我需要更改什么,以便FieldRef下的Name节点是FieldRef的属性,而不是子节点?
Suds目前生成以下肥皂:
<ns0:query>
<ns0:Where>
<ns0:Eq>
<ns0:FieldRef>
<ns0:Name>_ows_ID</ns0:Name>
</ns0:FieldRef>
<ns0:Value>66</ns0:Value>
</ns0:Eq>
</ns0:Where>
</ns0:query>
我需要的是:
<ns0:query>
<ns0:Where>
<ns0:Eq>
<ns0:FieldRef Name="_ows_ID">
</ns0:FieldRef>
<ns0:Value>66</ns0:Value>
</ns0:Eq>
</ns0:Where>
</ns0:query>
第一个xml结构是由以下代码中的suds生成的。
q = c.factory.create('GetListItems.query')
q['Where']=InstFactory.object('Where')
q['Where']['Eq']=InstFactory.object('Eq')
q['Where']['Eq']['FieldRef']=InstFactory.object('FieldRef')
q['Where']['Eq']['FieldRef'].Name='_ows_ID'
q['Where']['Eq']['Value']='66'
和print(q)
会产生
(query){
Where =
(Where){
Eq =
(Eq){
FieldRef =
(FieldRef){
Name = "_ows_ID"
}
Value = "66"
}
}
}
这是进行创建soap请求的WS调用的代码
c = client.Client(url='https://community.site.edu/_vti_bin/Lists.asmx?WSDL',
transport=WindowsHttpAuthenticated(username='domain\user',
password='password')
)
ll= c.service.GetListItems(listName="{BD59F6D9-AB4B-474D-BCC7-E4B4BEA7EB27}",
viewName="{407A6AB9-97CF-4E1F-8544-7DD67CEA997B}",
query=q
)
答案 0 :(得分:0)
from suds.sax.element import Element
#create the nodes
q = Element('query')
where=Element('Where')
eq=Element('Eq')
fieldref=Element('FieldRef')
fieldref.set('Name', '_ows_ID')
value=Element('Value')
value.setText('66')
#append them
eq.append(fieldref)
eq.append(value)
where.append(eq)
q.append(where)
https://fedorahosted.org/suds/wiki/TipsAndTricks
包含Literal XML
包含文字(非转义)XML 作为对象的参数值 属性,您需要设置值 对象的参数 属性为sax元素。该 marshaller的设计简单 附加和附加内容 已经是XML。
例如,你想通过 以XML作为参数:
<query> <name>Elmer Fudd</name>
<age unit="years">33</age>
<job>Wabbit Hunter</job> </query>
可以按如下方式完成:
from suds.sax.element import Element query = Element('query') name = Element('name').setText('Elmer Fudd') age = Element('age').setText('33') age.set('units', 'years') job = Element('job').setText('Wabbit Hunter') query.append(name) query.append(age) query.append(job) client.service.runQuery(query)