Search Results

Search found 15846 results on 634 pages for 'pom xml'.

Page 102/634 | < Previous Page | 98 99 100 101 102 103 104 105 106 107 108 109  | Next Page >

  • What MS technology to use for HTTP service returning XML?

    - by Borek
    I need to create a service that: accepts HTTP requests (with query string or HTTP POST parameters) does some processing on the requests (checking if the request is valid, authentication etc.) reads data from a custom store (another HTTP call in our case) returns the result as custom XML (defined with XSD) I'm trying to think of various MS technologies that could help me and how good they would be for this scenario (pretty standard one I guess). The tasks above are relatively separate, this is what comes to mind: HTTP front-end: ASP.NET Web Forms ASP.NET MVC (seems more appropriate here as I won't need server controls, view state etc.) WCF? Don't know much about it or how well it would suit my task. Custom logic on the server: this will probably be a generic C# code in all cases (sometimes "plugged into" or called from MVC controllers or some equivalent place in other technologies) Reading data from internal data stores: As said, this is another HTTP server in our case. Options that come to mind: Just read the data using something like WebClient (Just theoretically) implement a LINQ provider (Just even more theoretically) implement an EF provider Output the data as custom XML: Linq2XML Serialization? Is it flexible enough? Does WCF provide some tools for this? Some "OXM" - Object/XML mapper if there is something like that for .NET I may be wrong in many of my assumptions, this is just a quick list that comes to mind after a quick research. Some general notes / questions: Testing is important Solution with a clear domain model would be much preferred over the one without Can Entity Framework actually help somewhere in my scenario? If so, where and how? Would WCF be an appropriate technology for this? I don't know much about it.

    Read the article

  • Hibernate - Problem in parsing mapping file (.hbm.xml)

    - by Yatendra Goel
    I am new to Hibernate. I have an exception while running an Hibernate-based application. The exception is as follows: 16 [main] INFO org.hibernate.cfg.Environment - Hibernate 3.3.2.GA 16 [main] INFO org.hibernate.cfg.Environment - hibernate.properties not found 16 [main] INFO org.hibernate.cfg.Environment - Bytecode provider name : javassist 31 [main] INFO org.hibernate.cfg.Environment - using JDK 1.4 java.sql.Timestamp handling 94 [main] INFO org.hibernate.cfg.Configuration - configuring from resource: /hibernate.cfg.xml 94 [main] INFO org.hibernate.cfg.Configuration - Configuration resource: /hibernate.cfg.xml 219 [main] INFO org.hibernate.cfg.Configuration - Reading mappings from resource : app/data/City.hbm.xml 266 [main] ERROR org.hibernate.util.XMLHelper - Error parsing XML: XML InputStream(12) Attribute "coloumn" must be declared for element type "property". 266 [main] ERROR org.hibernate.util.XMLHelper - Error parsing XML: XML InputStream(13) Attribute "coloumn" must be declared for element type "property". 266 [main] ERROR org.hibernate.util.XMLHelper - Error parsing XML: XML InputStream(14) Attribute "coloumn" must be declared for element type "property". It seems that it is not finding coloumn attribute of the property element in the mappings file but my mappings file do have the coloumn attribute. Below is the mappings file (City.hbm.xml) <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd"> <hibernate-mapping package="app.data"> <class name="City" table="CITY"> <id column="CITY_ID" name="cityId"> <generator class="native"/> </id> <property name="cityDisplyaName" coloumn="CITY_DISPLAY_NAME" /> <property coloumn="CITY_MEANINGFUL_NAME" name="cityMeaningFulName" /> <property coloumn="CITY_URL" name="cityURL" /> </class> </hibernate-mapping>

    Read the article

  • XML Schema: How to specify an attribute with a custom 'simpleType' type?

