Search Results

Search found 18213 results on 729 pages for 'hibernate cfg xml'.

Page 13/729 | < Previous Page | 9 10 11 12 13 14 15 16 17 18 19 20  | Next Page >

  • C# encrypt whole XML File

    - by René
    I already have a solution for encrypting of several XML nodes or strings. But of course, you can open the local saved XML file and you should see the node tags. For some intelligent people it could be a reference for hidden informations. Is there any way to encrypt and decrypt the whole xml content including all node tags?

    Read the article

  • Code to read & write in XML

    - by user2954088
    I am trying to write code to read and write the XML but I am facing some errors, Could please someone provide the sample code to read XML in grid and can update/insert data from grid which will be saved in following sample xml file. Sample XML file: I am facing the below exception: The assembly with display name '...XmlSerializers' failed to load in the 'LoadFrom' binding context of the AppDomain with ID 1. The cause of the failure was: System.IO.FileNotFoundException: Could not load file or assembly '.....XmlSerializers, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null' or one of its dependencies. The system cannot find the file specified.

    Read the article

  • .NET XML Serialization without <?xml> root node

    - by Graphain
    Hi, I'm trying to generate XML like this: <?xml version="1.0"?> <!DOCTYPE APIRequest SYSTEM "https://url"> <APIRequest> <Head> <Key>123</Key> </Head> <ObjectClass> <Field>Value</Field </ObjectClass> </APIRequest> I have a class (ObjectClass) decorated with XMLSerialization attributes like this: [XmlRoot("ObjectClass")] public class ObjectClass { [XmlElement("Field")] public string Field { get; set; } } And my really hacky intuitive thought to just get this working is to do this when I serialize: ObjectClass inst = new ObjectClass(); XmlSerializer serializer = new XmlSerializer(inst.GetType(), ""); StringWriter w = new StringWriter(); w.WriteLine(@"<?xml version=""1.0""?>"); w.WriteLine("<!DOCTYPE APIRequest SYSTEM"); w.WriteLine(@"""https://url"">"); w.WriteLine("<APIRequest>"); w.WriteLine("<Head>"); w.WriteLine(@"<Field>Value</Field>"); w.WriteLine(@"</Head>"); XmlSerializerNamespaces ns = new XmlSerializerNamespaces(); ns.Add("", ""); serializer.Serialize(w, inst, ns); w.WriteLine("</APIRequest>"); However, this generates XML like this: <?xml version="1.0"?> <!DOCTYPE APIRequest SYSTEM "https://url"> <APIRequest> <Head> <Key>123</Key> </Head> <?xml version="1.0" encoding="utf-16"?> <ObjectClass> <Field>Value</Field> </ObjectClass> </APIRequest> i.e. the serialize statement is automatically adding a <?xml root element. I know I'm attacking this wrong so can someone point me in the right direction? As a note, I don't think it will make practical sense to just make an APIRequest class with an ObjectClass in it (because there are say 20 different types of ObjectClass that each needs this boilerplate around them) but correct me if I'm wrong.

    Read the article

  • Self-reference entity in Hibernate

    - by Marco
    Hi guys, I have an Action entity, that can have other Action objects as child in a bidirectional one-to-many relationship. The problem is that Hibernate outputs the following exception: "Repeated column in mapping for collection: DbAction.childs column: actionId" Below the code of the mapping: <?xml version="1.0"?> <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd"> <hibernate-mapping> <class name="DbAction" table="actions"> <id name="actionId" type="short" /> <property not-null="true" name="value" type="string" /> <set name="childs" table="action_action" cascade="all-delete-orphan"> <key column="actionId" /> <many-to-many column="actionId" unique="true" class="DbAction" /> </set> <join table="action_action" inverse="true" optional="false"> <key column="actionId" /> <many-to-one name="parentAction" column="actionId" not-null="true" class="DbAction" /> </join> </class> </hibernate-mapping>

    Read the article

  • What are the best workarounds for known problems with Hibernate's schema validation of floating poin

    - by Jason Novak
    I have several Java classes with double fields that I am persisting via Hibernate. For example, I have @Entity public class Node ... private double value; When Hibernate's org.hibernate.dialect.Oracle10gDialect creates the DDL for the Node table, it maps the value field to a "double precision" type. create table MDB.Node (... value double precision not null, ... It would appear that in Oracle, "double precision" is an alias for "float". So, when I try to verify the database schema using the org.hibernate.cfg.AnnotationConfiguration.validateSchema() method, Oracle appears to describe the value column as a "float". This causes Hibernate to throw the following Exception org.hibernate.HibernateException: Wrong column type in DBO.ACL_RULE for column value. Found: float, expected: double precision A very similar problem is listed in Hibernate's JIRA database as HHH-1961 (http://opensource.atlassian.com/projects/hibernate/browse/HHH-1961). I'd like to avoid doing anything that will break MySql, Postgres, and Sql Server support so extending the Oracle10gDialect appears to be the most promising of the workarounds mentioned in HHH-1961. But extending a Dialect is something I've never done before and I'm afraid there may be some nasty gotchas. What is the best workaround for this problem that won't break our compatibility with MySql, Postgres, and Sql Server? Thanks for taking the time to look at this!

