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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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  • 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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  • Using xsi:nil in XML

    - by Matt
    I am generating an XML file from a VB.NET app. The document was generating fine before I tried to add nillable elements. I am now testing putting in just one nil element as: <blah xsi:nil="true"></blah> Once this element is in place and I try to view the XML file in IE it is unable to display. I am receiving: > The XML page cannot be displayed > Cannot view XML input using XSL style > sheet. Please correct the error and > then click the Refresh button, or try > again later. > > -------------------------------------------------------------------------------- > > The operation completed successfully. > Error processing resource If I remove this one element it displays fine again. What am I missing here?

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  • Using Hibernate to do a query involving two tables

    - by Nathan Spears
    I'm inexperienced with sql in general, so using Hibernate is like looking for an answer before I know exactly what the question is. Please feel free to correct any misunderstandings I have. I am on a project where I have to use Hibernate. Most of what I am doing is pretty basic and I could copy and modify. Now I would like to do something different and I'm not sure how configuration and syntax need to come together. Let's say I have two tables. Table A has two (relevant) columns, user GUID and manager GUID. Obviously managers can have more than one user under them, so queries on manager can return more than one row. Additionally, a manager can be managing the same user on multiple projects, so the same user can be returned multiple times for the same manager query. Table B has two columns, user GUID and user full name. One-to-one mapping there. I want to do a query on manager GUID from Table A, group them by unique User GUID (so the same User isn't in the results twice), then return those users' full names from Table B. I could do this in sql without too much trouble but I want to use Hibernate so I don't have to parse the sql results by hand. That's one of the points of using Hibernate, isn't it? Right now I have Hibernate mappings that map each column in Table A to a field (well the get/set methods I guess) in a DAO object that I wrote just to hold that Table's data. I could also use the Hibernate DAOs I have to access each table separately and do each of the things I mentioned above in separate steps, but that would be less efficient (I assume) that doing one query. I wrote a Service object to hold the data that gets returned from the query (my example is simplified - I'm going to keep some other data from Table A and get multiple columns from Table B) but I'm at a loss for how to write a DAO that can do the join, or use the DAOs I have to do the join. FYI, here is a sample of my hibernate config file (simplified to match my example): <hibernate-mapping package="com.my.dao"> <class name="TableA" table="table_a"> <id name="pkIndex" column="pk_index" /> <property name="userGuid" column="user_guid" /> <property name="managerGuid" column="manager_guid" /> </class> </hibernate-mapping> So then I have a DAOImplementation class that does queries and returns lists like public List<TableA> findByHQL(String hql, Map<String, String> params) etc. I'm not sure how "best practice" that is either.

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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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  • Mapping a boolean[] PostgreSql column with Hibernate

    - by teabot
    I have a column in a PostgreSql database that is defined with type boolean[]. I wish to map this to a Java entity property using Hibernate 3.3.x. However, I cannot find a suitable Java type that Hibernate is happy to map to. I thought that the java.lang.Boolean[] would be the obvious choice, but Hibernate complains: Caused by: org.hibernate.HibernateException: Wrong column type in schema.table for column mycolumn. Found: _bool, expected: bytea at org.hibernate.mapping.Table.validateColumns(Table.java:284) at org.hibernate.cfg.Configuration.validateSchema(Configuration.java:1130) I have also tried the following property types without success: java.lang.String java.lang.boolean[] java.lang.Byte[] How can I map this column?

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  • Configure Hibernate validation for bean

    - by sergionni
    Hi. I need to perform validation based on SQL query result. Query is defined as annotation - as @NamedQuery in my entity bean. According to Hibernate documentation(doc), there is possibility to validate bean on following operations: pre-update pre-insert pre-delete looks like: <hibernate-configuration> <session-factory> ... <event type="pre-update"> <listener class="org.hibernate.cfg.beanvalidation.BeanValidationEventListener"/> </event> <event type="pre-insert"> <listener class="org.hibernate.cfg.beanvalidation.BeanValidationEventListener"/> </event> <event type="pre-delete"> <listener class="org.hibernate.cfg.beanvalidation.BeanValidationEventListener"/> </event> </hibernate-configuration> The question is how to connect my bean with the validation configuration, described above.

