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  • Stairway to XML: Level 3 - Working with Typed XML

    You can enforce the validation of an XML data type, variable or column by associating it with an XML Schema Collection. SQL Server validates a typed XML value against the rules defined in the schema collection so that INSERT or UPDATE operations will succeed only if the value being inserted or updated is valid as per the rules defined in the Schema Collection. NEW! Deployment Manager Early Access ReleaseDeploy SQL Server changes and .NET applications fast, frequently, and without fuss, using Deployment Manager, the new tool from Red Gate. Try the Early Access Release to get a 20% discount on Version 1. Download the Early Access Release.

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  • Convenience of mySQL over xml

    - by Bonechilla
    Currently I use XML to store specific information to correctly load a few things such as a list of specfied characters, scenes and music, Once more I use JAXB in combination with standard compression/decompression(ZIP) functionality to store a list of extrenous data. This data is called to add functionality to the character, somewhat like Skills in an RPG. Each skill is seperated into its own XML file with a grandlist which contains the names of each file with their extensions omitted and zipped in folder that gets encrypted. At first using xml was working fine however as the skill list grow i worry about its stability. I was wondering if I should begin storing the data in mySQL. Originally I planned to simply convert everything to JSON over xml but i think possibly mySQL would be a better move. Can anyone inform me of the key difference and pros and cons of each I guess i'm looking for the best way to store the data more conviently and would be easier to operate on. The data is mostly primatives and strings and the only arraylist of values i have i can just concat into a single field and parse later Edit: If I am going in the right direction with XML would it make sense to convert it to JSON and use maybe Kyro or EclipseLink JAXB (MOXy)

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  • Opening an XML in Unity3D when the game is built

    - by N0xus
    At the moment, my game can open up an XML file inside the editor when I run it. In my XMLReader.cs I'm loading in my file like so: _xmlDocument.Load(Application.dataPath + "\\HV_Settings\\Settings.xml"); This class also deals with what the XML should do once it has been read in. However, when I build the game and run the exe, this file isn't called. I know that I can store this file in the C drive, but I want to keep everything in one place so when I start to release what I'm working on, the user doesn't need to do anything. Am I doing something silly which is causing the XML not to be read?

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  • CAM XML Editor version 2.2.1 now available.

    - by drrwebber
    CAM Editor v2.2.1 release is now available. Lots of nice enhancements, CAMV performance boost and important bug fixes for DoD, NIEM and LEXS schema. Download is available from the CAM XML Editor Resource Site. The CAM editor is the leading open source XML Editor/Validation/Schema designer for rapidly building and deploying complete XML information exchanges. Provides a visual WYSIWYG structure with rule entry wizards and drag and drop dictionary components. Will import, analyze and refactor existing XML Schema. Oracle is a proud sponsor of the project and its use on the NIEM.gov initiative.Creates XSD schema + JAXB bindings, Mindmap or UML models (XMI), XML test suite examples, HTML documentation + spreadsheets (NIEM IEPDs). XSD schema export in default, flatten, NIEM, and OASIS modes. Generates canonical component dictionaries from schema sets, ERwin models, or spreadsheets.

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  • Problem deserializing xml file

