#Root
public class DataSet {
#Namespace()
#Path("")
private String WordKey;
private String Pron;
private String Orig;
private String Trans;
}
I want use retrofit2 with simplexml to transform http request. It's android code. I have trouble with transform xml. How to transform this ? How can i write the class ?
<DataSet xmlns="http://WebXml.com.cn/">
<xs:schema xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata" id="Dictionary">...</xs:schema>
<diffgr:diffgram xmlns:msdata="urn:schemas-microsoft-com:xml-msdata" xmlns:diffgr="urn:schemas-microsoft-com:xml-diffgram-v1">
<Dictionary xmlns="">
<Trans diffgr:id="Trans1" msdata:rowOrder="0">
<WordKey>hello</WordKey>
<Pron>'heləu, he'ləu</Pron>
<Info/>
<Translation>int.(见面打招呼或打电话用语)喂,哈罗</Translation>
<Mp3>1059.mp3</Mp3>
</Trans>
<Sentence diffgr:id="Sentence1" msdata:rowOrder="0">
<Orig>
She actually condescended to say hello to me in the street today.
</Orig>
<Trans>她今天在街上竟能屈尊跟我打招呼.</Trans>
</Sentence>
<Sentence diffgr:id="Sentence2" msdata:rowOrder="1">
<Orig>
I said hello to her, but she ignored me completely!
</Orig>
<Trans>我向她打招呼, 可她根本不理我!</Trans>
</Sentence>
<Sentence diffgr:id="Sentence3" msdata:rowOrder="2">
<Orig>Hello there, what a coincidence!</Orig>
<Trans>你好,真巧啊!</Trans>
</Sentence>
</Dictionary>
</diffgr:diffgram>
</DataSet>
Found solution by skipping xs:schema and diffgr:diffgram like this:
#Root(name = "DataSet", strict = false)
#Namespace(reference = "http://WebXml.com.cn/")
public class DataSet {
#Path("diffgr:diffgram[1]")
#Element(name = "Dictionary", required = false)
private Dictionary dictionary;
}
Related
I am trying to create a xml whose first element is:
<speak version="1.0"
xmlns="http://www.w3.org/2001/10/synthesis" xml:lang="en-US">
</speak>
I am able to add the first attributes with...
from lxml.etree import Element, SubElement, QName, tostring
root = Element('speak', version="1.0",
xmlns="http://www.w3.org/2001/10/synthesis")
...but not the namespace xml:lang="en-US". Based on several tuto/question like this and this I tried many solutions but none worked.
For example, I tried this :
class XMLNamespaces:
xml = 'http://www.w3.org/2001/10/synthesis'
root.attrib[QName(XMLNamespaces.xml, 'lang')] = "en-US"
But the ouput is
<speak xmlns:ns0="http://www.w3.org/2001/10/synthesis" version="1.0" xmlns="http://www.w3.org/2001/10/synthesis" ns0:lang="en-US">
How can I create the xml:lang="en-US" of my first xml element?
The special xml: prefix is associated with the http://www.w3.org/XML/1998/namespace URI.
The following code adds xml:lang="en-US" to the root element:
root.attrib[QName("http://www.w3.org/XML/1998/namespace", "lang")] = "en-US"
I use an XML Schema Definition and JAXB to generate Java classes with proper #XmlElement or #XmlRootElement annotations.
Since the schema has some deep nesting in it (not my choice), I'd rather use jxpath to access deeply buried classes using an XPath (rather than cumbersome daisy-chain of .getThat() and that != null).
The problem is that some of the XML element names contain dashes, e.g. foo-bar. When I try to access elements using org.apache.jxpath, I need to rewrite my XPath so that such names are camel-cased instead (fooBar) like the name of the actual Java objects. Is there any way to tell jxpath to find the elements using the XPath corresponding to the XML element names (instead of the camel-cased Bean names)?
I think it is related to this question, however in my case I don't actually case what kind of tricks and decorations are used on the auto-generated classes, as long as xjc can do it.
Here is a simple example to illustrate the issue.
First, a small XSD file:
<?xml version="1.0" encoding="UTF-8"?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" elementFormDefault="qualified">
<xs:element name="record">
<xs:complexType>
<xs:sequence>
<xs:element name="foo-bar" type="xs:string"/>
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema>
Compile it...
xjc -p org.pd.jx example.xsd
Here is the xjc-generated Record.java class (minus comments):
#XmlAccessorType(XmlAccessType.FIELD)
#XmlType(name = "", propOrder = {"fooBar"})
#XmlRootElement(name = "record")
public class Record {
#XmlElement(name = "foo-bar", required = true)
protected String fooBar;
public String getFooBar() {return fooBar;}
public void setFooBar(String value) {this.fooBar = value;}
}
Then, trying to access the data via jxpath (in practice I have to deal with lots of deeply nested classes), one can see below that the correct XPath ("foo-bar") doesn't work, but a camel-cased version does.