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  • Screen flickers after resuming from hibernate with Intel GMA 3600 (Acer D270 with Intel N2600)

    - by Cameron
    Fresh system install, with correct display driver from Intel (8.14.8.1065). Clicking "Update driver" merely results in a message saying the driver was already up-to-date. After resuming from hibernate the entire screen flickers. This is especially noticeable while using IE9 (which has hardware acceleration enabled by default) on Google maps, in particular while typing in an address in the map search field (the flickering is much worse under these circumstances). Note that sleep worked fine, only hibernate causes this issue. Restarting "fixes" the problem temporarily until the next hibernate-resume. Aero is enabled. This is on Windows 7 (Pro) 32-bit, on an Acer Aspire One D270-1998, with the Intel N2600 (which has the Intel GMA 3600 built-in).

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  • JNDI Datasource definition in Tomcat 6.0

    - by romaintaz
    I want to define a DataSource to an Oracle database on my Tomcat 6.0. So, in conf/server.xml (yes, I know that this DataSource will be available for all the webapps in Tomcat, but it's not a problem here), I've set this Resource: <GlobalNamingResources> <Resource name="hibernate/HibernateDS" auth="Container" type="javax.sql.DataSource" url="jdbc:oracle:thin:@myserver:1542:foo" username="foo" password="bar" driverClassName="oracle.jdbc.OracleDriver" maxActive="50" maxIdle="10" validationQuery="select 1 from dual"/> Then, in the web.xml of my application, I set a resource-ref element: <resource-ref> <description>Hibernate Datasource</description> <res-ref-name>hibernate/HibernateDS</res-ref-name> <res-type>javax.sql.DataSource</res-type> <res-auth>Container</res-auth> </resource-ref> Finally, as Hibernate is used to manage the database connection, I have a webapps/mywebapp/WEB-INF/classes/hibernate.cfg.xml that creates a session-factory using the JNDI DataSource: <hibernate-configuration> <session-factory> <property name="connection.datasource">java:comp/env/hibernate/HibernateDS</property> ... However, when I start my Tomcat server, I get an error that says it could not create the INFO [net.sf.hibernate.util.NamingHelper] JNDI InitialContext properties:{} INFO [net.sf.hibernate.connection.DatasourceConnectionProvider] Using datasource: java:comp/env/hibernate/HibernateDS INFO [net.sf.hibernate.transaction.TransactionFactoryFactory] Transaction strategy: net.sf.hibernate.transaction.JDBCTransactionFactory INFO [net.sf.hibernate.transaction.TransactionManagerLookupFactory] No TransactionManagerLookup configured (in JTA environment, use of process level read-write cache is not recommended) WARN [net.sf.hibernate.cfg.SettingsFactory] Could not obtain connection metadata org.apache.tomcat.dbcp.dbcp.SQLNestedException: Cannot create JDBC driver of class '' for connect URL 'null' at org.apache.tomcat.dbcp.dbcp.BasicDataSource.createDataSource(BasicDataSource.java:1150) at org.apache.tomcat.dbcp.dbcp.BasicDataSource.getConnection(BasicDataSource.java:880) at net.sf.hibernate.connection.DatasourceConnectionProvider.getConnection(DatasourceConnectionProvider.java:59) at net.sf.hibernate.cfg.SettingsFactory.buildSettings(SettingsFactory.java:84) at net.sf.hibernate.cfg.Configuration.buildSettings(Configuration.java:1172) ... Caused by: java.lang.NullPointerException at sun.jdbc.odbc.JdbcOdbcDriver.getProtocol(JdbcOdbcDriver.java:507) at sun.jdbc.odbc.JdbcOdbcDriver.knownURL(JdbcOdbcDriver.java:476) at sun.jdbc.odbc.JdbcOdbcDriver.acceptsURL(JdbcOdbcDriver.java:307) at java.sql.DriverManager.getDriver(DriverManager.java:253) at org.apache.tomcat.dbcp.dbcp.BasicDataSource.createDataSource(BasicDataSource.java:1143) ... 11 more Do you have any idea why Hibernate is not able to construct the session-factory? What is wrong in my configuration?