    - by Andy
    I auto generated an xsd file from the below xml and used xsd2code to get a c# class. The problem is the entire xml doesn't deserialize. Here is how I'm attempting to deserialize: static void Main(string[] args) { using (TextReader textReader = new StreamReader("config.xml")) { // string temp = textReader.ReadToEnd(); XmlSerializer deserializer = new XmlSerializer(typeof(project)); project p = (project)deserializer.Deserialize(textReader); } } here is the actual XML: <?xml version='1.0' encoding='UTF-8'?> <project> <scm class="hudson.scm.SubversionSCM"> <locations> <hudson.scm.SubversionSCM_-ModuleLocation> <remote>https://svn.xxx.com/test/Validation/CPS DRTest DLL/trunk</remote> </hudson.scm.SubversionSCM_-ModuleLocation> </locations> <useUpdate>false</useUpdate> <browser class="hudson.scm.browsers.FishEyeSVN"> <url>http://fisheye.xxxx.net/browse/Test/</url> <rootModule>Test</rootModule> </browser> <excludedCommitMessages></excludedCommitMessages> </scm> <openf>Hello there</openf> <buildWrappers/> </project> When I run the above, the locations node remains null. Here is the xsd that I'm using: <?xml version="1.0" encoding="utf-8"?> <xs:schema id="NewDataSet" xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata"> <xs:element name="project"> <xs:complexType> <xs:all> <xs:element name="openf" type="xs:string" minOccurs="0" /> <xs:element name="buildWrappers" type="xs:string" minOccurs="0" /> <xs:element name="scm" minOccurs="0"> <xs:complexType> <xs:sequence> <xs:element name="useUpdate" type="xs:string" minOccurs="0" msdata:Ordinal="1" /> <xs:element name="excludedCommitMessages" type="xs:string" minOccurs="0" msdata:Ordinal="2" /> <xs:element name="locations" minOccurs="0"> <xs:complexType> <xs:sequence> <xs:element name="hudson.scm.SubversionSCM_-ModuleLocation" minOccurs="0"> <xs:complexType> <xs:sequence> <xs:element name="remote" type="xs:string" minOccurs="0" /> </xs:sequence> </xs:complexType> </xs:element> </xs:sequence> </xs:complexType> </xs:element> <xs:element name="browser" minOccurs="0"> <xs:complexType> <xs:sequence> <xs:element name="url" type="xs:string" minOccurs="0" msdata:Ordinal="0" /> <xs:element name="rootModule" type="xs:string" minOccurs="0" msdata:Ordinal="1" /> </xs:sequence> <xs:attribute name="class" type="xs:string" /> </xs:complexType> </xs:element> </xs:sequence> <xs:attribute name="class" type="xs:string" /> </xs:complexType> </xs:element> </xs:all> </xs:complexType> </xs:element> <xs:element name="NewDataSet" msdata:IsDataSet="true" msdata:UseCurrentLocale="true"> <xs:complexType> <xs:choice minOccurs="0" maxOccurs="unbounded"> <xs:element ref="project" /> </xs:choice> </xs:complexType> </xs:element> </xs:schema>

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  • How do I require that an element has either one set of attributes or another in an XSD schema?

    - by Eli Courtwright
    I'm working with an XML document where a tag must either have one set of attributes or another. For example, it needs to either look like <tag foo="hello" bar="kitty" /> or <tag spam="goodbye" eggs="world" /> e.g. <root> <tag foo="hello" bar="kitty" /> <tag spam="goodbye" eggs="world" /> </root> So I have an XSD schema where I use the xs:choice element to choose between two different attribute groups: <xsi:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema" attributeFormDefault="unqualified" elementFormDefault="qualified"> <xs:element name="root"> <xs:complexType> <xs:sequence> <xs:element maxOccurs="unbounded" name="tag"> <xs:choice> <xs:complexType> <xs:attribute name="foo" type="xs:string" use="required" /> <xs:attribute name="bar" type="xs:string" use="required" /> </xs:complexType> <xs:complexType> <xs:attribute name="spam" type="xs:string" use="required" /> <xs:attribute name="eggs" type="xs:string" use="required" /> </xs:complexType> </xs:choice> </xs:element> </xs:sequence> </xs:complexType> </xs:element> </xsi:schema> However, when using lxml to attempt to load this schema, I get the following error: >>> from lxml import etree >>> etree.XMLSchema( etree.parse("schema_choice.xsd") ) Traceback (most recent call last): File "<stdin>", line 1, in <module> File "xmlschema.pxi", line 85, in lxml.etree.XMLSchema.__init__ (src/lxml/lxml.etree.c:118685) lxml.etree.XMLSchemaParseError: Element '{http://www.w3.org/2001/XMLSchema}element': The content is not valid. Expected is (annotation?, ((simpleType | complexType)?, (unique | key | keyref)*))., line 7 Since the error is with the placement of my xs:choice element, I've tried putting it in different places, but no matter what I try, I can't seem to use it to define a tag to have either one set of attributes (foo and bar) or another (spam and eggs). Is this even possible? And if so, then what is the correct syntax?

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  • XSD Restrictions based on target xml elements

    - by ??????
    Is it possible in xsd to create restriction based on elements of some type in target (processed) document? For example I have XML like this: <Pets> <Pet name="Murka" /> <Pet name="Browko" /> <Pet name="Tuzik" /> </Pets> <Children> <Child name="Petruk" favoritePet="Browko" /> </Children> so what I want to restrict the attribute "favoritePet" of element "Child" based on existing "Pet" elements. How can I do this?