Record record = new Record();
record.setFooBar("hello world");
JXPathContext context = JXPathContext.newContext(record);
context.setLenient(true);
String a = (String)context.getValue("foo-bar", String.class); // is null
String b = (String)context.getValue("fooBar", String.class); // "hello world"
I believe JXPath operates on an objects properties and not the XML elements they correspond to. I do not believe that JXPath parses any of the JAXB metadata so it wouldn't know the XML nodea you have mapped. What you are seeing appears to be the expected behaviour.
for example i have this xml:
<xml>
<a>1</a>
<b>2</b>
</xml>
and this class:
#XmlRootElement(name = "xml")
public class xml{
int aPlusb;
....
}
i know how to create an XmlAdapter but i would like it to set the filed aPlusb to be the plus between the value of element a and the value of filed b.
is there a way to do this in JAXB without making an XmlTransient field and calculate it separately?
I have the schema like this :
<xsd:complexType name="ContentType" mixed="true">
<xsd:annotation>
<xsd:documentation><![CDATA[
The content type is a broad base type allowing any content.
]]></xsd:documentation>
</xsd:annotation>
<xsd:complexContent>
<xsd:extension base="BaseContentType">
<xsd:sequence>
<xsd:any minOccurs="0" maxOccurs="unbounded" namespace="##any" processContents="lax"
/>
</xsd:sequence>
<xsd:attribute name="orientation" type="OrientationEnum" use="optional"
default="portrait">
<xsd:annotation>
<xsd:documentation><![CDATA[
The #orientation attribute is used to specify a "landscape"
orientation for the published form. This is primarily used
for schedules or for tables.
]]></xsd:documentation>
</xsd:annotation>
</xsd:attribute>
</xsd:extension>
</xsd:complexContent>
</xsd:complexType>
I use xjc command line tool to generate the Java classes from the above schema and the classes are generated as follows:
public abstract class BaseContentType {
#XmlMixed
protected List<Serializable> content;
#XmlAttribute(name = "id")
#XmlJavaTypeAdapter(CollapsedStringAdapter.class)
#XmlID
#XmlSchemaType(name = "ID")
protected String id;
and
public class ContentType
extends BaseContentType
{
#XmlMixed
#XmlAnyElement(lax = true)
#OverrideAnnotationOf
protected List<Object> contentOverrideForContentType;
When I unmarshall, all the nested xml elements get populated in the List contentOverrideForContentType of the ContentType object and all the text elements get populated in the List content of BaseContentType.
How do I determine the order of the text elements with respect to the nested elements and construct the entire text?
I am trying to get the entire text within the ContentType for which I have to look at top level text and the text of all the nested tags and combine them all(here is where I need to know the order). Is there a better way to just extract all the text from ContentType?
Thanks!
EDIT
This is related to this question.
If you declare a propOrder on the specific jaxB class you can specify the order that objects will appear in the xml
e.g.
#XmlType(propOrder = {
"firstObjectName",
"fsecondObjectName"
})
note: when using a propOrder all elements in the object must be added to the propOrder.
I had the same issue and I found out that the elements will be ordered following the order in which the elements had been added to the list ... this is assuming you are using an ordered list (List, ArrayList...)
I have XML file which needs 3 attributes in an element. How can make the order of street, zip and city attribute as I wanted?
<address street="Big Street" zip="2012" city="Austin">
</address>
#XmlType(name="Street)
#XmlRootElement(name = "Street")
public class Street {
#XmlAttribute
private String name;
#XmlAttribute
private String type;
... set and get method
}
Anecdotally, the attributes seem to be in reverse order than they are mentioned in code. In my case, I'm using two variables (name & value) and I had to declare them as:
// The inverse order of name & value seems to make them render in XML in name/value order
#XmlAttribute
protected String value;
#XmlAttribute
protected String name;
When the XML is generated, it results in the following:
<attribute name="nameValue" value="valueValue"/>
You can use #XmlAccessorOrder(has predefined values) or #XmlType(Only works for properties) to govern the ordering.
Samples
Edit :
For custom ordering JAXB specification doesnt provide anything, but you can do if your JAXB provider provides you some features.
Found this link where it speaks about ordering using EclipseLink JAXB.