    - by mackenir
    In my XML schema definition, I'm trying to restrict the value of an attribute to be an integer between 0 and 100. With reference to the sample schema below, I want attribute 'attr' on element 'root' to have this restriction. To achieve this I define a simpleType 'Percentage' and set this as the 'type' of 'attr'. However, my XML schema editor (VS 2008) flags the attribute up as having a problem: "Type 'Percentage' is not declared or is not a simple type". <?xml version="1.0" encoding="utf-8"?> <xs:schema elementFormDefault="qualified" id="test" xmlns:xs="http://www.w3.org/2001/XMLSchema" targetNamespace="http://testtttt"> <xs:simpleType name="Percentage"> <xs:restriction base="xs:integer"> <xs:minInclusive value="0"/> <xs:maxInclusive value="100"/> </xs:restriction> </xs:simpleType> <xs:element name="root"> <xs:complexType> <xs:attribute name="attr" type="Percentage" use="optional" /> </xs:complexType> </xs:element>

    Read the article

  • Parsing log files in a folder in ColdFusion

    - by Simon Guo
    The problem is there is a folder ./log/ containing the files like: jan2010.xml, feb2010.xml, mar2010.xml, jan2009.xml, feb2009.xml, mar2009.xml ... each xml file would like: <root><record name="bob" spend="20"></record>...(more records)</root> I want to write a piece of ColdFusion code (log.cfm) that simply parsing those xml files. For the front end I would let user to choose a year, then the click submit button. All the content in that year will be show up in separate table by month. Each table shows the total money spent for each person. like: person cost bob 200 mike 300 Total 500 Thanks.

    Read the article

  • How to define multiple elements in XML Schema with the same name and different attribute value allow

    - by David Skyba
    I would like to create XML Schema for this chunk of xml, I would like to restrict the values of "name" attribute, so that in output document on and only one instance of day is allowed for each week day: <a> <day name="monday" /> <day name="tuesday" /> <day name="wednesday" /> </a> I have tried to use this: <xs:complexType name="a"> <xs:sequence> <xs:element name="day" minOccurs="1" maxOccurs="1"> <xs:complexType> <xs:attribute name="name" use="required"> <xs:simpleType> <xs:restriction base="xs:string"> <xs:enumeration value="monday" /> </xs:restriction> </xs:simpleType> </xs:attribute> </xs:complexType> </xs:element> <xs:element name="day" minOccurs="1" maxOccurs="1"> <xs:complexType> <xs:attribute name="name" use="required"> <xs:simpleType> <xs:restriction base="xs:string"> <xs:enumeration value="tuesday" /> </xs:restriction> </xs:simpleType> </xs:attribute> </xs:complexType> </xs:element> </xs:sequence> </xs:complexType> but XML Schema validator in eclipse says error "Multiple elements with name 'day', with different types, appear in the model group.". Is there any other way?

    Read the article

  • How to syntax-highlight XML in CDATA elements in Vim?

    - by Jim Hurne
    Vim's syntax highlighting for XML/XSL is great, except it turns off all syntax highlighting in CDATA regions. Is there a way to turn on syntax highlighting on in CDATA regions? At work, we have a lot of XSL code embedded within other XML documents. It would be great if I could get all of the goodness of XML editing for the embedded XSL code as well without having to temporarily remove the CDATA tags, or copy the CDATA content into a temporary file. Example: <root> <someTag><![CDATA[ <xsl:template match="/"> <!-- XSL content here --> </xsl:template> ]]> </someTag> </root> Note that the name of the tag (in the example, someTag) containing the content could be anything. We also sometimes embed Javascript inside CDATA regions as well, and again, it would be nice to turn on Javascript syntax highlighting for those regions. Again, the tag the data is embedded in is usually arbitrary and can be anything.

    Read the article

  • How can I create/update an XML node that may or may not exist?

    - by ScottBai
    Is there a method available (i.e. without me creating my own recursive method), for a given xpath (or other method of identifying the hierarchical position) to create/update an XML node, where the node would be created if it does not exist? This would need to create the parent node if it does not exist as well. I do have an XSD which includes all possible nodes. i.e. Before: <employee> <name>John Smith</name> </employee> Would like to call something like this: CoolXmlUpdateMethod("/employee/address/city", "Los Angeles"); After: <employee> <name>John Smith</name> <address> <city>Los Angeles</city> </address> </employee> Or even a method to create a node, given an xpath, wherein it will recursively create the parent node(s) if they do not exist? As far as the application (if it matters), this is taking an existing XML doc that contains only populated nodes, and adding data to it from another system. The new data may or may not already have values populated in the source XML. Surely this is not an uncommon scenario?