    Read the article

  • Using Hibernate with MS ACCESS 2007 Database (Free JDBC Driver)

    - by Quentin T.
    1. I want to do a reverse engineering action with the Hibernate plugin of Eclipse on a MS Access 2007 Database. I'm forced to use a existing MS Access 2007 db. A easy solution is to buy the HXTT. But I want to use a free driver to do my work. So I tried to apply this post : http://www.programmingforfuture.com/2011/06/how-to-use-ms-access-with-hibernate.html (That uses the SQL Server dialect and the driver sun.jdbc.odbc.JdbcOdbcDriver) Unfortunately I have an error that nobody seems to have been on the internet: Exception while generating code Reason : org.hibernate.exception.GenericJDBCException: Error while reading primary key meta data for `c:/myaccessdb.mdb`.TableTest1 I have try to change the primary key on my MS Access DB (deleting all primary key) or to try the reverse engineering on a MS ACCESS with only one table without primary key, but I got all times the problems. 2. The purpose of my job is to transfer daily (weekly) an Oracle 11g database with data from an existing database MS ACCESS 2007. And I thought to use a procedure (Hibernate EJB) Java to be launched automatically every week to do the data transfer. Is this is the best solution ? Configuration : sun.jdbc.odbc.JdbcOdbcDriver v??? Hibernate v3.4 Eclipse ps: If you are a HXTT developer or seller please be indulgent with my post ;). Making money by making people believe that you help, it's bad ! A solution is to use Derby Client driver, as the solution in the post: Does anyone know if Hibernate and java will work effectively with Access? But a clarification of the answer of Rich Seller is required. Could you explain your answer and explain your configuration (hibernate.cfg.xml, persistence.xml and what URL you use in the property name="hibernate.connection.url") without using paying HXTT driver but with the free Derby driver.

    Read the article

  • SSIS - XML Source Script

    - by simonsabin
    The XML Source in SSIS is great if you have a 1 to 1 mapping between entity and table. You can do more complex mapping but it becomes very messy and won't perform. What other options do you have? The challenge with XML processing is to not need a huge amount of memory. I remember using the early versions of Biztalk with loaded the whole document into memory to map from one document type to another. This was fine for small documents but was an absolute killer for large documents. You therefore need a streaming approach. For flexibility however you want to be able to generate your rows easily, and if you've ever used the XmlReader you will know its ugly code to write. That brings me on to LINQ. The is an implementation of LINQ over XML which is really nice. You can write nice LINQ queries instead of the XMLReader stuff. The downside is that by default LINQ to XML requires a whole XML document to work with. No streaming. Your code would look like this. We create an XDocument and then enumerate over a set of annoymous types we generate from our LINQ statement XDocument x = XDocument.Load("C:\\TEMP\\CustomerOrders-Attribute.xml");   foreach (var xdata in (from customer in x.Elements("OrderInterface").Elements("Customer")                        from order in customer.Elements("Orders").Elements("Order")                        select new { Account = customer.Attribute("AccountNumber").Value                                   , OrderDate = order.Attribute("OrderDate").Value }                        )) {     Output0Buffer.AddRow();     Output0Buffer.AccountNumber = xdata.Account;     Output0Buffer.OrderDate = Convert.ToDateTime(xdata.OrderDate); } As I said the downside to this is that you are loading the whole document into memory. I did some googling and came across some helpful videos from a nice UK DPE Mike Taulty http://www.microsoft.com/uk/msdn/screencasts/screencast/289/LINQ-to-XML-Streaming-In-Large-Documents.aspx. Which show you how you can combine LINQ and the XmlReader to get a semi streaming approach. I took what he did and implemented it in SSIS. What I found odd was that when I ran it I got different numbers between using the streamed and non streamed versions. I found the cause was a little bug in Mikes code that causes the pointer in the XmlReader to progress past the start of the element and thus foreach (var xdata in (from customer in StreamReader("C:\\TEMP\\CustomerOrders-Attribute.xml","Customer")                                from order in customer.Elements("Orders").Elements("Order")                                select new { Account = customer.Attribute("AccountNumber").Value                                           , OrderDate = order.Attribute("OrderDate").Value }                                ))         {             Output0Buffer.AddRow();             Output0Buffer.AccountNumber = xdata.Account;             Output0Buffer.OrderDate = Convert.ToDateTime(xdata.OrderDate);         } These look very similiar and they are the key element is the method we are calling, StreamReader. This method is what gives us streaming, what it does is return a enumerable list of elements, because of the way that LINQ works this results in the data being streamed in. static IEnumerable<XElement> StreamReader(String filename, string elementName) {     using (XmlReader xr = XmlReader.Create(filename))     {         xr.MoveToContent();         while (xr.Read()) //Reads the first element         {             while (xr.NodeType == XmlNodeType.Element && xr.Name == elementName)             {                 XElement node = (XElement)XElement.ReadFrom(xr);                   yield return node;             }         }         xr.Close();     } } This code is specifically designed to return a list of the elements with a specific name. The first Read reads the root element and then the inner while loop checks to see if the current element is the type we want. If not we do the xr.Read() again until we find the element type we want. We then use the neat function XElement.ReadFrom to read an element and all its sub elements into an XElement. This is what is returned and can be consumed by the LINQ statement. Essentially once one element has been read we need to check if we are still on the same element type and name (the inner loop) This was Mikes mistake, if we called .Read again we would advance the XmlReader beyond the start of the Element and so the ReadFrom method wouldn't work. So with the code above you can use what ever LINQ statement you like to flatten your XML into the rowsets you want. You could even have multiple outputs and generate your own surrogate keys.        

