Search Results

Search found 15718 results on 629 pages for 'for xml'.

Page 37/629 | < Previous Page | 33 34 35 36 37 38 39 40 41 42 43 44  | Next Page >

  • Spring validation has error in XML document from ServletContext resource

    - by user1441404
    I applied spring validation in my registration page .but the follwing error are shown in my server log of my app engine server. javax.servlet.UnavailableException: org.springframework.beans.factory.xml.XmlBeanDefinitionStoreException: Line 22 in XML document from ServletContext resource [/WEB-INF/spring/appServlet/servlet-context.xml] is invalid; nested exception is org.xml.sax.SAXParseException; lineNumber: 22; columnNumber: 30; cvc-complex-type.2.4.c: The matching wildcard is strict, but no declaration can be found for element 'property'. My code is given below : <?xml version="1.0" encoding="UTF-8"?> <beans:beans xmlns="http://www.springframework.org/schema/mvc" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:beans="http://www.springframework.org/schema/beans" xmlns:context="http://www.springframework.org/schema/context" xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.0.xsd http://www.springframework.org/schema/jee http://www.springframework.org/schema/jee/spring-jee-3.0.xsd > <beans:bean name="/register" class="com.my.registration.NewUserRegistration"> <property name="validator"> <bean class="com.my.validation.UserValidator" /> </property> <beans:property name="formView" value="newuser"></beans:property> <beans:property name="successView" value="home"></beans:property> </beans:bean> <beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver"> <beans:property name="prefix" value="/WEB-INF/views/" /> <beans:property name="suffix" value=".jsp" /> </beans:bean> </beans:beans>

    Read the article

  • How to convert Xml files to Text Files

    - by John
    Hi all, I have around 8000 xml files that needs to be converted into text files. The text file must contain title, description and keywords of the xml file without the tags and removing other elements and attributes as well. In other words, i need to create 8000 text files containing the title,description and keywords of the xml file. I need codings for this to be done systematically. Any help would be greatly appreciated. Thanks in advance.

    Read the article

  • PHP File System or XML : Security Issue

    - by jasmine
    I want to make a news portal(php) site with minimum mysql force. :create a cron, fetch data from mysql and write to a php file . (I dont know is it right way) But Can I use xml instead of php file? Write mysql data to xml. Is this a secure way? What is the best way? XML or php file? Thanks in advance

    Read the article

  • How to make a Stored Procedure that takes in XML and uses that xml as an Update + call this stored p

    - by chobo2
    Hi I am using ms sql server 2005 and I want to do a mass update. I am thinking that I might be able to do it with sending an xml document to a stored procedure. So I seen many examples on how to do it for insert CREATE PROCEDURE [dbo].[spTEST_InsertXMLTEST_TEST](@UpdatedProdData XML) AS INSERT INTO dbo.UserTable(CreateDate) SELECT @UpdatedProdData.value('(/ArrayOfUserTable/UserTable/CreateDate)[1]', 'DATETIME') But I am not sure how it would look like for an update. I am also unsure how do I pass in the xml through ado.net? Do I pass it as a string through a parameter or what? I know sqlDataApater has a batch update method but I am using linq to sql. So I rather keep using it. So if this works I would be able to grab all records with linq to sql and have them as objects. Then manipulate the objects and use xml seralization. Finally I could just use ado.net simple to send the xml to the server. This might be slower then the sqlDataAdapter but I am willing to take that hit if I can keep using objects.

    Read the article

  • Why does "xsd:date" of XML Schema Type mapped "javax.xml.datatype.XMLGregorianCalendar" When Schema-to-Java Mapping of JAXB does.

    - by Take
    I don't know why does "xsd:date" of XML Schema Type mapped "javax.xml.datatype.XMLGregorianCalendar" When Schema-to-Java Mapping of JAXB does. Why does "xsd:date" of XML Schema Type mapped "java.util.Date" ? I guess that JAXB intentionally does its mapping. I want to know that reason if any. And if exists it, how to change "xsd:date" of XML Schema Type to "java.util.Date" of Java class without using annotation(ex.@XmlJavaTypeAdapter). I want to do mashalling and unmarshalling without all annotations.