    Read the article

  • How to parse XML to flex Data Grid contents.

    - by Jeeva
    My xml file which is in a webserver is show below. <root> <userdetails> <username>raja</username> <status>offline</status> </userdetails> <userdetails> <username>Test</username> <status>online</status> </userdetails> </root> How can i parse this to flex data grid contents. I tried with below coding < ?xml version="1.0" encoding="utf-8"? < mx:Application xmlns:mx="http://www.adobe.com/2006/mxml" layout="absolute" creationComplete="initApp()" < mx:HTTPService id="userList" result="handleData(event)" resultFormat="object" url="http://apps.facebook.com/ajparkin/user_list.xml" / <mx:Script> <![CDATA[ import mx.collections.ArrayCollection; import mx.rpc.events.ResultEvent; import mx.controls.Alert; public function initApp():void { userList.send(); } [Bindable] var userdetailsArray:ArrayCollection; private function handleData(evt:ResultEvent):void { this.userdetailsArray= evt.result.userdetails; } ]]> </mx:Script> <mx:DataGrid dataProvider="{userdetailsArray}"> <mx:columns> <mx:DataGridColumn dataField="username" headerText="User Name"/> <mx:DataGridColumn dataField="status" headerText="Status" /> </mx:columns> </mx:DataGrid> </mx:Application> I'm getting only the field names not the data.

    Read the article

  • Parse large XML file w/ script or use BioPython API ?

    - by jeremy04
    Hey guys this is my first question on here. I'm trying to make a local copy of the UniprotKB in SQL. The UniprotKB is 2.1GB, and it comes in XML and a special text format used by SwissProt Here are my options: 1) Use a SAX parser (XML) - I chose Ruby, and Nokogiri. I started writing the parser, but my initial reaction: how would I map the XML schema to the SAX parser? 2) BioPython - I already have BioSQL/Biopython installed, which literally created my SQL schema for me, and I was able to successfully insert one SwissProt/Uniprot txt file into the database. I'm running it right now (crosses fingers) on the entire 2.1gb. Here is the code I'm running: from Bio import SeqIO from BioSQL import BioSeqDatabase from Bio import SwissProt server = BioSeqDatabase.open_database(driver = "MySQLdb", user = "root", passwd = "", host="localhost", db = "bioseqdb") db = server["uniprot"] iterator = SeqIO.parse(open("/path/to/uniprot_sprot.dat", "r"), "swiss") db.load(iterator) server.commit() Edit: it's now crashing because the transactions are getting locked (since the tables are Innodb) Error Number: 1205 Lock wait timeout exceeded; try restarting transaction. I'm using MySQL version: 5.1.43 Should I switch my database to Postgrelsql ?

    Read the article

  • jquery how to access the an xml node by index?

    - by DS
    Hi, say I've an xml returned from server like this: <persons> <person> <firstname>Jon</firstname> </person> <person> <firstname>Jack</firstname> </person> <person> <firstname>James</firstname> </person> </persons> If I want to access the 3rd firstname node (passed dynamically and stored in i, assumed to be 3 here), how do I do that? My weird attempt follows: var i=3; $(xml).find('firstname').each(function(idx){ if (idx==i) alert($(this).text()); }); It does fetch me the right content... but it just feels wrong to me especially the looping part. Basically I'm looping through the whole tree using .each()! Is there any better approach than this? Something that'd take me to the nth node directly like: alert( $(xml).find('firstname')[idx].text() ); // where idx=n I'm new to jquery so please excuse my jquery coding approach.

    Read the article

  • How to fix Eclipse validation error "No grammar constraints detected for the document"?

    - by Casey
    Eclipse 3.5.2 is throwing an XML schema warning message: No grammar constraints (DTD or XML schema) detected for the document. The application.xml file: <?xml version="1.0" encoding="UTF-8"?> <application xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/application_5.xsd" version="5"> </application> I do not want to disable the warning. How can I get Eclipse to correctly validate the XML document?