    Read the article

  • Parse an XML file

    - by karan@dotnet
    The following code shows a simple method of parsing through an XML file/string. We can get the parent name, child name, attributes etc from the XML. The namespace System.Xml would be the only additional namespace that we would be using. string myXMl = "<Employees>" + "<Employee ID='1' Name='John Mayer'" + "Address='12th Street'" + "City='New York' Zip='10004'>" + "</Employee>" + "</Employees>"; XmlDocument xDoc = new XmlDocument();xDoc.LoadXml(myXMl);XmlNodeList xNodeList = xDoc.SelectNodes("Employees/child::node()");foreach (XmlNode xNode in xNodeList){ if (xNode.Name == "Employee") { string ID = xNode.Attributes["ID"].Value; //Outputs: 1 string Name = xNode.Attributes["Name"].Value;//Outputs: John Mayer string Address = xNode.Attributes["Address"].Value;//Outputs: 12th Street string City = xNode.Attributes["City"].Value;//Outputs: New York string Zip = xNode.Attributes["Zip"].Value; //Outputs: 10004 }} Lets look at another XML: string myXMl = "<root>" + "<parent1>..some data</parent1>" + "<parent2>" + "<Child1 id='1' name='Adam'>data1</Child1>" + "<Child2 id='2' name='Stanley'>data2</Child2>" + "</parent2>" + "</root>"; XmlDocument xDoc = new XmlDocument();xDoc.LoadXml(myXMl);XmlNodeList xNodeList = xDoc.SelectNodes("root/child::node()"); //Traverse the entire XML nodes.foreach (XmlNode xNode in xNodeList) { //Looks for any particular nodes if (xNode.Name == "parent1") { //some traversing.... } if (xNode.Name == "parent2") { //If the parent node has child nodes then //traverse the child nodes foreach (XmlNode xNode1 in xNode.ChildNodes) { string childNodeName = xNode1.Name; //Ouputs: Child1 string childNodeData = xNode1.InnerText; //Outputs: data1 //Loop through each attribute of the child nodes foreach (XmlAttribute xAtt in xNode1.Attributes) { string attrName = xAtt.Name; //Outputs: id string attrValue = xAtt.Value; //Outputs: 1 } } }}  

    Read the article

  • IIS MIME type for XML content

    - by Rodolfo
    recently a third party plugin I'm using to display online magazines stopped working on mobile devices. According to their help page, this happens for people serving with IIS. Their solution is to set the MIME type .xml to "application/xml". It's by default set to "text/xml". Changing it does work, but would that have unintended side effects or is it actually the correct way and IIS just set it wrong?

    Read the article

  • .XML Sitemaps and HTML Sitemaps Clarification

    - by MSchumacher
    I've got a website with about 170 pages and I want to create an effective Sitemap for it as it is long due. The website is internally linked very well but I still want to take advantage of creating a sitemap to allow SE's to crawl my site easier and to hopefully increase my websites PR. Though I am slightly confused to what I must do: Is it necessary to create a .xml sitemap AND a HTML Sitemap (both)? ... Because I've never worked with .xml ... where do I put this file once it's created? In the Root folder? So I assume that this sitemap.xml is ONLY to be read by spiders and NOT by website visitors. IE: No visitor on my website is going to visit the page sitemap.xml, am I correct? ... Hence why I should also create an HTML sitemap (sitemap.htm)?

    Read the article

  • 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)

    Read the article

  • 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?