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  • JNDI Datasource definition in Tomcat 6.0

    - by romaintaz
    Hi all, I want to define a DataSource to an Oracle database on my Tomcat 6.0. So, in conf/server.xml (yes, I know that this DataSource will be available for all the webapps in Tomcat, but it's not a problem here), I've set this Resource: <GlobalNamingResources> <Resource name="hibernate/HibernateDS" auth="Container" type="javax.sql.DataSource" url="jdbc:oracle:thin:@myserver:1542:foo" username="foo" password="bar" driverClassName="oracle.jdbc.OracleDriver" maxActive="50" maxIdle="10" validationQuery="select 1 from dual"/> Then, in the web.xml of my application, I set a resource-ref element: <resource-ref> <description>Hibernate Datasource</description> <res-ref-name>hibernate/HibernateDS</res-ref-name> <res-type>javax.sql.DataSource</res-type> <res-auth>Container</res-auth> </resource-ref> Finally, as Hibernate is used to manage the database connection, I have a webapps/mywebapp/WEB-INF/classes/hibernate.cfg.xml that creates a session-factory using the JNDI DataSource: <hibernate-configuration> <session-factory> <property name="connection.datasource">java:comp/env/hibernate/HibernateDS</property> ... However, when I start my Tomcat server, I get an error that says it could not create the INFO [net.sf.hibernate.util.NamingHelper] JNDI InitialContext properties:{} INFO [net.sf.hibernate.connection.DatasourceConnectionProvider] Using datasource: java:comp/env/hibernate/HibernateDS INFO [net.sf.hibernate.transaction.TransactionFactoryFactory] Transaction strategy: net.sf.hibernate.transaction.JDBCTransactionFactory INFO [net.sf.hibernate.transaction.TransactionManagerLookupFactory] No TransactionManagerLookup configured (in JTA environment, use of process level read-write cache is not recommended) WARN [net.sf.hibernate.cfg.SettingsFactory] Could not obtain connection metadata org.apache.tomcat.dbcp.dbcp.SQLNestedException: Cannot create JDBC driver of class '' for connect URL 'null' at org.apache.tomcat.dbcp.dbcp.BasicDataSource.createDataSource(BasicDataSource.java:1150) at org.apache.tomcat.dbcp.dbcp.BasicDataSource.getConnection(BasicDataSource.java:880) at net.sf.hibernate.connection.DatasourceConnectionProvider.getConnection(DatasourceConnectionProvider.java:59) at net.sf.hibernate.cfg.SettingsFactory.buildSettings(SettingsFactory.java:84) at net.sf.hibernate.cfg.Configuration.buildSettings(Configuration.java:1172) ... Caused by: java.lang.NullPointerException at sun.jdbc.odbc.JdbcOdbcDriver.getProtocol(JdbcOdbcDriver.java:507) at sun.jdbc.odbc.JdbcOdbcDriver.knownURL(JdbcOdbcDriver.java:476) at sun.jdbc.odbc.JdbcOdbcDriver.acceptsURL(JdbcOdbcDriver.java:307) at java.sql.DriverManager.getDriver(DriverManager.java:253) at org.apache.tomcat.dbcp.dbcp.BasicDataSource.createDataSource(BasicDataSource.java:1143) ... 11 more Do you have any idea why Hibernate is not able to construct the session-factory? What is wrong in my configuration?