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  • XMl Data Structure

    - by metdos
    Which one of two XML structures below do you prefer? Why? Any other suggestion is welcome :) <Parameters> <Parameter id=username>metdos</Parameter> <Parameter id=password>123</Parameter> </Parameters> or <Parameters> <username>metdos</username> <password>123</password> </Parameters>

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  • XML deserializer (Iserialzable)

    - by user311130
    Hey everybody, I have a class in c# that implements Iserialzable. I'm using a XMLSerializer that produces a XML from instance of that class. I get the following XML: <?xml version="1.0"?> <Configuration xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <SessionConfiguration> <RemoteMachineName>HV-BENDA</RemoteMachineName> </SessionConfiguration> <SessionsCredentialsList> <CredentialsItem> <User>test0</User> <Password>Pa$$word1</Password> </CredentialsItem> <CredentialsItem> <User>test1</User> <Password>Pa$$word1</Password> </CredentialsItem> <CredentialsItem> <User>test2</User> <Password>Pa$$word1</Password> </CredentialsItem> <CredentialsItem> <User>test3</User> <Password>Pa$$word1</Password> </CredentialsItem> <CredentialsItem> <User>test4</User> <Password>Pa$$word1</Password> </CredentialsItem> </SessionsCredentialsList> <TIME_OUT /> <LOCAL_USERS_NUM>5</LOCAL_USERS_NUM> </Configuration> At some later point in the code I use a XMLSerializer again to deserial that XML document. and I get the following error: {"There is an error in XML document (1, 1)."} Inner exception: {"Data at the root level is invalid. Line 1, position 1."} Do someone knows wat could be the problem? All the best

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  • How do I serialize an object to xml but not have it be the root element of the xml document

    - by mezoid
    I have the following object: public class MyClass { public int Id { get; set;} public string Name { get; set; } } I'm wanting to serialize this to the following xml string: <MyClass> <Id>1</Id> <Name>My Name</Name> </MyClass> Unfortunately, when I use the XMLSerializer I get a string which looks like: <?xml version="1.0" encoding="utf-8"?> <MyClass xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <Id>1</Id> <Name>My Name</Name> </MyClass> I'm not wanting MyClass to be the root element the document, rather I'm eventually wanting to add the string with other similar serialized objects which will be within a larger xml document. i.e. Eventually I'll have a xml string which looks like this: <Classes> <MyClass> <Id>1</Id> <Name>My Name</Name> </MyClass> <MyClass> <Id>1</Id> <Name>My Name</Name> </MyClass> </Classes>" My first thought was to create a class as follows: public class Classes { public List<MyClass> MyClasses { get; set; } } ...but that just addes an additional node called MyClasses to wrap the list of MyClass.... My gut feeling is that I'm approaching this the wrong way and that my lack of experience with creating xml files isn't helping to point me to some part of the .NET framework or some other library that simplifies this.

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  • How to change the extension of a processed xml file (using eXist & cocoon)

    - by Carsten C.
    Hi all, I'm really new to this whole web stuff, so please be nice if I missed something important to post. Short: Is there a possibility to change the name of a processed file (eXist-DB) after serialization? Here my case, the following request to my eXist-db: http://localhost:8080/exist/cocoon/db/caos/test.xml and I want after serialization the follwing (xslt is working fine): http://localhost:8080/exist/cocoon/db/caos/test.html I'm using the followong sitemap.xmap with cocoon (hoping this is responsible for it) <map:match pattern="db/caos/**"> <!-- if we have an xpath query --> <map:match pattern="xpath" type="request-parameter"> <map:generate src="xmldb:exist:///db/caos/{../1}/#{1}"/> <map:act type="request"> <map:parameter name="parameters" value="true"/> <map:parameter name="default.howmany" value="1000"/> <map:parameter name="default.start" value="1"/> <map:transform type="filter"> <map:parameter name="element-name" value="result"/> <map:parameter name="count" value="{howmany}"/> <map:parameter name="blocknr" value="{start}"/> </map:transform> <map:transform src=".snip./webapp/stylesheets/db2html.xsl"> <map:parameter name="block" value="{start}"/> <map:parameter name="collection" value="{../../1}"/> </map:transform> </map:act> <map:serialize type="html" encoding="UTF-8"/> </map:match> <!-- if the whole file will be displayed --> <map:generate src="xmldb:exist:/db/caos/{1}"/> <map:transform src="..snip../stylesheets/caos2soac.xsl"> <map:parameter name="collection" value="{1}"/> </map:transform> <map:transform type="encodeURL"/> <map:serialize type="html" encoding="UTF-8"/> </map:match> So my Question is: How do I change the extension of the test.xml to test.html after processing the xml file? Background: I'm generating some information out of some xml-dbs, this infos will be displayed in html (which is working), but i want to change some entrys later, after I generated the html site. To make this confortable, I want to use Jquery & Jeditable, but the code does not work on the xml files. Saving the generated html is not an option. tia for any suggestions [and|or] help CC Edit: After reading all over: could it be, that the extension is irrelevant and that this is only a problem of port 8080? I'm confused...

