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  • Switch XML value for another in google maps

    - by JeffP
    all thanks for taking a look. Problem: Using Google maps I need to switch catid value to another value. How would I approach this and display it in a the window. E.G var catid = { 2: { newcatid: 'london' }, 4: { newcatid: 'newyork' } }; function load() { var map = new google.maps.Map(document.getElementById("map"), { center: new google.maps.LatLng(47.6145, -122.3418), zoom: 5, mapTypeId: 'roadmap' }); var infoWindow = new google.maps.InfoWindow; // Change this depending on the name of your PHP file downloadUrl("index_xml.php", function(data) { var xml = data.responseXML; var markers = xml.documentElement.getElementsByTagName("marker"); for (var i = 0; i < markers.length; i++) { var name = markers[i].getAttribute("title"); var address = markers[i].getAttribute("address"); var catid = markers[i].getAttribute("catid"); var point = new google.maps.LatLng( parseFloat(markers[i].getAttribute("latitude")), parseFloat(markers[i].getAttribute("langtitude"))); var html = "<b>" + name + "</b> <br/>" + address + "<br/>" + **catid** + "<br/> <a href='@'>Info</a>" ;

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  • C# XML Documentation Compiler Warning

    - by ImperialLion
    I am curious as to why I get a compiler warning in the following situation. /// <summary>This is class A /// </summary> public class A { /// <summary>This is the documentation for Method A /// </summary> public void MethodA() { //Do something } } /// <summary>This is class B /// </summary> public class B : A { /// <summary>This does something that I want to /// reference <see cref="MethodA"/> /// </summary> public void MethodB() { //Do something } } The warning states that "XML comment on 'B.MethodB()' has cref attribute 'MethodA' that could not be resolved." If B inherits from A shouldn't the compiler be able to see that method when generating the documentation without me specifying the parent class in the cref? If I change the cref to be cref="A.MethodA()" it works fine, but it seems like that's unnecessary and is a pain to do, especially if I have to go up more than one level. As a note to anyone testing this you have to be sure to "XML documentation file" checked in the Properties - Build in order to see the warning.

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  • Query decendants in XML to Linq

    - by Gordon
    I have the following xml data: <portfolio> <item> <title>Site</title> <description>Site.com is a </description> <url>http://www.site.com</url> <photos> <photo url="http://www.site.com/site/thumbnail.png" thumbnail="true" description="Main" /> <photo url="http://www.site.com/site/1.png" thumbnail="false" description="Main" /> </photos> </item> </portfolio> In c# I am using the following link query: List list = new List(); XDocument xmlDoc = XDocument.Load(HttpContext.Current.Server.MapPath("~/app_data/portfolio.xml")); list = (from portfolio in xmlDoc.Descendants("item") select new PortfolioItem() { Title = portfolio.Element("title").Value, Description = portfolio.Element("description").Value, Url = portfolio.Element("url").Value }).ToList(); How do I go about querying the photos node? In the PortfolioItem class I have a property: List Photos {get;set;} Any ideas would be greatly appreciated!

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  • Linq Return node level of hierarchical xml

    - by Ryan
    In a treeview you can retrieve the level of an item. I am trying to accomplish the same thing with the given input being an object. The XML data I will use for this example would be something like the following <?xml version="1.0" encoding="utf-8" ?> <Testing> <Numbers> <Number val="1"> <Number val="1.1"> <Number val="1.1.1"> <Number val="1.1.2" /> <Number val="1.1.3" /> <Number val="1.1.4" /> </Number> </Number> <Number val="1.2" /> <Number val="1.3" /> <Number val="1.4" /> </Number> <Number val="2" /> <Number val="3" /> <Number val="4" /> </Numbers> <Numbers> <Number val="5" /> <Number val="6" /> <Number val="7" /> <Number val="8" /> </Numbers> </Testing> This one is kicking my butt!