    Read the article

  • org.hibernate.MappingException: Unknown entity:

    - by tsegay
    I tried to see all the questions on this topic but none of them helped me. And I really want to understand what is going on with my code. I have a standalone application which uses spring and Hibernate as JPA and I am trying to run the test using a single main Class My main class package edu.acct.tsegay.common; import edu.acct.tsegay.model.User; import edu.acct.tsegay.business.IUserBusinessObject; import org.springframework.context.ApplicationContext; import org.springframework.context.support.ClassPathXmlApplicationContext; public class App { public static void main(String[] args) { try { ApplicationContext context = new ClassPathXmlApplicationContext( "Spring3AndHibernate-servlet.xml"); IUserBusinessObject userBusinessObject = (IUserBusinessObject) context .getBean("userBusiness"); User user = (User) context.getBean("user1"); user.setPassword("pass"); user.setUsername("tsegay"); System.out.println(user.getPassword()); userBusinessObject.delete(user); User user2 = new User(); user2.setUsername("habest"); user2.setPassword("pass1"); System.out.println(user2.getPassword()); /* * userBusinessObject.save(user2); * * User user3 = userBusinessObject.searchUserbyId("tsegay"); * System.out.println("Search Result: " + user3.getUsername()); */ System.out.println("Success"); } catch (Exception e) { e.printStackTrace(); } } } my application context is: <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd"> <!-- data source --> <bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource"> <property name="driverClassName" value="com.mysql.jdbc.Driver" /> <property name="url" value="jdbc:mysql://localhost:3306/test" /> <property name="username" value="test" /> <property name="password" value="password" /> </bean> <!-- session factory --> <bean id="sessionFactory" class="org.springframework.orm.hibernate3.LocalSessionFactoryBean"> <property name="dataSource"> <ref bean="dataSource" /> </property> <property name="hibernateProperties"> <props> <prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</prop> <prop key="hibernate.show_sql">true</prop> </props> </property> </bean> <!-- exposed person business object --> <bean id="userBusiness" class="edu.acct.tsegay.business.UserBusinessObject"> <property name="userDao" ref="userDao" /> </bean> <bean id="user1" class="edu.acct.tsegay.model.User"> <property name="username" value="tse" /> <property name="password" value="pass" /> </bean> <!-- Data Access Object --> <bean id="userDao" class="edu.acct.tsegay.dao.UserDao"> <property name="sessionFactory" ref="sessionFactory" /> </bean> </beans> My User Model is: package edu.acct.tsegay.model; import java.io.Serializable; import javax.persistence.Entity; import javax.persistence.Id; import javax.persistence.Version; import org.hibernate.annotations.NaturalId; @Entity public class User implements Serializable { /** * */ private static final long serialVersionUID = 1L; private String username; private String password; private Integer VERSION; @Version public Integer getVERSION() { return VERSION; } public void setVERSION(Integer vERSION) { VERSION = vERSION; } @NaturalId public String getUsername() { return username; } public void setUsername(String username) { this.username = username; } public String getPassword() { return password; } public void setPassword(String password) { this.password = password; } } My DAO is: package edu.acct.tsegay.dao; import edu.acct.tsegay.model.User; import org.hibernate.SessionFactory; import org.springframework.beans.factory.annotation.Autowired; import org.springframework.orm.hibernate3.HibernateTemplate; import org.springframework.stereotype.Repository; @Repository public class UserDao implements IUserDao { private SessionFactory sessionFactory; private HibernateTemplate hibernateTemplate; public SessionFactory getSessionFactory() { return sessionFactory; } @Autowired public void setSessionFactory(SessionFactory sessionFactory) { this.sessionFactory = sessionFactory; this.hibernateTemplate = new HibernateTemplate(sessionFactory); } public void save(User user) { // TODO Auto-generated method stub // getHibernateTemplate().save(user); this.hibernateTemplate.save(user); } public void delete(User user) { // TODO Auto-generated method stub this.hibernateTemplate.delete(user); } public User searchUserbyId(String username) { // TODO Auto-generated method stub return this.hibernateTemplate.get(User.class, username); } } And this my stacktrace error when i run the program: pass org.springframework.orm.hibernate3.HibernateSystemException: Unknown entity: edu.acct.tsegay.model.User; nested exception is org.hibernate.MappingException: Unknown entity: edu.acct.tsegay.model.User at org.springframework.orm.hibernate3.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:679) at org.springframework.orm.hibernate3.HibernateAccessor.convertHibernateAccessException(HibernateAccessor.java:412) at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:411) at org.springframework.orm.hibernate3.HibernateTemplate.executeWithNativeSession(HibernateTemplate.java:374) at org.springframework.orm.hibernate3.HibernateTemplate.delete(HibernateTemplate.java:837) at org.springframework.orm.hibernate3.HibernateTemplate.delete(HibernateTemplate.java:833) at edu.acct.tsegay.dao.UserDao.delete(UserDao.java:34) at edu.acct.tsegay.business.UserBusinessObject.delete(UserBusinessObject.java:30) at edu.acct.tsegay.common.App.main(App.java:23) Caused by: org.hibernate.MappingException: Unknown entity: edu.acct.tsegay.model.User at org.hibernate.impl.SessionFactoryImpl.getEntityPersister(SessionFactoryImpl.java:580) at org.hibernate.impl.SessionImpl.getEntityPersister(SessionImpl.java:1365) at org.hibernate.event.def.DefaultDeleteEventListener.onDelete(DefaultDeleteEventListener.java:100) at org.hibernate.event.def.DefaultDeleteEventListener.onDelete(DefaultDeleteEventListener.java:74) at org.hibernate.impl.SessionImpl.fireDelete(SessionImpl.java:793) at org.hibernate.impl.SessionImpl.delete(SessionImpl.java:771) at org.springframework.orm.hibernate3.HibernateTemplate$25.doInHibernate(HibernateTemplate.java:843) at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:406) ... 6 more Please let me know if you need any more of my configuration. Any help is much appreciated..