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  • c# linq to xml to list

    - by WtFudgE
    I was wondering if there is a way to get a list of results into a list with linq to xml. If I would have the following xml for example: <?xml version="1.0"?> <Sports xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <SportPages> <SportPage type="test"> <LinkPage> <IDList> <string>1</string> <string>2</string> </IDList> </LinkPage> </SportPage> </SportPages> </Sports> How could I get a list of strings from the IDList? I'm fairly new to linq to xml so I just tried some stuff out, I'm currently at this point: var IDs = from sportpage in xDoc.Descendants("SportPages").Descendants("SportPage") where sportpage.Attribute("type").Value == "Karate" select new { ID = sportpage.Element("LinkPage").Element("IDList").Elements("string") }; But the var is to chaotic to read decently. Isn't there a way I could just get a list of strings from this? Thanks

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  • Best approach to convert XML to RDF/XML using an ontology

    - by krisvandenbergh
    I have an XML which uses the XPDL standard (which has an XML schema). What I'm trying to do now is to convert its content to RDF format (serialized in XML), in terms of a certain ontology. Clearly, there needs to be some sort of mapping here. I would like to do this using PHP. The thing is, I have no idea how to do this best. I know how to read an XML file, but how would the mappings occur? What would be a good approach?

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  • How to generate XML with attributes in c#.

    - by user292815
    I have that code: ... request data = new request(); data.username = formNick; xml = data.Serialize(); ... [System.Serializable] public class request { public string username; public string password; static XmlSerializer serializer = new XmlSerializer(typeof(request)); public string Serialize() { StringBuilder builder = new StringBuilder(); XmlWriterSettings settings = new XmlWriterSettings(); settings.OmitXmlDeclaration = true; settings.Encoding = Encoding.UTF8; serializer.Serialize( System.Xml.XmlWriter.Create(builder, settings ), this); return builder.ToString(); } public static request Deserialize(string serializedData) { return serializer.Deserialize(new StringReader(serializedData)) as request; } } I want to add attributes to some nodes and create some sub-nodes. Also how to parse xml like that: <answer> <player id="2"> <coordinate axis="x"></coordinate> <coordinate axis="y"></coordinate> <coordinate axis="z"></coordinate> <action name="nothing"></action> </player> <player id="3"> <coordinate axis="x"></coordinate> <coordinate axis="y"></coordinate> <coordinate axis="z"></coordinate> <action name="boom"> <1>1</1> <2>2</2> </action> </player> </answer> p.s. it is not a xml file, it's answer from http server.

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  • How to generate XML with attributes in .NET?

    - by user292815
    I have that code: ... request data = new request(); data.username = formNick; xml = data.Serialize(); ... [System.Serializable] public class request { public string username; public string password; static XmlSerializer serializer = new XmlSerializer(typeof(request)); public string Serialize() { StringBuilder builder = new StringBuilder(); XmlWriterSettings settings = new XmlWriterSettings(); settings.OmitXmlDeclaration = true; settings.Encoding = Encoding.UTF8; serializer.Serialize( System.Xml.XmlWriter.Create(builder, settings ), this); return builder.ToString(); } public static request Deserialize(string serializedData) { return serializer.Deserialize(new StringReader(serializedData)) as request; } } I want to add attributes to some nodes and create some sub-nodes. Also how to parse xml like that: <answer> <player id="2"> <coordinate axis="x"></coordinate> <coordinate axis="y"></coordinate> <coordinate axis="z"></coordinate> <action name="nothing"></action> </player> <player id="3"> <coordinate axis="x"></coordinate> <coordinate axis="y"></coordinate> <coordinate axis="z"></coordinate> <action name="boom"> <1>1</1> <2>2</2> </action> </player> </answer> p.s. it is not a xml file, it's answer from http server.