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  • Deleting items from datagrid (xml)

    - by Tonz
    Hello, I have a datagrid buttoncolumn which acts as delete buttons for my xml nodes. The elements are simply displayed in a boundcolumn, so there names get displayed. Each item generated gets a unique id (each time one is made id+++). My question his how can i remove a item (the entire element node with that certain id) when i click on one of the buttons in the bound column? <root> <element id="0"> <name>One</name> </element> <element id="1"> <name>Two</name> </element> </root> protected void dg_DeleteCommand(object sender, DataGridCommandEventArgs e) { XmlFunctions.Remove(index); }/*dg_DeleteCommand*/ (function on other class, where all my xml methods are written) public static void Remove(string index) { XmlDocument XMLDoc = new XmlDocument(); XMLDoc.Load(XMLFile); XPathNavigator nav = XMLDoc.CreateNavigator(); var node = nav.SelectSingleNode("/test/one[@id='" +???+ "']"); node.DeleteSelf(); XMLDoc.Save(XMLFile); } Edit: added datagrid <asp:View ID="viewDelete" runat="server"> <asp:DataGrid ID="dgDelete runat="server" AutoGenerateColumns="False" OnDeleteCommand="dg_DeleteCommand"> <Columns> <asp:BoundColumn DataField="name" HeaderText="names" /> <asp:ButtonColumn ButtonType="PushButton" Text="Delete" CommandName="Delete" ></asp:ButtonColumn> </Columns> </asp:DataGrid> </asp:View>

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  • Properly obsoleting old members in an XML Serializable class in C# VB .NET

    - by George
    Hi! Some time ago I defined a class that was serialized using XML. That class contained a serializable propertyA of integer type. Now I have extended and updated this class, whereby a new propertyB was added, whose type is another class that also has several serializable properties. The new propertyB is now supposed to play the role of propertyA, that is since type of propertyB is another class, one of its members would contain the value that previously propertyA contained, thus making peroptyA obsolete. What I am trying to figure out is how do I make sure that when I desireliaze the OLD version of this class (without propertyB in it), I make sure that the desreializer would take the value of propertyA from the old calss and set it as a value of one of the members of propertyB in a new class? Private WithEvents _Position As Position = New Position(Alignment.MiddleMiddle, 0, True, 0, True) Public Property Position() As Position 'NEW composite property that holds the value of the obsolted property, i.e. Alignment Get Return _Position End Get Set(ByVal value As Position) _Position = value End Set End Property Private _Alignment As Alignment = Alignment.MiddleMiddle <Xml.Serialization.XmlIgnore(), Obsolete("Use Position property instead.")> _ Public Property Alignment() As Alignment'The old, obsoleted property that I guess must be left for compliance with deserializing the old version of this class Get Return _Alignment End Get Set(ByVal value As Alignment) _Alignment = value End Set End Property Can you help me, please?

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  • Copy existing XML, duplicate element and modify

    - by Robert
    Hi, I have a tricky XSL problem at the moment. I need to copy the existing XML, copy a certain element (plus its child elements) and modify the value of two child-elements. The modifications are: divide value of the 'value' element by 110 and edit the value of the 'type' element from 'normal' to 'discount'. This is currently what I have: Current XML: <dataset> <data> <prices> <price> <value>50.00</value> <type>normal</type> </price> </prices> </data> </dataset> Expected result <dataset> <data> <prices> <price> <value>50.00</value> <type>normal</type> </price> <price> <value>45.00</value> <type>discount</type> </price> </prices> </data> </dataset> Any takers? I've gotten as far as copying the desired 'price' element using copy-of, but I'm stuck as to how to modify it next.

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  • Is it possible to create the following XML using Xdocument(C#3.0)

    - by Newbie
    <?xml version='1.0' encoding='UTF-8'?> <StockMarket> <StockDate Day = "02" Month="06" Year="2010"> <Stock> <Symbol>ABC</Symbol> <Amount>110.45</Amount> </Stock> <Stock> <Symbol>XYZ</Symbol> <Amount>366.25</Amount> </Stock> </StockDate> <StockDate Day = "03" Month="06" Year="2010"> <Stock> <Symbol>ABC</Symbol> <Amount>110.35</Amount> </Stock> <Stock> <Symbol>XYZ</Symbol> <Amount>369.70</Amount> </Stock> </StockDate> </StockMarket> My approach so far is XDocument doc = new XDocument( new XElement("StockMarket", new XElement("StockDate", new XAttribute("Day", "02"),new XAttribute("Month","06"),new XAttribute("Year","2010")), new XElement("Stock") ) ); Since I am new to Linq to XML, I am presently struggling a lot and henceforth seeking for help. Using C#3.0 . Thanks

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  • How to deal with the Hibernate hql multi-join query result in an Object-Oriented Way?