    Read the article

  • Not recognizing second monitor after hibernate (Windows 7, Dell D630 laptop)

    - by Brooks Moses
    I have a Dell Latitude D630 laptop which I've recently updated to Windows 7 64-bit. (The Dell site confirms that it's Windows-7-compatible.) Normally it lives in a docking station with a second monitor connected to the DVI port on the docking station, and I use the second monitor in a multi-monitor configuration with the laptop screen. Sometimes I undock the laptop and use it separately. Here's the problem: If I hibernate the laptop while undocked, and then power it back up in the docking station, it does not recognize the second monitor. By which I mean that not only does it not share the desktop onto the second monitor, but if I go into the control panel for display settings and press "Detect", it does not even detect the existence of the second monitor. I can tell it to "use the VGA port anyway" for a second monitor, but the monitor is connected to a DVI port on the docking station, so that doesn't do anything useful. If I entirely reboot the laptop while it's connected to the docking station, it has no problem recognizing the second monitor and using it. But then, if I hibernate, undock, de-hibernate while undocked and rehibernate, and then re-dock and de-hibernate, it's back to not recognizing the second monitor again. I'm reasonably certain that this is not a limitation of the hardware; this worked fine on Windows XP. I'm currently using the Windows 7 driver for my video card. I attempted to use the video driver from the Dell website for this laptop, but Dell only provides Vista 64-bit drivers, not Windows 7 64-bit drivers. Their "Windows 7 compatibility" page suggests that the Vista drivers should work, but when I attempted to install the driver, it gave me a "this operating system not supported" error and refused to install. Any further ideas?