    Read the article

  • how to get xml html after transpose and databind()

    - by rlee923
    Hi. I have some code that uses xsl and xml. The Xml control is on the design page. The xml control id is xmlApplication The xmlstring is generated and xsl has the format with all the tables and cells etc. Here is a part of thecode of a page which generates the final product which shows the xml in a certain format. xmlApplication.Document = xmlDoc; xmlApplication.Transform = transApp; xmlApplication.DataBind(); I am guessing after xmlApplication.Databind(), xmlApplication will be converted into something that can be put inside . Is it possible to grab as a string? Please let me know if I have a wrong idea abut this. Thanks a lot.

    Read the article

  • libxml2 or NSXMLDocument for create a XML

    - by zp26
    Hi, I have a question for my iPhone app. I looking for a documentation for libxml2. I have find more example for reading by XML, but my program must create a XML and i don't find anything (example or tutorial) for this. I can create a XML with libxml2 or i must use NSXMLDocument? Can you take me a example? Thanks so much.

    Read the article

  • Inheriting XML files and modifying values

    - by Veehmot
    This is a question about concept. I have an XML file, let's call it base: <base id="default"> <tags> <tag>tag_one</tag> <tag>tag_two</tag> <tag>tag_three</tag> </tags> <data> <data_a>blue</data_a> <data_b>3</data_b> </data> </base> What I want to do is to be able to extend this XML in another file, modifying individual properties. For example, I want to inherit that file and make a new one with a different data/data_a node: <base id="green" import="default"> <data> <data_a>green</data_a> </data> </base> So far it's pretty simple, it replaces the old data/data_a with the new one. I even can add a new node: <base id="ext" import="default"> <moredata> <data>extended version</data> </moredata> </base> And still it's pretty simple. The problem comes when I want to delete a node or deal with XML Lists (like the tags node). How should I reference a particular index on a list? I was thinking doing something like: <base id="diffList" import="default"> <tags> <tag index="1">this is not anymore tag_two</tag> </tags> </base> And for deleting a node / array index: <base id="deleting" import="default"> <tags> <tag index="2"/> </tags> <data/> </base> <!-- This will result in an XML containing these values: --> <base> <tag>tag_one</tag> <tag>tag_two</tag> </base> But I'm not happy with my solutions. I don't know anything about XSLT or other XML transformation tools, but I think someone must have done this before. The key goal I'm looking for is ease to write the XML by hand (both the base and the "extended"). I'm open to new solutions besides XML, if they are easy to write manually. Thanks for reading.

    Read the article

  • Comparing XML in SSRS

    - by silves89
    I'm new to SSRS. We'll have two slightly different chunks of XML in a single row of an SQL Server database table. In an SSRS report we'll want to show only the differences between the XML chunks. I don't know how to do this, but I suspect the XML Type in SQLServer 2005 might be useful, or XSLT transformations in SSRS. Could anyone point me in the right direction?