    - by EugeneP
    How to deal with the Hibernate hql multi-join query result in an Object-Oriented Way? As I see it returns a list of Objects. yes, it is tricky and only you who write the query know what should the query return (what objects). But are there ways to simplify things, so that it returned specific objects with no need in casting Object to a specific class according to its position in the query ? Maybe Spring can simplify things here? It has the similar functionality for JDBC, but I don't see if it can help in a similar way with Hibernate.

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  • out-of-the-box way to get an idmap from hibernate for a given entity?

    - by Geert-Jan
    Over and over again I notive myself getting a list from hibernate, and the first thing next is put it in an idmap like: List<House> entities = s.createCriteria(House.class).list(); Map<String,House> entitymap = new HashMap<String,House>(); for(TA_entity e:entities){ entitymap.put(e.getId(), e); } Is there a way to get this directly out of hibenerate? afterall Hibernate is familiar with the id.

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  • creating xml from database

    - by Prady
    hi, I am creating an xml from salesforce database, Everything works fine except when there is a & in the data which is been fetched. <apex:page contenttype="text/xml" > controller="Test2ab" > <data > wiki-section="Timeline"> <apex:repeat > value="{!lsttask}" var="e" > <event > start="{!e.ActivityDate}" title= > "{!e.Subject}"> <apex:outputText > value="{!e.Subject}" /> </event> > </apex:repeat> </data></apex:page> and in the controller i am just querying > lsttask =[Select OwnerId,WhoId,Status,Subject,ActivityDate from Task where Status = 'Completed' Order By ActivityDate Desc]; How can i use an escape for the value retrieved from the database Thanks Prady

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  • Admob in xml not showing in Linear

    - by NoobMe
    i am implementing admob on my app it appears when the parent is in relative layout but i must not use the alignparentbottom so i am changing it to linear but it doesnt show when i change it to linear.. any tips? help? thanks in advance here it is in xml: <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="match_parent" android:layout_height="match_parent" android:orientation="vertical" > <RelativeLayout android:id="@+id/banner_holder" android:layout_width="match_parent" android:layout_height="wrap_content" > <ImageView android:id="@+id/offline_banner" android:layout_width="match_parent" android:layout_height="wrap_content" android:layout_centerInParent="true" android:background="@color/black" android:src="@drawable/offline_banner" /> <com.google.ads.AdView xmlns:ads="http://schemas.android.com/apk/lib/com.google.ads" android:id="@+id/adView" android:layout_width="wrap_content" android:layout_height="wrap_content" android:layout_centerInParent="true" ads:adSize="SMART_BANNER" ads:adUnitId="@string/unit_id" ads:loadAdOnCreate="true" /> </RelativeLayout> <FrameLayout android:id="@+id/fragmentContainer" android:layout_width="match_parent" android:layout_height="wrap_content" /> </LinearLayout> i want the admob to be at the bottom part of the screen without using the alignparentbottom of relative layout thanks~

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  • A typical lifecycle of a Hibernate object in a web app - ?

    - by EugeneP
    Describe please a typical lifecycle of a Hibernate object (that maps to a db table) in a web app. Suppose, you create a new instance of an object and persist in the db. But during the app lifetime you'll be working on a detached object and finally you need to update it in the database, for example on exit. How does it look like with hibernate and spring? p.s. Can transactions and sessions live between servlet transitions? So that we opened 1 session and use it in all servlets without a need to reopen it?