    Read the article

  • 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>

    Read the article

  • Can't get running JPA2 with Hibernate and Maven

    - by erlord
    Have been trying the whole day long and googled the ** out of the web ... in vain. You are my last hope: Here's my code: The Entity: package sas.test.model; import javax.persistence.Entity; import javax.persistence.Id; @Entity public class Employee { @Id private int id; private String name; private long salary; public Employee() {} public Employee(int id) { this.id = id; } public int getId() { return id; } public void setId(int id) { this.id = id; } public String getName() { return name; } public void setName(String name) { this.name = name; } public long getSalary() { return salary; } public void setSalary (long salary) { this.salary = salary; } } The service class: package sas.test.dao; import sas.test.model.Employee; import javax.persistence.*; import java.util.List; public class EmployeeService { protected EntityManager em; public EmployeeService(EntityManager em) { this.em = em; } public Employee createEmployee(int id, String name, long salary) { Employee emp = new Employee(id); emp.setName(name); emp.setSalary(salary); em.persist(emp); return emp; } public void removeEmployee(int id) { Employee emp = findEmployee(id); if (emp != null) { em.remove(emp); } } public Employee raiseEmployeeSalary(int id, long raise) { Employee emp = em.find(Employee.class, id); if (emp != null) { emp.setSalary(emp.getSalary() + raise); } return emp; } public Employee findEmployee(int id) { return em.find(Employee.class, id); } } And the main class: package sas.test.main; import javax.persistence.*; import java.util.List; import sas.test.model.Employee; import sas.test.dao.EmployeeService; public class ExecuteMe { public static void main(String[] args) { EntityManagerFactory emf = Persistence.createEntityManagerFactory("EmployeeService"); EntityManager em = emf.createEntityManager(); EmployeeService service = new EmployeeService(em); // create and persist an employee em.getTransaction().begin(); Employee emp = service.createEmployee(158, "John Doe", 45000); em.getTransaction().commit(); System.out.println("Persisted " + emp); // find a specific employee emp = service.findEmployee(158); System.out.println("Found " + emp); // find all employees // List<Employee> emps = service.findAllEmployees(); // for (Employee e : emps) // System.out.println("Found employee: " + e); // update the employee em.getTransaction().begin(); emp = service.raiseEmployeeSalary(158, 1000); em.getTransaction().commit(); System.out.println("Updated " + emp); // remove an employee em.getTransaction().begin(); service.removeEmployee(158); em.getTransaction().commit(); System.out.println("Removed Employee 158"); // close the EM and EMF when done em.close(); emf.close(); } } Finally my confs. pom.xml: <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId>Test_JPA_CRUD</groupId> <artifactId>Test_JPA_CRUD</artifactId> <packaging>jar</packaging> <version>1.0</version> <name>Test_JPA_CRUD</name> <url>http://maven.apache.org</url> <repositories> <repository> <id>maven2-repository.dev.java.net</id> <name>Java.net Repository for Maven</name> <url>http://download.java.net/maven/2/ </url> <layout>default</layout> </repository> <repository> <id>maven.org</id> <name>maven.org Repository</name> <url>http://repo1.maven.org/maven2</url> <releases> <enabled>true</enabled> </releases> <snapshots> <enabled>true</enabled> </snapshots> </repository> </repositories> <dependencies> <dependency> <groupId>junit</groupId> <artifactId>junit</artifactId> <version>4.8.1</version> <scope>test</scope> </dependency> <!-- <dependency> <groupId>javax</groupId> <artifactId>javaee-api</artifactId> <version>6.0</version> </dependency> --> <!-- <dependency> <groupId>javax.persistence</groupId> <artifactId>persistence-api</artifactId> <version>1.0</version> </dependency> --> <!-- JPA2 provider --> <dependency> <groupId>org.hibernate</groupId> <artifactId>hibernate-entitymanager</artifactId> <version>3.4.0.GA</version> </dependency> <!-- JDBC driver --> <dependency> <groupId>org.apache.derby</groupId> <artifactId>derby</artifactId> <version>10.5.3.0_1</version> </dependency> <dependency> <groupId>org.hibernate</groupId> <artifactId>hibernate-core</artifactId> <version>3.3.2.GA</version> </dependency> <dependency> <groupId>org.hibernate</groupId> <artifactId>ejb3-persistence</artifactId> <version>3.3.2.Beta1</version> </dependency> <dependency> <groupId>org.hibernate</groupId> <artifactId>hibernate-annotations</artifactId> <version>3.4.0.GA</version> </dependency> <dependency> <groupId>org.slf4j</groupId> <artifactId>slf4j-log4j12</artifactId> <version>1.5.2</version> </dependency> <dependency> <groupId>log4j</groupId> <artifactId>log4j</artifactId> <version>1.2.14</version> </dependency> </dependencies> <build> <plugins> <!-- compile with mvn assembly:assembly --> <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-jar-plugin</artifactId> <version>2.2</version> </plugin> <!-- compile with mvn assembly:assembly --> <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-assembly-plugin</artifactId> <version>2.2-beta-2</version> <configuration> <descriptorRefs> <descriptorRef>jar-with-dependencies</descriptorRef> </descriptorRefs> <archive> <manifest> <mainClass>sas.test.main.ExecuteMe</mainClass> </manifest> </archive> </configuration> <executions> <execution> <phase>package</phase> </execution> </executions> </plugin> <plugin> <!-- Force UTF-8 & Java-Version 1.6 --> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-compiler-plugin</artifactId> <configuration> <source>1.6</source> <target>1.6</target> <!--<encoding>utf-8</encoding>--> </configuration> </plugin> </plugins> </build> </project> and the persistence.xml, which, I promise, is in the classpath of the target: <?xml version="1.0" encoding="UTF-8"?> <persistence version="1.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd http://java.sun.com/xml/ns/persistence "> <persistence-unit name="EmployeeService" transaction-type="RESOURCE_LOCAL"> <provider>org.hibernate.ejb.HibernatePersistence</provider> <class>sas.test.model.Employee</class> <properties> <property name="javax.persistence.jdbc.driver" value="org.apache.derby.jdbc.EmbeddedDriver"/> <property name="hibernate.dialect" value="org.hibernate.dialect.DerbyDialect"/> <property name="hibernate.show_sql" value="true"/> <property name="javax.persistence.jdbc.url" value="jdbc:derby:webdb;create=true"/> </properties> </persistence-unit> </persistence> As you may have noticed from some commented code, I tried both, the Hibernate and the J2EE 6 implementation of JPA2.0, however, both failed. The above-mentioned code ends up with following error: log4j:WARN No appenders could be found for logger (org.hibernate.cfg.annotations.Version). log4j:WARN Please initialize the log4j system properly. Exception in thread "main" java.lang.UnsupportedOperationException: The user must supply a JDBC connection at org.hibernate.connection.UserSuppliedConnectionProvider.getConnection(UserSuppliedConnectionProvider.java:54) at org.hibernate.jdbc.ConnectionManager.openConnection(ConnectionManager.java:446) at org.hibernate.jdbc.ConnectionManager.getConnection(ConnectionManager.java:167) at org.hibernate.jdbc.JDBCContext.connection(JDBCContext.java:142) Any idea what's going wrong? Any "Hello World" maven/JPA2 demo that actually runs? I couldn't get any of those provided by google's search running. Thanx in advance.