    Read the article

  • php loop xml data with xsd schema - how do get the data

    - by miholzi
    i try to geht the data from a xml file and i have troubles to get data, for example, how can i get the caaml:locRef value or the caaml:beginPosition value? here is the code so far: /* a big thank you to helderdarocha */ /* – he already helped me yesterday with a part of this code */ $doc = new DOMDocument(); $doc->load('xml/test.xml'); $xpath = new DOMXpath($doc); $xpath->registerNamespace("caaml", "http://caaml.org/Schemas/V5.0/Profiles/BulletinEAWS"); if ($doc->schemaValidate('http://caaml.org/Schemas/V5.0/Profiles/BulletinEAWS/CAAMLv5_BulletinEAWS.xsd')) { foreach ($xpath->query('//caaml:DangerRating') as $key) { echo $key->nodeValue; print_r($key); } } and here ist the print_r from $key DOMElement Object ( [tagName] => caaml:DangerRating [schemaTypeInfo] => [nodeName] => caaml:DangerRating [nodeValue] => 2014-03-03+01:00 2 [nodeType] => 1 [parentNode] => (object value omitted) [childNodes] => (object value omitted) [firstChild] => (object value omitted) [lastChild] => (object value omitted) [previousSibling] => (object value omitted) [nextSibling] => (object value omitted) [attributes] => (object value omitted) [ownerDocument] => (object value omitted) [namespaceURI] => http://caaml.org/Schemas/V5.0/Profiles/BulletinEAWS [prefix] => caaml [localName] => DangerRating [baseURI] => /Applications/MAMP/htdocs/lola/xml/test.xml [textContent] => 2014-03-03+01:00 2 ) 2014-03-04+01:00 2 and here a part of the xml <caaml:DangerRating> <caaml:locRef xlink:href="AT7R1"/> <caaml:validTime> <caaml:TimePeriod> <caaml:beginPosition>2014-03-06T00:00:00+01:00</caaml:beginPosition> <caaml:endPosition>2014-03-06T11:59:59+01:00</caaml:endPosition> </caaml:TimePeriod> </caaml:validTime> <caaml:validElevation> <caaml:ElevationRange uom="m"> <caaml:beginPosition>2200</caaml:beginPosition> </caaml:ElevationRange> </caaml:validElevation> <caaml:mainValue>2</caaml:mainValue> </caaml:DangerRating> <caaml:DangerRating> <caaml:locRef xlink:href="AT7R1"/> <caaml:validTime> <caaml:TimePeriod> thanks!

    Read the article

  • How to assign XML attribute values to drop down list using XSL

    - by Vijay
    Hi, I have a sample xml as; <?xml version="1.0" encoding="iso-8859-9"?> <DropDownControl id="dd1" name="ShowValues" choices="choice1,choice2,choice3,choice4"> </DropDownControl > I need to create a UI representation of this XML using XSL. I want to fill the drop down list with values specified in choices attribute. Does anyone have any idea about this ? Thanks in advance :)

    Read the article

  • Loading xml with encoding UTF 16 using XDocument

    - by Sangram
    Hi, I am trying to read the xml document using XDocument method . but i am getting an error when xml has <?xml version="1.0" encoding="utf-16"?> When i removed encoding manually.It works perfectly. I am getting error " There is no Unicode byte order mark. Cannot switch to Unicode. " i tried searching and i landed up here-- Why does C# XmlDocument.LoadXml(string) fail when an XML header is included? But could not solve my problem. My code : XDocument xdoc = XDocument.Load(path); Any suggestions ?? thank you.

    Read the article

  • [java] how to parse XML document?

    - by user32167
    I have xml document in variable (not in file). How can i get data storaged in that? I don't have any additional file with that, i have it 'inside' my sourcecode. When i use DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance(); DocumentBuilder db = dbf.newDocumentBuilder(); Document doc = db.parse(XML); (XML is my xml variable), i get an error java.io.FileNotFoundException: C:\netbeans\app-s7013\<network ip_addr="10.0.0.0\8" save_ip="true"> File not found.