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  • Cannot get list of elements using Linq to XML

    - by Blackator
    Sample XML: <CONFIGURATION> <Files> <File>D:\Test\TestFolder\TestFolder1\TestFile.txt</File> <File>D:\Test\TestFolder\TestFolder1\TestFile01.txt</File> <File>D:\Test\TestFolder\TestFolder1\TestFile02.txt</File> <File>D:\Test\TestFolder\TestFolder1\TestFile03.txt</File> <File>D:\Test\TestFolder\TestFolder1\TestFile04.txt</File> </Files> <SizeMB>3</SizeMB> <BackupLocation>D:\Log backups\File backups</BackupLocation> </CONFIGURATION> I've been doing some tutorials but I am unable to get all the list of file inside the files element. It only shows the first element and doesn't display the rest. This is my code: var fileFolders = from file in XDocument.Load(@"D:\Hello\backupconfig1.xml").Descendants("Files") select new { File = file.Element("File").Value }; foreach (var fileFolder in fileFolders) { Console.WriteLine("File = " + fileFolder.File); } How do I display all the File in the Files element, the SizeMB and BackupLocation? Thanks

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  • Retrieving XML node from a path specified in an attribute value of another node

    - by Olivier PAYEN
    From this XML source : <?xml version="1.0" encoding="utf-8" ?> <ROOT> <STRUCT> <COL order="1" nodeName="FOO/BAR" colName="Foo Bar" /> <COL order="2" nodeName="FIZZ" colName="Fizz" /> </STRUCT> <DATASET> <DATA> <FIZZ>testFizz</FIZZ> <FOO> <BAR>testBar</BAR> <LIB>testLib</LIB> </FOO> </DATA> <DATA> <FIZZ>testFizz2</FIZZ> <FOO> <BAR>testBar2</BAR> <LIB>testLib2</LIB> </FOO> </DATA> </DATASET> </ROOT> I want to generate this HTML : <html> <head> <title>Test</title> </head> <body> <table border="1"> <tr> <td>Foo Bar</td> <td>Fizz</td> </tr> <tr> <td>testBar</td> <td>testFizz</td> </tr> <tr> <td>testBar2</td> <td>testFizz2</td> </tr> </table> </body> </html> Here is the XSLT I currently have : <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"> <xsl:output method="html" indent="yes"/> <xsl:template match="/ROOT"> <html> <head> <title>Test</title> </head> <body> <table border="1"> <tr> <!--Generate the table header--> <xsl:apply-templates select="STRUCT/COL"> <xsl:sort data-type="number" select="@order"/> </xsl:apply-templates> </tr> <xsl:apply-templates select="DATASET/DATA" /> </table> </body> </html> </xsl:template> <xsl:template match="COL"> <!--Template for generating the table header--> <td> <xsl:value-of select="@colName"/> </td> </xsl:template> <xsl:template match="DATA"> <xsl:variable name="pos" select="position()" /> <tr> <xsl:for-each select="/ROOT/STRUCT/COL"> <xsl:sort data-type="number" select="@order"/> <xsl:variable name="elementName" select="@nodeName" /> <td> <xsl:value-of select="/ROOT/DATASET/DATA[$pos]/*[name() = $elementName]" /> </td> </xsl:for-each> </tr> </xsl:template> </xsl:stylesheet> It almost works, the problem I have is to retrieve the correct DATA node from the path specified in the "nodeName" attribute value of the STRUCT block.

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  • placing the matched 2 different child elements xml values in a single line from xslt2.0