    Read the article

  • Recursion Problem in PHP

    - by streetparade
    I need to create a valid xml from a given array(); My Method looks like this, protected function array2Xml($array) { $xml = ""; if(is_array($array)) { foreach($array as $key=>$value) { $xml .= "<$key>"; if(is_array($value)) { $xml .= $this->array2Xml($value); } $xml .= "</$key>"; } return $xml; } else { throw new Exception("in valid"); } } protected function createValidXMLfromArray($array,$node) { $xml = '<?xml version="1.0" encoding="ISO-8859-1"?>'; $xmlArray = $this->array2Xml($array); $xml .= "<$node>$xmlArray</$node>"; return $xml; } if i execute the above i just get keys with empty values; like <node> <name></name> </node> What i need is if i pass this array("name"=>"test","value"=>array("test1"=>33,"test2"=>40)); that it return this <node> <name>test</name> <value> <test1>33</test1> <test2>40</test2> </value> </node> Where is the error what did i wrong in the above recursion?

    Read the article

  • How to wrtie a XML License Line in C#?

    - by Nano HE
    My want to write a XML file as this: <?xml version="1.0" encoding="UTF-8"?> <Equipment xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <License licenseId="" licensePath="" /> Some piece of my code attached here // Create a new file in D:\\ and set the encoding to UTF-8 XmlTextWriter textWriter = new XmlTextWriter("D:\\books.xml", System.Text.Encoding.UTF8); // Format automatically textWriter.Formatting = Formatting.Indented; // Opens the document textWriter.WriteStartDocument(); // Write the namespace declaration. textWriter.WriteStartElement("books", null); // Write the genre attribute. textWriter.WriteAttributeString("xmlns", "xsd", null, "http://www.w3.org/2001/XMLSchema"); textWriter.WriteAttributeString("xmlns", "xsi", null, "http://www.w3.org/2001/XMLSchema-instance"); And now I need to write the License Line below in C# <License licenseId="" licensePath="" /> But I don't know how to move on for I found the Line ended with the special / .Thank you.

    Read the article

  • C# XML parsing with LINQ storing directly to a struct?

    - by Luke
    Say I have the following XML schema: <root> <version>2.0</version> <type>fiction</type> <chapters> <chapter>1</chapter> <title>blah blah</title> </chapter> <chapters> <chapter>2</chapter> <title>blah blah</title> </chapters> </root> Would it be possibly to parse the elements which I know will not be repeated in the XML and store them directly into the struct using LINQ? For example, could I do something like this for "version" and "type" //setup structs Book book = new Book(); book.chapter = new Chapter(); //use LINQ to parse the xml var bookData = from b in xmlDoc.Decendants("root") select new { book.version = b.Element("version").Value, book.type = b.Element("type").Value }; //then for "chapters" since I know there are multiple I can do: var chapterData = from c in xmlDoc.Decendants("root" select new { chapter = c.Element("chapters") }; foreach (var ch in chapterData) { book.chapter.Add(getChapterData(ch.chapter)); }

    Read the article

  • Store Business Rules in XML Document, Validate afterwards in Java, how?

    - by JavaPete
    Example XML Rules document: <user> <username> <not-null/> <capitals value="false"/> <max-length value="15"/> </username> <email> <not-null/> <isEmail/> <max-length value="40"/> </email> </user> How do I implement this? I'm starting from scratch, what I currently have is a User-class, and a UserController which saves the User object in de DB (through a Service-layer and Dao-layer), basic Spring MVC. I can't use Spring MVC Validation however in our Model-classes, I have to use an XML document so an Admin can change the rules I think I need a pattern which dynamically builds an algorithm based on what is provided by the XML Rules document, but I can't seem to think of anything other than a massive amount of if-statements. I also have nothing for the parsing yet and I'm not sure how I'm gonna (de)couple it from the actual implementation of the validation-process.

    Read the article

  • Maven: Re-use a POM file in every project.

    - by Zombies
    My goal is pretty simple actually but since there are multiple (and seemingly complex ways to do this) I wonder what I need to do... So I have certain runtime libraries (ADF libraries in particular) that are needed to be added to every project. This parent pom file will just have JAR dependencies in it. How can I use this pom file from a child pom file?