    Read the article

  • How to use second level cache for lazy loaded collections in Hibernate?

    - by Chandru
    Let's say I have two entities, Employee and Skill. Every employee has a set of skills. Now when I load the skills lazily through the Employee instances the cache is not used for skills in different instances of Employee. Let's Consider the following data set. Employee - 1 : Java, PHP Employee - 2 : Java, PHP When I load Employee - 2 after Employee - 1, I do not want hibernate to hit the database to get the skills and instead use the Skill instances already available in cache. Is this possible? If so how? Hibernate Configuration <session-factory> <property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property> <property name="hibernate.connection.password">pass</property> <property name="hibernate.connection.url">jdbc:mysql://localhost/cache</property> <property name="hibernate.connection.username">root</property> <property name="hibernate.dialect">org.hibernate.dialect.MySQLInnoDBDialect</property> <property name="hibernate.cache.use_second_level_cache">true</property> <property name="hibernate.cache.use_query_cache">true</property> <property name="hibernate.cache.provider_class">net.sf.ehcache.hibernate.EhCacheProvider</property> <property name="hibernate.hbm2ddl.auto">update</property> <property name="hibernate.show_sql">true</property> <mapping class="org.cache.models.Employee" /> <mapping class="org.cache.models.Skill" /> </session-factory> The Entities with imports, getters and setters Removed @Entity @Table(name = "employee") @Cache(usage = CacheConcurrencyStrategy.READ_WRITE) public class Employee { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) private int id; private String name; public Employee() { } @ManyToMany @JoinTable(name = "employee_skills", joinColumns = @JoinColumn(name = "employee_id"), inverseJoinColumns = @JoinColumn(name = "skill_id")) @Cache(usage = CacheConcurrencyStrategy.READ_WRITE) private List<Skill> skills; } @Entity @Table(name = "skill") @Cache(usage = CacheConcurrencyStrategy.READ_WRITE) public class Skill { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) private int id; private String name; } SQL for Loading the Second Employee and his Skills Hibernate: select employee0_.id as id0_0_, employee0_.name as name0_0_ from employee employee0_ where employee0_.id=? Hibernate: select skills0_.employee_id as employee1_1_, skills0_.skill_id as skill2_1_, skill1_.id as id1_0_, skill1_.name as name1_0_ from employee_skills skills0_ left outer join skill skill1_ on skills0_.skill_id=skill1_.id where skills0_.employee_id=? In that I specifically want to avoid the second query as the first one is unavoidable anyway.

    Read the article

  • XSD Schema for XML with multiple structures

    - by Xetius
    I am attempting to write an XML Schema to cover a number of XML conditions which I may encounter. I have the same root element (serviceRequest) with different child elements. I was trying to use the xs:extension element to define multiple versions, but it is complaining about unexpected element orderInclusionCriteria etc. Am I going about this the right way, or is there a better way to define this? The other way I thought about this was to have a single xs:choice with all the options inside it, but this seemed somewhat inelegant. These XSD files are for use within XMLBeans if that makes any difference. I have Given the following 2 examples of XML: 1) <?xml version="1.0" encoding="utf-8"?> <serviceRequest method="GOO" debug="NO"> <sessionId sID="ABC1234567" /> <orderInclusionCriteria accountId="1234567" accountNum="1234567890" /> </serviceRequest> 2) <?xml version="1.0" encoding="utf-8"?> <serviceRequest method="GOO" debug="NO"> <sessionId sID="ABC1234567" /> <action aType='MakePayment'> <makePayment accountID='CH91015165S' amount='5.00' /> </action> </serviceRequest> I thought I could use the following schema file: <?xml version="1.0" encoding="UTF-8"?> <xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema"> <xs:element name="serviceRequest" type="ServiceRequestType" /> <xs:element name="session" type="SessionType" /> <xs:attribute name="method" type="xs:string" /> <xs:attribute name="debug" type="xs:string" /> <xs:complexType name="SessionType"> <xs:attribute name="sID" use="required"> <xs:simpleType> <xs:restriction base="xs:string"/> </xs:simpleType> </xs:attribute> </xs:complexType> <xs:complexType name="ServiceRequestType"> <xs:sequence> <xs:element ref="session" /> </xs:sequence> <xs:attribute ref="method" /> <xs:attribute ref="debug" /> </xs:complexType> <xs:complexType name="OrderTrackingServiceRequest"> <xs:complexContent> <xs:extension base="ServiceRequestType"> <xs:complexType> <xs:sequence> <xs:element name="OrderInclusionCriteria" type="xs:string" /> </xs:sequence> </xs:complexType> </xs:extension> </xs:complexContent> </xs:complexType> <xs:complexType name="Action"> <xs:complexContent> <xs:extension base="ServiceRequestType"> <xs:complexType> <xs:sequence> <xs:element name="makePayment"> <xs:complexType> <xs:attribute name="accountID" type="xs:string" /> <xs:attribute name="amount" type="xs:string" /> <xs:complexType> </xs:element> </xs:sequence> <xs:attribute name="aType" type="xs:string" /> </xs:complexType> </xs:extension> </xs:complexContent> </xs:complexType> </xs:schema>