    - by Girikumar Mathivanan
    I have the below input xml, <GSKProductHierarchy> <GlobalBusinessIdentifier>ZGB001</GlobalBusinessIdentifier> <Hierarchy> <Material>335165140779</Material> <Level1>02</Level1> <Level2>02AQ</Level2> <Level3>02AQ006</Level3> <Level4>02AQ006309</Level4> <Level5>02AQ006309</Level5> <Level6>02AQ006309</Level6> <Level7>02AQ006309</Level7> <Level8>02AQ006309</Level8> </Hierarchy> <Hierarchy> <Material>335165140780</Material> <Level1>02</Level1> <Level2>02AQ</Level2> <Level3>02AQ006</Level3> <Level4>02AQ006309</Level4> <Level5>02AQ006309</Level5> <Level6>02AQ006309</Level6> <Level7>02AQ006309</Level7> <Level8>02AQ006310</Level8> </Hierarchy> <Texts> <ProductHierarchy>02AQ006310</ProductHierarchy> <Language>A</Language> <Description>CREAM</Description> </Texts> <Texts> <ProductHierarchy>02AQ006309</ProductHierarchy> <Language>B</Language> <Description>CREAM</Description> </Texts> as per the requirement, xsl should check the matched value of GSKProductHierarchy/Hierarchy/Level8 in the GSKProductHierarchy/Texts/ProductHierarchy elements...and its should result as below flat file. 335165140779|02|02AQ|02AQ006|02AQ006309|02AQ006309|02AQ006309|02AQ006309|02AQ006309|02AQ006309|A|CREAM| 335165140780|02|02AQ|02AQ006|02AQ006309|02AQ006309|02AQ006309|02AQ006309|02AQ006310|02AQ006310|B|CREAM| Right now I have the below xslt, <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:exsl="http://exslt.org/common" xmlns:set="http://exslt.org/sets" xmlns:str="http://exslt.org/strings" xmlns:java="http://xml.apache.org/xslt/java" xmlns:saxon="http://saxon.sf.net/" exclude-result-prefixes="exsl set str java saxon"> <xsl:output method="text" indent="yes"/> <xsl:variable name="VarPipe" select="'|'"/> <xsl:variable name="VarBreak" select="'&#xa;'"/> <xsl:template match="/"> <xsl:for-each select="GSKProductHierarchy/Hierarchy"> <xsl:variable name="currentIndex" select="position()"/> <xsl:variable name="Level8" select="Level8"/> <xsl:variable name="ProductHierarchy" select="../Texts[$currentIndex]/ProductHierarchy"/> <xsl:if test="$Level8=$ProductHierarchy"> <xsl:value-of select="Material"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level1"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level2"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level3"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level4"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level5"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level6"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level7"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level8"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="../Texts[$currentIndex]/ProductHierarchy"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="../Texts[$currentIndex]/Language"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="../Texts[$currentIndex]/Description"/> <xsl:value-of select="$VarPipe"/> <xsl:if test="not(position() = last())"> <xsl:value-of select="$VarBreak"/> </xsl:if> </xsl:if> </xsl:for-each> </xsl:template> can anyone please suggest what function should i need to use to get the desired result. Regards, Giri

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  • Can I make Windows to open Excel XML files with Excel without opening Explorer?

    - by Sorin Sbarnea
    I want to be able to open Excel XML files in Excel but without assigning XML directly to Excel. There are lots of XML files that are not Excel files and I don't want to open all of them in Excel. The file has proper header for opening in Excel but currently it does open Internet Explorer that asks me if I want to open the file with Excel, save or cancel. I just want to open it without two another annoying windows.

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  • What is a good XML Editor for Mac OS X?

    - by g.
    I am looking for a good lightweight XML viewer/editor for Mac OS X. It would only be for occasional use, so free options are preferable though paid options aren't out of the question. It would be used primarily for reviewing and making small changes to XML files and would require the following basic features. easily create a new file from clipboard (copy/paste) re-format (pretty print) xml syntax highlighting validation find

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  • Windows 7 unidentified network (or limited access) after hibernate.

    - by null
    My windows 7 network will show limited access or unidentified network after coming up from hibernation. In the office I normally use LAN connection, I turn-off my wireless card (DELL Latitude has on/off switch for the wireless card). When I back at home I will turn on the wireless card, but it will take about 15 seconds to detect my home WIFI and then show limited access. I will have to restart the notebook and it will be able to connect to my WIFI and internet. The problem will be solved if I restart the notebook, but that defeats the purpose of hibernation doesn't it? I have tried uninstalling the wireless card driver but still does not solve it. I also tried updating my network card driver but windows says I am using the latest driver. On support.dell.com also showing I am using the latest driver.

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  • XML-RPC with PHP and GoDaddy? Confusion sets in.

