ReportAuth = {'username': username, 'cookie': YBY},
    ReportRequest= Types.stringType('', attrs={'startRow': 0, 'shownRows': 200}))

生成的请求包含如下标签:

<ReportRequest xsi:type="xsd:string" shownRows="200" startRow="0"></ReportRequest>