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  • Android programming - How to acces [to draw on] XML view in main.xml layout, using code

    - by user556248
    Ok I'm a newbie at Android programming, have a hard time with the graphics part. Understand the beauty of creating layout in XML file, but lost how to access various elements, especially a View element to draw on it. See example of my layout main.xml here; <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/root" android:orientation="vertical" android:layout_width="fill_parent" android:layout_height="fill_parent"> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="wrap_content"> <TextView xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/Title" android:text="App title" android:layout_width="fill_parent" android:layout_height="wrap_content" android:textColor="#000000" android:background="#A0A0FF"/> </LinearLayout> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/PaperLayout" android:layout_width="fill_parent" android:layout_height="0dp" android:layout_weight="1" android:orientation="horizontal" android:focusable="true"> <View android:id="@+id/Paper" android:layout_width="fill_parent" android:layout_height="fill_parent" /> </LinearLayout> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="wrap_content"> <Button android:id="@+id/File" android:layout_width="fill_parent" android:layout_weight="1" android:layout_height="34dp" android:layout_alignParentTop="true" android:layout_centerInParent="true" android:clickable="true" android:textSize="10sp" android:text="File" /> <Button android:id="@+id/Edit" android:layout_width="fill_parent" android:layout_weight="1" android:layout_height="34dp" android:clickable="true" android:textSize="10sp" android:text="Edit" /> </LinearLayout> </LinearLayout> As you can see I have a custom app title bar, then a View filling middle, and finally two buttons in bottom. Catching button events and responding to, for example changing title bar text and changing View background color works fine, but how the heck do I access and more importantly draw on the view defined in main.xml UPDATE: That for your suggestion, however besides that I need a View, not ImageView and you are missing a parameter on canvas.drawText and an ending bracket, it does not work. Now this is most likely because you missed the fact that I am a newbie and assuming I can fill in any blanks. Now first of all I do NOT understand why in my main.xml layout file I can create a View or even a SurfaceView element, which is what I need, but according to your solution I don't even specify the View like Anyways I edited my main.xml according to your solution, and slimmed it down a bit for simplicity; <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/root" android:orientation="vertical" android:layout_width="fill_parent" android:layout_height="fill_parent"> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="wrap_content"> <TextView xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/Title" android:text="App title" android:layout_width="fill_parent" android:layout_height="wrap_content" android:textColor="#000000" android:background="#A0A0FF"/> </LinearLayout> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/PaperLayout" android:layout_width="fill_parent" android:layout_height="0dp" android:layout_weight="1" android:orientation="horizontal" android:focusable="true"> <com.example.MyApp.CustomView android:id="@+id/Paper" android:layout_width="fill_parent" android:layout_height="fill_parent" /> <com.example.colorbook.CustomView/> </LinearLayout> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="wrap_content"> <Button android:id="@+id/File" android:layout_width="fill_parent" android:layout_weight="1" android:layout_height="34dp" android:layout_alignParentTop="true" android:layout_centerInParent="true" android:clickable="true" android:textSize="10sp" android:text="File" /> </LinearLayout> </LinearLayout> In my main java file, MyApp.java, I added this after OnCreate; public class CustomView extends ImageView { @Override protected void onDraw(Canvas canvas) { super.onDraw(canvas); canvas.drawText("Your Text", 1, 1, null); } } But I get error on the "CustomView" part; "Implicit super constructor ImageView() is undefined for default constructor.Must define an explicit constructor" Eclipse suggests 3 quick fixes about adding constructor, but none helps, well it removes error but gives error on app when running. I hope somebody can break this down for me and provide a solution, and perhaps explain why I can't just create a View element in my main.xml layotu file and draw on it in code.

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  • jaxb unmarshaling with schema validation in runtime

    - by ekeren
    I am using jaxb for my application configurations I feel like I am doing something really crooked and I am looking for a way to not need an actual file or this transaction. As you can see in code I: 1.create a schema into a file from my JaxbContext (from my class annotation actually) 2.set this schema file in order to allow true validation when I unmarshal Schema mySchema = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI).newSchema(schemaFile); jaxbContext.generateSchema(new MySchemaOutputResolver()); // ultimately creates schemaFile Unmarshaller u = m_context.createUnmarshaller(); u.setSchema(mySchema); u.unmarshal(...); do any of you know how I can validate jaxb without needing to create a schema file that sits in my computer? Do I need to create a schema for validation, it looks redundant when I get it by JaxbContect.generateSchema ? How do you do this?