    Read the article

  • how to fetch data from XML and update database table.

    - by ppp
    I am passing serialized collection(XML) to stored procedure. My XML structure is- <ArrayofDepartmentEntity> <Department> <id>1004</id> <budget>2500.oo</budget> </Department> <Department> <id>1080</id> <budget>3500.oo</budget> </Department> <Department> <id>1029</id> <budget>4500.00</budget> </Department> </ArrayofDepartmentEntity> How can I UPDATE corresponding budget column where department IDs are in above XML?? can any body write down the sql syntax. my SP syntax- ALTER PROCEDURE [dbo].[usp_SaveDepartentBudget] ( @departmentBudgetXML ntext = NULL ) AS BEGIN DECLARE @ptrHandle int EXEC sp_xml_preparedocument @ptrHandle OUTPUT, @departmentBudgetXML ....Here I want to update Budget foreach departmentID in @departmentBudgetXML

    Read the article

  • How to copy child nodes to another xml document?

    - by Alex
    Below is my xml XML1 <?xml version="1.0" encoding="ISO-8859-1" ?> <CATALOG> <CD> <TITLE>1</TITLE> <ARTIST>Bob Dylan</ARTIST> <COUNTRY>USA</COUNTRY> <COMPANY>Columbia</COMPANY> <PRICE>10.90</PRICE> <YEAR>1985</YEAR> </CD> <CD> <TITLE>2</TITLE> <ARTIST>Bonnie Tyler</ARTIST> <COUNTRY>UK</COUNTRY> <COMPANY>CBS Records</COMPANY> <PRICE>9.90</PRICE> <YEAR>1988</YEAR> </CD> </CATALOG> XML2 <?xml version="1.0" encoding="ISO-8859-1" ?> <CATALOG> <CD> <TITLE>3</TITLE> <ARTIST>Dolly Parton</ARTIST> <COUNTRY>USA</COUNTRY> <COMPANY>RCA</COMPANY> <PRICE>9.90</PRICE> <YEAR>1982</YEAR> </CD> </CATALOG> i need output like this <?xml version="1.0" encoding="ISO-8859-1" ?> <CATALOG> <CD> <TITLE>1</TITLE> <ARTIST>Bob Dylan</ARTIST> <COUNTRY>USA</COUNTRY> <COMPANY>Columbia</COMPANY> <PRICE>10.90</PRICE> <YEAR>1985</YEAR> </CD> <CD> <TITLE>2</TITLE> <ARTIST>Bonnie Tyler</ARTIST> <COUNTRY>UK</COUNTRY> <COMPANY>CBS Records</COMPANY> <PRICE>9.90</PRICE> <YEAR>1988</YEAR> </CD> <CD> <TITLE>3</TITLE> <ARTIST>Dolly Parton</ARTIST> <COUNTRY>USA</COUNTRY> <COMPANY>RCA</COMPANY> <PRICE>9.90</PRICE> <YEAR>1982</YEAR> </CD> </CATALOG> How i write this in classic asp ?

