xmlroot

Can't rename XML & Json Root Element in Spring 3.2.2, Jackson 2.2 and Xstream

可紊 提交于 2019-12-11 06:07:49
问题 Please help me to understand what is going on with name of XML&Json root element in Spring in JSon I get {"addressList": [{"StreetName":"Boul. Rene-Levesque Ouest","AddressID":1,"StreetNumber":1350}, in XML I get <list> <Address2> <StreetName>Boul. Rene-Levesque Ouest</StreetName> <AddressID>1</AddressID> <StreetNumber>1350</StreetNumber> </Address2> but I want to get {"AddressList": [{"StreetName":"Boul. Rene-Levesque Ouest","AddressID":1,"StreetNumber":1350}, XML: <AddressList> <Address>

PYTHON : How to add root node to an XML

淺唱寂寞╮ 提交于 2019-12-08 06:36:01
问题 I have an xml file looks something like this <A> <B> <C> .... </C> </B> </A> I want to add root on top of element 'A'. I found out a way to add elements to root. But How to change existing root and add on top of it using python. After adding root to the xml it should look like this <ROOT> <A> <B> <C> .... </C> </B> </A> </ROOT> 回答1: import lxml.etree as ET tree = ET.parse('data') root = tree.getroot() newroot = ET.Element("root") newroot.insert(0, root) print(ET.tostring(newroot, pretty_print

PYTHON : How to add root node to an XML

房东的猫 提交于 2019-12-06 15:34:18
I have an xml file looks something like this <A> <B> <C> .... </C> </B> </A> I want to add root on top of element 'A'. I found out a way to add elements to root. But How to change existing root and add on top of it using python. After adding root to the xml it should look like this <ROOT> <A> <B> <C> .... </C> </B> </A> </ROOT> import lxml.etree as ET tree = ET.parse('data') root = tree.getroot() newroot = ET.Element("root") newroot.insert(0, root) print(ET.tostring(newroot, pretty_print=True)) yields <root> <A> <B> <C> .... </C> </B> </A> </root> But really, unless you need to add something

XmlRoot() for Xml Serilization does not work

不羁的心 提交于 2019-11-29 09:25:23
I'm trying to get my httphandler to print out an XML file with the format: <ScheduledShows> <ScheduledShowElement>...</ScheduledShowElement> <ScheduledShowElement>...</ScheduledShowElement> <ScheduledShowElement>...</ScheduledShowElement> </ScheduledShows> But for some reason, the attribute XmlRoot("ScheduledShowElement") in ScheduledShow.cs is not working the way i want it to work. Instead, the output i get is: <ScheduledShows> <ScheduledShow>...<ScheduledShow> <ScheduledShow>...<ScheduledShow> <ScheduledShow>...<ScheduledShow </ScheduledShows> Basically, the name of the node is not being

XmlRoot() for Xml Serilization does not work

前提是你 提交于 2019-11-28 02:55:38
问题 I'm trying to get my httphandler to print out an XML file with the format: <ScheduledShows> <ScheduledShowElement>...</ScheduledShowElement> <ScheduledShowElement>...</ScheduledShowElement> <ScheduledShowElement>...</ScheduledShowElement> </ScheduledShows> But for some reason, the attribute XmlRoot("ScheduledShowElement") in ScheduledShow.cs is not working the way i want it to work. Instead, the output i get is: <ScheduledShows> <ScheduledShow>...<ScheduledShow> <ScheduledShow>...