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  • Unable to load huge XML document (incorrectly suppose it's due to the XSLT processing)

    - by krisvandenbergh
    I'm trying to match certain elements using XSLT. My input document is very large and the source XML fails to load after processing the following code (consider especially the first line). <xsl:template match="XMI/XMI.content/Model_Management.Model/Foundation.Core.Namespace.ownedElement/Model_Management.Package/Foundation.Core.Namespace.ownedElement"> <rdf:RDF> <rdf:Description rdf:about=""> <xsl:for-each select="Foundation.Core.Class"> <xsl:for-each select="Foundation.Core.ModelElement.name"> <owl:Class rdf:ID="@Foundation.Core.ModelElement.name" /> </xsl:for-each> </xsl:for-each> </rdf:Description> </rdf:RDF> </xsl:template> Apparently the XSLT fails to load after "Model_Management.Model". The PHP code is as follows: if ($xml->loadXML($source_xml) == false) { die('Failed to load source XML: ' . $http_file); } It then fails to perform loadXML and immediately dies. I think there are two options now. 1) I should set a maximum executing time. Frankly, I don't know how that I do this for the built-in PHP 5 XSLT processor. 2) Think about another way to match. What would be the best way to deal with this? The input document can be found at http://krisvandenbergh.be/uml_pricing.xml Any help would be appreciated! Thanks.

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  • Python XML + Java XML interoperability.

    - by erb
    Hello. I need a recommendation for a pythonic library that can marshall python objects to XML(let it be a file). I need to be able read that XML later on with Java (JAXB) and unmarshall it. I know JAXB has some issues that makes it not play nice with .NET XML libraries so a recommendation on something that actually works would be great.

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  • how can i unmarshall in jaxb and enjoy the schema validation without using an explicit schema file

    - by ekeren
    I am using jaxb for my application configurations I feel like I am doing something really crooked and I am looking for a way to not need an actual file or this transaction. As you can see in code I: 1.create a schema into a file from my JaxbContext (from my class annotation actually) 2.set this schema file in order to allow true validation when I unmarshal JAXBContext context = JAXBContext.newInstance(clazz); Schema mySchema = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI).newSchema(schemaFile); jaxbContext.generateSchema(new MySchemaOutputResolver()); // ultimately creates schemaFile Unmarshaller u = m_context.createUnmarshaller(); u.setSchema(mySchema); u.unmarshal(...); do any of you know how I can validate jaxb without needing to create a schema file that sits in my computer? Do I need to create a schema for validation, it looks redundant when I get it by JaxbContect.generateSchema ? How do you do this?

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  • Hibernate schema parameter doesn't work in @SequenceGenerator annotation

    - by tabdulin
    I hav the following code: @Entity @Table(name = "my_table", schema = "my_schema") @SequenceGenerator(name = "my_table_id_seq", sequenceName = "my_table_id_seq", schema = "my_schema") public class MyClass { @Id @GeneratedValue(generator = "my_table_id_seq", strategy = GenerationType.SEQUENCE) private int id; } Database: Postgresql 8.4, Hibernate annotations 3.5.0-Final. When saving the object of MyClass it generates the following SQL query: select nextval('my_table_id_seq') So there is no schema prefix and therefore the sequence cannot be found. When I write the sequenceName like sequenceName = "my_schema.my_table_id_seq" everything works. Do I have misunderstandings for meaning of schema parameter or is it a bug? Any ideas how to make schema parameter working?

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  • Writing to XML issue Unity3D C#