    Read the article

  • how get xml responce using JAX-WS SOAP handler

    - by khris
    I have implemented web service: @WebServiceClient(//parameters//) @HandlerChain(file = "handlers.xml") public class MyWebServiceImpl {...} Also I have implemented ObjectFactory with list of classes for creating my requests and responses. For Example class Test. I need to get xml of response. I try to use JAX-WS SOAP handler, so I add this @HandlerChain(file = "handlers.xml") anotation. My handlers.xml looks like: <?xml version="1.0" encoding="UTF-8"?> <handler-chains xmlns="http://java.sun.com/xml/ns/javaee"> <handler-chain> <handler> <handler-class>java.com.db.crds.ws.service.LoggingHandler</handler-class> </handler> </handler-chain> </handler-chains> My LoggingHandler class is: import java.io.PrintWriter; import java.util.Set; import javax.xml.namespace.QName; import javax.xml.soap.SOAPMessage; import javax.xml.ws.handler.MessageContext; import javax.xml.ws.handler.soap.SOAPMessageContext; public class LoggingHandler implements javax.xml.ws.handler.soap.SOAPHandler<SOAPMessageContext> { public void close(MessageContext messagecontext) { } public Set<QName> getHeaders() { return null; } public boolean handleFault(SOAPMessageContext messagecontext) { return true; } public boolean handleMessage(SOAPMessageContext smc) { Boolean outboundProperty = (Boolean) smc.get (MessageContext.MESSAGE_OUTBOUND_PROPERTY); if (outboundProperty.booleanValue()) { System.out.println("\nOutbound message:"); } else { System.out.println("\nInbound message:"); } SOAPMessage message = smc.getMessage(); try { PrintWriter writer = new PrintWriter("soap_responce" + System.currentTimeMillis(), "UTF-8"); writer.println(message); writer.close(); message.writeTo(System.out); System.out.println(""); // just to add a newline } catch (Exception e) { System.out.println("Exception in handler: " + e); } return outboundProperty; } } I have test class which creates request, here are part of code: MyWebServiceImpl impl = new MyWebServiceImpl(url, qName); ws = impl.getMyWebServicePort(); Test req = new Test(); I suppose to get xml response in file "soap_responce" + System.currentTimeMillis(). But such file isn't even created. Please suggest how to get xml response, I'm new to web services and may do something wrong. Thanks

    Read the article

  • Deploying a PHP Library project with Maven

    - by Marco
    Hi, I've created a PHP Library project using Maven, and I'm now ready for its deployment. Following the instructions at http://www.php-maven.org/deploy.html, something went wrong. The configuration is set to: <descriptorRef>php-lib</descriptorRef> During the execution of mvn deploy I get a list of errors for unfound dependencies in the repository: [INFO] [jar:jar {execution: default-jar}] [INFO] Building jar: /home/marco/projects/php/my-app/target/my-app-1.0-SNAPSHOT.jar [INFO] [plugin:addPluginArtifactMetadata {execution: default-addPluginArtifactMetadata}] Downloading: http://repo1.php-maven.org/release/org/phpmaven/maven-php-plugin/2.2-beta-2/maven-php-plugin-2.2-beta-2.pom [INFO] Unable to find resource 'org.phpmaven:maven-php-plugin:pom:2.2-beta-2' in repository release-repo1.php-maven.org (http://repo1.php-maven.org/release) Downloading: http://repo1.maven.org/maven2/org/phpmaven/maven-php-plugin/2.2-beta-2/maven-php-plugin-2.2-beta-2.pom [INFO] Unable to find resource 'org.phpmaven:maven-php-plugin:pom:2.2-beta-2' in repository central (http://repo1.maven.org/maven2) Downloading: http://repo1.php-maven.org/release/org/phpmaven/maven-php-plugin/2.2-beta-2/maven-php-plugin-2.2-beta-2.pom [INFO] Unable to find resource 'org.phpmaven:maven-php-plugin:pom:2.2-beta-2' in repository release-repo1.php-maven.org (http://repo1.php-maven.org/release) Downloading: http://repo1.maven.org/maven2/org/phpmaven/maven-php-plugin/2.2-beta-2/maven-php-plugin-2.2-beta-2.pom [INFO] Unable to find resource 'org.phpmaven:maven-php-plugin:pom:2.2-beta-2' in repository central (http://repo1.maven.org/maven2) Downloading: http://repo1.php-maven.org/release/org/apache/maven/wagon/wagon-http-shared/1.0-beta-6/wagon-http-shared-1.0-beta-6.pom [INFO] Unable to find resource 'org.apache.maven.wagon:wagon-http-shared:pom:1.0-beta-6' in repository release-repo1.php-maven.org (http://repo1.php-maven.org/release) Downloading: http://repo1.php-maven.org/release/org/apache/maven/wagon/wagon-http-shared/1.0-beta-6/wagon-http-shared-1.0-beta-6.pom [INFO] Unable to find resource 'org.apache.maven.wagon:wagon-http-shared:pom:1.0-beta-6' in repository release-repo1.php-maven.org (http://repo1.php-maven.org/release) Downloading: http://repo1.maven.org/maven2/org/apache/maven/wagon/wagon-http-shared/1.0-beta-6/wagon-http-shared-1.0-beta-6.pom Downloading: http://repo1.php-maven.org/release/nekohtml/xercesMinimal/1.9.6.2/xercesMinimal-1.9.6.2.pom [INFO] Unable to find resource 'nekohtml:xercesMinimal:pom:1.9.6.2' in repository release-repo1.php-maven.org (http://repo1.php-maven.org/release) Downloading: http://repo1.php-maven.org/release/nekohtml/xercesMinimal/1.9.6.2/xercesMinimal-1.9.6.2.pom [INFO] Unable to find resource 'nekohtml:xercesMinimal:pom:1.9.6.2' in repository release-repo1.php-maven.org (http://repo1.php-maven.org/release) Downloading: http://repo1.maven.org/maven2/nekohtml/xercesMinimal/1.9.6.2/xercesMinimal-1.9.6.2.pom And this is my settings.xml file: <settings> <profiles> <profile> <id>profile-php-maven</id> <pluginRepositories> <pluginRepository> <id>release-repo1.php-maven.org</id> <name>PHP-Maven 2 Release Repository</name> <url>http://repo1.php-maven.org/release</url> <releases> <enabled>true</enabled> </releases> </pluginRepository> <pluginRepository> <id>snapshot-repo1.php-maven.org</id> <name>PHP-Maven 2 Snapshot Repository</name> <url>http://repo1.php-maven.org/snapshot</url> <releases> <enabled>false</enabled> </releases> <snapshots> <enabled>true</enabled> </snapshots> </pluginRepository> </pluginRepositories> <repositories> <repository> <id>release-repo1.php-maven.org</id> <name>PHP-Maven 2 Release Repository</name> <url>http://repo1.php-maven.org/release</url> <releases> <enabled>true</enabled> </releases> </repository> <repository> <id>snapshot-repo1.php-maven.org</id> <name>PHP-Maven 2 Snapshot Repository</name> <url>http://repo1.php-maven.org/snapshot</url> <releases> <enabled>false</enabled> </releases> <snapshots> <enabled>true</enabled> </snapshots> </repository> </repositories> </profile> </profiles> <activeProfiles> <activeProfile>profile-php-maven</activeProfile> </activeProfiles> </settings> For every step I've followed the documentation (which is poor, though). Any tips? Thanks