    - by Chris Cooper
    Hey folks, I am attempting to work with XML-RPC via PHP on a GoDaddy server. This same server is hosting a Wordpress Blog that makes use of XML-RPC and is functioning, though that may be unrelated... Whenever I attempt to use any functions that are integrated into PHP for use with XML-RPC, I get an error (function list here: http://us3.php.net/manual/en/ref.xmlrpc.php) e.g.: Fatal error: Class 'xmlrpc_client' not found Is this because XML-RPC's PHP functions are not enabled on my server? If so, how do I go about enabling those - it would seem I would have to install the XML-RPC library to do so and of course I cannot do that on a shared server. Doesn't Wordpress use the same batch of XML-RPC functions though (it works fine)? I think I have managed to thoroughly confuse myself. I have zero experience with XML-RPC.

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  • What is your prefered way to return XML from an ActionMethod in Asp.net MVC?

    - by serbrech
    I am displaying charts that load the data asynchronously because the searches are the work to fetch the data is quite heavy. The data has to be return as XML to make the chart library happy. My ActionMethods return a ContentResult with the type set as text/xml. I build my Xml using Linq to XML and call ToString. This works fine but it's not ideal to test. I have another idea to achieve this which would be to return a view that builds my XML using the XSLT View engine. I am curious and I always try to do the things "the right way". So how are you guys handling such scenarios? Do you implement a different ViewEngine (like xslt) to build your XML or do you Build your XML inside your controller (Or the service that serves your controller)?

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  • I am using jquery ajax using text() but any html written in the XML doesn't render?

    - by Xtian
    So I have Jquery Ajax working real nice, but an issue I am having is in my XML if I want to bold a work or Italicize a sentence, if I do it in the XML using HTML tags it will not show up. I am pretty sure it is due to using the .text(). Any suggestions on a work around for this? $(document).ready(function(){ $.ajax({ type: "GET", url: "xml/sites.xml", dataType: "xml", success: function(xml) { $(xml).find('site').each(function(){ $(this).find('desc').each(function(){ var brief = $(this).find('brief').text(); var long = $(this).find('long').text(); var url = $(this).find('url').text(); $('<div class="brief"></div>').html(brief).appendTo('#link_'+id);

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  • Delete all previous records and insert new ones

    - by carlos
    When updating an employee with id = 1 for example, what is the best way to delete all previous records in the table certificate for this employee_id and insert the new ones?. create table EMPLOYEE ( id INT NOT NULL auto_increment, first_name VARCHAR(20) default NULL, last_name VARCHAR(20) default NULL, salary INT default NULL, PRIMARY KEY (id) ); create table CERTIFICATE ( id INT NOT NULL auto_increment, certificate_name VARCHAR(30) default NULL, employee_id INT default NULL, PRIMARY KEY (id) ); Hibernate mapping <?xml version="1.0" encoding="utf-8"?> <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD//EN" "http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd"> <hibernate-mapping> <class name="Employee" table="EMPLOYEE"> <id name="id" type="int" column="id"> <generator class="sequence"> <param name="sequence">employee_seq</param> </generator> </id> <set name="certificates" lazy="false" cascade="all"> <key column="employee_id" not-null="true"/> <one-to-many class="Certificate"/> </set> <property name="firstName" column="first_name"/> <property name="lastName" column="last_name"/> <property name="salary" column="salary"/> </class> <class name="Certificate" table="CERTIFICATE"> <id name="id" type="int" column="id"> <param name="sequence">certificate_seq</param> </id> <property name="employee_id" column="employee_id" insert="false" update="false"/> <property name="name" column="certificate_name"/> </class> </hibernate-mapping>

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  • Is there a way to use a dictionary or xml in the Application Settings?

    - by Shimmy
    I have to store a complex type in the application settings. I thought that storing it as XML would work best. The problem is I don't know how store XML. I prefer to store it as a managed XML rather than using just a string of raw XML having to parse it on each access. I managed to set the Type column of the setting to XDocument, but I was unable to set its value. Is there a way to use XDocument or XML in application settings? Update I found a way, simply by editing the .settings file with the xml editor. I changed it to a custom serializable dictionary, but I get the following error when I try to access the setting-property (I set it to a serialized representation of the default value). The property 'Setting' could not be created from it's default value. Error message: There is an error in XML document (1, 41). Any ideas will be appreciated.

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