    - by N0xus
    I'm trying to create a tool using Unity to generate an XML file for use in another project. Now, please, before someone suggests I do it in something, the reason I am using Unity is that it allows me to easily port this to an iPad or other device with next to no extra development. So please. Don't suggest to use something else. At the moment, I am using the following code to write to my XML file. public void WriteXMLFile() { string _filePath = Application.dataPath + @"/Data/HV_DarkRideSettings.xml"; XmlDocument _xmlDoc = new XmlDocument(); if (File.Exists(_filePath)) { _xmlDoc.Load(_filePath); XmlNode rootNode = _xmlDoc.CreateElement("Settings"); // _xmlDoc.AppendChild(rootNode); rootNode.RemoveAll(); XmlElement _cornerNode = _xmlDoc.CreateElement("Screen_Corners"); _xmlDoc.DocumentElement.PrependChild(_cornerNode); #region Top Left Corners XYZ Values // indent top left corner value to screen corners XmlElement _topLeftNode = _xmlDoc.CreateElement("Top_Left"); _cornerNode.AppendChild(_topLeftNode); // set the XYZ of the top left values XmlElement _topLeftXNode = _xmlDoc.CreateElement("TopLeftX"); XmlElement _topLeftYNode = _xmlDoc.CreateElement("TopLeftY"); XmlElement _topLeftZNode = _xmlDoc.CreateElement("TopLeftZ"); // indent these values to the top_left value in XML _topLeftNode.AppendChild(_topLeftXNode); _topLeftNode.AppendChild(_topLeftYNode); _topLeftNode.AppendChild(_topLeftZNode); #endregion #region Bottom Left Corners XYZ Values // indent top left corner value to screen corners XmlElement _bottomLeftNode = _xmlDoc.CreateElement("Bottom_Left"); _cornerNode.AppendChild(_bottomLeftNode); // set the XYZ of the top left values XmlElement _bottomLeftXNode = _xmlDoc.CreateElement("BottomLeftX"); XmlElement _bottomLeftYNode = _xmlDoc.CreateElement("BottomLeftY"); XmlElement _bottomLeftZNode = _xmlDoc.CreateElement("BottomLeftZ"); // indent these values to the top_left value in XML _bottomLeftNode.AppendChild(_bottomLeftXNode); _bottomLeftNode.AppendChild(_bottomLeftYNode); _bottomLeftNode.AppendChild(_bottomLeftZNode); #endregion #region Bottom Left Corners XYZ Values // indent top left corner value to screen corners XmlElement _bottomRightNode = _xmlDoc.CreateElement("Bottom_Right"); _cornerNode.AppendChild(_bottomRightNode); // set the XYZ of the top left values XmlElement _bottomRightXNode = _xmlDoc.CreateElement("BottomRightX"); XmlElement _bottomRightYNode = _xmlDoc.CreateElement("BottomRightY"); XmlElement _bottomRightZNode = _xmlDoc.CreateElement("BottomRightZ"); // indent these values to the top_left value in XML _bottomRightNode.AppendChild(_bottomRightXNode); _bottomRightNode.AppendChild(_bottomRightYNode); _bottomRightNode.AppendChild(_bottomRightZNode); #endregion _xmlDoc.Save(_filePath); } } This generates the following XML file: <Settings> <Screen_Corners> <Top_Left> <TopLeftX /> <TopLeftY /> <TopLeftZ /> </Top_Left> <Bottom_Left> <BottomLeftX /> <BottomLeftY /> <BottomLeftZ /> </Bottom_Left> <Bottom_Right> <BottomRightX /> <BottomRightY /> <BottomRightZ /> </Bottom_Right> </Screen_Corners> </Settings> Which is exactly what I want. However, each time I press the button that has the WriteXMLFile() method attached to it, it's write the entire lot again. Like so: <Settings> <Screen_Corners> <Top_Left> <TopLeftX /> <TopLeftY /> <TopLeftZ /> </Top_Left> <Bottom_Left> <BottomLeftX /> <BottomLeftY /> <BottomLeftZ /> </Bottom_Left> <Bottom_Right> <BottomRightX /> <BottomRightY /> <BottomRightZ /> </Bottom_Right> </Screen_Corners> <Screen_Corners> <Top_Left> <TopLeftX /> <TopLeftY /> <TopLeftZ /> </Top_Left> <Bottom_Left> <BottomLeftX /> <BottomLeftY /> <BottomLeftZ /> </Bottom_Left> <Bottom_Right> <BottomRightX /> <BottomRightY /> <BottomRightZ /> </Bottom_Right> </Screen_Corners> <Screen_Corners> <Top_Left> <TopLeftX /> <TopLeftY /> <TopLeftZ /> </Top_Left> <Bottom_Left> <BottomLeftX /> <BottomLeftY /> <BottomLeftZ /> </Bottom_Left> <Bottom_Right> <BottomRightX /> <BottomRightY /> <BottomRightZ /> </Bottom_Right> </Screen_Corners> </Settings> Now, I've written XML files before using winforms, and the WriteXMLFile function is exactly the same. However, in my winform, no matter how much I press the WriteXMLFile button, it doesn't write the whole lot again. Is there something that I'm doing wrong or should change to stop this from happening?

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