    Read the article

  • Book Review: The Art of XSD - SQL Server XML schemas

    The 14 chapters of "The Art of XSD”, written by MVP Jacob Sebastian, will take the reader step-by–step all the way from the basics of XML Schema design all the way to advanced topics on SQL Server XML Schema Collections. Reviewer Hima Bindu Vejella gives it an 8/10 rating, and gives us an excellent distilled description of what the book has to offer.

    Read the article

  • Book Review: The Art of XSD - SQL Server XML schemas

    The 14 chapters of "The Art of XSD, written by MVP Jacob Sebastian, will take the reader step-bystep all the way from the basics of XML Schema design all the way to advanced topics on SQL Server XML Schema Collections. Reviewer Hima Bindu Vejella gives it an 8/10 rating, and gives us an excellent distilled description of what the book has to offer....Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

    Read the article

  • remap an xml feed to the address of a wordpress rss feed

    - by cboettig
    I used to have a blog based on Wordpress and moved to one based on Jekyll. I can create a new feed in Jekyll by building an atom page in XML with a bit of Liquid code, like this The trouble is, the location of the new feed is http://carlboettiger.info/atom.xml, while the old feed from the wordpress site is http://carlboettiger.info/feed, with no extension. how can I configure the Jekyll-created feed such that followers who have pointed their readers to the old feed address from wordpress will start to get the new content? (Site's Jekyll source here)

    Read the article

  • Android XML Parser Performance

    Shane Conder will show us how different XML parsers affect performance with Android and the answers might surprise you. The article provides developers with data for choosing a particular XML parser and Android code that demonstrates all three parsers.

    Read the article

< Previous Page | 98 99 100 101 102 103 104 105 106 107 108 109  | Next Page >