    Read the article

  • JPA persistence.xml share same jar file

    - by user213855
    Hi! I'm wondering if I can share the same jar file for several persistence units.. I mean: I have two persistence units described in my persistence.xml file and the entities are not in the same JAR. Entities are in a separated JAR file but, in that one, there are entites for both persistence units. I think I red somewhere that I coould use tag something like this: externalEntities.jar#com.mycompany.EntityA so I could separate them. I tried this solution and it doesn't work. Now I guess that it couldn't be done to package all entities (for two different persistence units) in the same JAR file. What do yo think?

    Read the article

  • Lost with hibernate - OneToMany resulting in the one being pulled back many times..

    - by Andy
    I have this DB design: CREATE TABLE report ( ID MEDIUMINT PRIMARY KEY NOT NULL AUTO_INCREMENT, user MEDIUMINT NOT NULL, created TIMESTAMP NOT NULL, state INT NOT NULL, FOREIGN KEY (user) REFERENCES user(ID) ON UPDATE CASCADE ON DELETE CASCADE ); CREATE TABLE reportProperties ( ID MEDIUMINT NOT NULL, k VARCHAR(128) NOT NULL, v TEXT NOT NULL, PRIMARY KEY( ID, k ), FOREIGN KEY (ID) REFERENCES report(ID) ON UPDATE CASCADE ON DELETE CASCADE ); and this Hibernate Markup: @Table(name="report") @Entity(name="ReportEntity") public class ReportEntity extends Report{ @Id @GeneratedValue(strategy = GenerationType.AUTO) @Column(name="ID") private Integer ID; @Column(name="user") private Integer user; @Column(name="created") private Timestamp created; @Column(name="state") private Integer state = ReportState.RUNNING.getLevel(); @OneToMany(mappedBy="pk.ID", fetch=FetchType.EAGER) @JoinColumns( @JoinColumn(name="ID", referencedColumnName="ID") ) @MapKey(name="pk.key") private Map<String, ReportPropertyEntity> reportProperties = new HashMap<String, ReportPropertyEntity>(); } and @Table(name="reportProperties") @Entity(name="ReportPropertyEntity") public class ReportPropertyEntity extends ReportProperty{ @Embeddable public static class ReportPropertyEntityPk implements Serializable{ /** * long#serialVersionUID */ private static final long serialVersionUID = 2545373078182672152L; @Column(name="ID") protected int ID; @Column(name="k") protected String key; } @EmbeddedId protected ReportPropertyEntityPk pk = new ReportPropertyEntityPk(); @Column(name="v") protected String value; } And i have inserted on Report and 4 Properties for that report. Now when i execute this: this.findByCriteria( Order.asc("created"), Restrictions.eq("user", user.getObject(UserField.ID)) ) ); I get back the report 4 times, instead of just the once with a Map with the 4 properties in. I'm not great at Hibernate to be honest, prefer straight SQL but I must learn, but i can't see what it is that is wrong.....? Any suggestions?

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

  • 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...

    Read the article

  • Problem updating BLOB with Hibernate?

    - by JohnSmith
    hi, i am having problem updating a blob with hibernate. (i am using Hiberante 3.3.1-GA) my model have these getters/setters for hibernate, i.e. internally i deal with byte[] so any getter/setter convert the byte[] to blog. I can create an initial object without problem, but if I try to change the content of the blob, the database column is not updated. I do not get any error message, everything looks fine, except that the database is not updated. /** do not use, for hibernate only */ public Blob getLogoBinaryBlob() { if(logoBinary == null){ return null; } return Hibernate.createBlob(logoBinary); } /** do not use, for hibernate only */ public void setLogoBinaryBlob(Blob logoBinaryBlob) { ByteArrayOutputStream baos = new ByteArrayOutputStream(); try { logoBinary = toByteArrayImpl(logoBinaryBlob, baos); } catch (Exception e) { } } my hibernate mapping for the blob looks like <property name="logoBinaryBlob" column="LOGO_BINARY" type="blob" />

    Read the article

  • 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

    Read the article

< Previous Page | 9 10 11 12 13 14 15 16 17 18 19 20  | Next Page >