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  • #1146 - Table 'phpmyadmin.pma_recent' doesn't exist

    - by Mumin Ali
    Solution Guys... FYI i am using xampp to use phpmyadmin. and this error happens during the process of creating a database on localhost. Below is the code for config.inc file under phpmyadmin directory: <?php /* * This is needed for cookie based authentication to encrypt password in * cookie */ $cfg['blowfish_secret'] = 'xampp'; /* YOU SHOULD CHANGE THIS FOR A MORE SECURE COOKIE AUTH! */ /* * Servers configuration */ $i = 0; /* * First server */ $i++; /* Authentication type and info */ $cfg['Servers'][$i]['auth_type'] = 'HTTP'; $cfg['Servers'][$i]['user'] = 'root'; $cfg['Servers'][$i]['password'] = 'password'; $cfg['Servers'][$i]['extension'] = 'mysql'; $cfg['Servers'][$i]['AllowNoPassword'] = true; $cfg['Lang'] = ''; /* Bind to the localhost ipv4 address and tcp */ $cfg['Servers'][$i]['host'] = 'localhost'; $cfg['Servers'][$i]['connect_type'] = 'tcp'; /* User for advanced features */ $cfg['Servers'][$i]['controluser'] = 'pma'; $cfg['Servers'][$i]['controlpass'] = ''; /* Advanced phpMyAdmin features */ $cfg['Servers'][$i]['pmadb'] = 'phpmyadmin'; $cfg['Servers'][$i]['bookmarktable'] = 'pma_bookmark'; $cfg['Servers'][$i]['relation'] = 'pma_relation'; $cfg['Servers'][$i]['table_info'] = 'pma_table_info'; $cfg['Servers'][$i]['table_coords'] = 'pma_table_coords'; $cfg['Servers'][$i]['pdf_pages'] = 'pma_pdf_pages'; $cfg['Servers'][$i]['column_info'] = 'pma_column_info'; $cfg['Servers'][$i]['history'] = 'pma_history'; $cfg['Servers'][$i]['designer_coords'] = 'pma_designer_coords'; $cfg['Servers'][$i]['tracking'] = 'pma_tracking'; $cfg['Servers'][$i]['userconfig'] = 'pma_userconfig'; $cfg['Servers'][$i]['recent'] = 'pma_recent'; $cfg['Servers'][$i]['table_uiprefs'] = 'pma_table_uiprefs'; /* * End of servers configuration */ ?>

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  • AWS RDS Mysql with benstalk Hibernate app: Character encoding issue

    - by TeraTon
    I'm running a webapp from amazon rds with tomcat 7 and spring, which uses hibernate as the persistence layer. The application and utf-8 encoding work properly on localhost, but for some reason when I deploy to amazon, the UTF-8 encoding breaks. I use mysql 5.5.27 on amazon rds and the table that we wish to update has collation set to utf8 - utf8_unicode_ci And in hibernate I have set: < prop key="hibernate.connection.charSet"UTF-8 UTF-8 characters get replaced by ??? and this is of course especially bad for passwords and usernames + email as it basically kills them. Anyone else encountered character encoding breaking when deploying to amazon?

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  • hibernate transaction safety (with, without JBoss)

    - by Andy Nuss
    Hi, I am currently using just Hibernate and tomcat (no JBoss), and have hibernate transactions which I'm not clear on what transaction safety level I'm actually using, especially for those which read and get values and then update them). Thus I might be getting dirty reads? So I'm going to start having to study my transactions that require non-dirty reads, and make sure that (1) hibernate controls the transaction safety level of those transactions properly, and (2) be able to still have those transactions where dirty reads are ok. Do I need to install Hibernate with JBoss to control transaction safety levels? If so, what's the easiest way to do this without dramatically changing my application to use the J2EE apis, as I am currently using the basic Hibernate apis. Or better, can I get JTA control with Hibernate without using JBoss? Andy

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  • Windows 7 Hibernate Problem

    - by goygoycu
    I cannot hibernate windows. When I click "hibernate", my laptop(windows) just locks and the screen goes black. I can unlock without any problem. I do not have any problem with other options such as "sleep" or "shut down". I updated the chipset drivers but it did not help. There is not any option in BIOS about the sleep modes. "Hibernate" is "on" on Windows. Any advice? My Laptop specifications: MSI A5000 3gb system memory, Windows 7 Home Premium 32bit installed, Gentoo linux installed, Grub bootloader(MBR). Hard drive: Around 4gb of free space in windows partition.

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  • hibernate pagination mechanism

    - by haicnpmk44
    I am trying to use Hibernate pagination for my query (PostgreSQL ) i set setFirstResult(0), setMaxResults(20) for my sql query. My code like below: Session session = getSessionFactory().getCurrentSession(); session.beginTransaction(); Query query = session.createQuery("select id , customer_name , address from tbl_customers "); query.setFirstResult(0); query.setMaxResults(20); List<T> entities = query.list(); session.getTransaction().commit(); but when viewing SQL hibernate log, i still see full sql query: Hibernate: select customer0_.id as id9_, customer0_.customer_name as dst2_9_, customer0_.addres as dst3_9_ from tbl_customers customer0_ Why there is no LIMIT OFFSET in query of Hibernate pagination SQL log? Does anyone know about Hibernate pagination mechanism? I guess that Hibernate will select all data, put data into Resultset, and then paging in Resultset, right?

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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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  • XML Signature in a Web application

    - by OpenDevSoft
    Hi, We are developing an e-Banking web application for a small bank (up to 20000 clients/users). We have to implement digital signatures with X509 certificates (issued by CA on USB tokens) for signing payment information. We tried using CAPICOM but it seems that it is not working well with Windows Vista (have not tried it with Win 7). The other problem is that core banking system can process only Xml digital signatures, so we have to sign XML documents (not just a bulk-formatted text data like with CAPICOM and Win32 Crypto API). So my questions here are: Does anyone of you have similar problem and how did they solved it? Is there a plug-in, library, component or external tool (for Internet Explorer and/or Fire Fox) that supports XML Digital Signatures in a web application? Can you please recommend some of these products and write something about your experience with them? Thank you very much.

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  • parse Linq To Xml with attribute nodes

    - by Manoj
    I am having xml with following structure <ruleDefinition appId="3" customerId = "acf"> <node alias="element1" id="1" name="department"> <node alias="element2" id="101" name="mike" /> <node alias="element2" id="102" name="ricky" /> <node alias="element2" id="103" name="jim" /> </node> </ruleDefinition> Here nodes are differentiated using alias and not with node tag. As you can see top level node element1 has same node name "node" as element2. I want to parse this XML based on attribute alias. What should be the Linq-To-Xml code (using C#)to acheive this?

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  • XML deserialization doesn't read in second level

    - by Andy
    Sorry if the title doesn't make much sense, but I'm not too familiar with the terminology. My question is: I have a config.xml file for a program I'm working on. I created an xsd file by 'xsd.exe config.xml'. I then took this xsd and added it to the solution in visual studio. My last step used a program called xsd2code that turned that xsd file into a class I can serialize too. The problem is it doesn't read more then a layer deep in the xml tree. By this I mean the elements in the root node get deserialized into my object, but those that are in a node inside the root node are not. I found this out by putting a breakpoint after the deserialization and looking at my object. Any Ideas? Let me know if this needs some clarification or you need a snippet of something.

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  • Element Based XML Parsing

    - by demos
    I have an XML document which reads like this: <xml> <web:Web> <web:Total>4000</web:Total> <web:Offset>0</web:Offset> </web:Web> </xml> my question is how do I access them using a library like BeautifulSoup in python? xmlDom.web["Web"].Total ? does not work?

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