    Read the article

  • Getting 404 in Android app while trying to get xml from localhost

    - by Patrick
    This must be something really stupid, trying to solve this issue for a couple of days now and it's really not working. I searched everywhere and there probably is someone with the same problem, but I can't seem to find it. I'm working on an Android app and this app pulls some xml from a website. Since this website is down a lot, I decided to save it and run it locally. Now what I did; - I downloaded the kWs app for hosting the downloaded xml file - I put the file in the right directory and could access it through the mobile browser, but not with my app (same code as I used with pulling it from some other website, not hosted by me, only difference was the URL obviously) So I tried to host it on my PC and access it with my app from there. Again the same results, the mobile browsers had no problem finding it, but the app kept saying 404 Not Found: "The requested URL /test.xml&parama=Someone&paramb= was not found on this server." Note: Don't mind the 2 parameters I am sending, I needed that to get the right stuff from the website that wasn't hosted by me. My code: public String getStuff(String name){ String URL = "http://10.0.0.8/test.xml"; ArrayList<NameValuePair> params = new ArrayList<NameValuePair>(2); params.add(new BasicNameValuePair("parama", name)); params.add(new BasicNameValuePair("paramb", "")); APIRequest request = new APIRequest(URL, params); try { RequestXML rxml = new RequestXML(); AsyncTask<APIRequest, Void, String> a = rxml.execute(request); ... } catch(Exception e) { e.printStackTrace(); } return null; } That should be working correctly. Now the RequestXML class part: class RequestXML extends AsyncTask<APIRequest, Void, String>{ @Override protected String doInBackground(APIRequest... uri) { HttpClient httpclient = new DefaultHttpClient(); String completeUrl = uri[0].url; // ... Add parameters to URL ... HttpGet request = null; try { request = new HttpGet(new URI(completeUrl)); } catch (URISyntaxException e1) { e1.printStackTrace(); } HttpResponse response; String responseString = ""; try { response = httpclient.execute(request); StatusLine statusLine = response.getStatusLine(); if(statusLine.getStatusCode() == HttpStatus.SC_OK){ // .. It crashes here, because statusLine.getStatusCode() returns a 404 instead of a 200. The xml is just plain xml, nothing special about it. I changed the contents of the .htaccess file into "ALLOW FROM ALL" (works, cause the browser on my mobile device can access it and shows the correct xml). I am running Android 4.0.4 and I am using the default browser AND chrome on my mobile device. I am using MoWeS to host the website on my PC. Any help would be appreciated and if you need to know anything before you can find an answer to this problem, I'll be more than happy to give you that info. Thank you for you time! Cheers.

    Read the article

  • WebMethod Return xml

    - by BabelFish
    I keep reading how everyone states to return XmlDocument when you want to return XML. Is there a way to return raw XML as a string? I have used many web services (written by others) that return string and the string contains XML. If you return XmlDocument how is that method consumed by users that are not on .Net? What is the method to just return the raw XML as string without it having being wrapped with <string></string> Thanks!

    Read the article

  • JAVA: XML parsers gives null element

    - by Johan
    When I try to parse a XML-file, it gives sometimes a null element by the title. I think it has to do with HTML-tags &#039; How can I solve this problem? I have the follow XML-file: <item> <title>&#039; Nieuwe DVD &#039;</title> <description>tekst, tekst tekst</description> <link>dvd.html</link> <category>nieuws</category> <pubDate>Sat, 1 Jan 2011 9:24:00 +0000</pubDate> </item> And the follow code to parse the xml-file: //DocumentBuilderFactory, DocumentBuilder are used for //xml parsing DocumentBuilderFactory dbf = DocumentBuilderFactory .newInstance(); DocumentBuilder db = dbf.newDocumentBuilder(); //using db (Document Builder) parse xml data and assign //it to Element Document document = db.parse(is); Element element = document.getDocumentElement(); //take rss nodes to NodeList element.normalize(); NodeList nodeList = element.getElementsByTagName("item"); if (nodeList.getLength() > 0) { for (int i = 0; i < nodeList.getLength(); i++) { //take each entry (corresponds to <item></item> tags in //xml data Element entry = (Element) nodeList.item(i); entry.normalize(); Element _titleE = (Element) entry.getElementsByTagName( "title").item(0); Element _categoryE = (Element) entry .getElementsByTagName("category").item(0); Element _pubDateE = (Element) entry .getElementsByTagName("pubDate").item(0); Element _linkE = (Element) entry.getElementsByTagName( "link").item(0); String _title = _titleE.getFirstChild().getNodeValue(); String _category = _categoryE.getFirstChild().getNodeValue(); Date _pubDate = new Date(_pubDateE.getFirstChild().getNodeValue()); String _link = _linkE.getFirstChild().getNodeValue(); //create RssItemObject and add it to the ArrayList RssItem rssItem = new RssItem(_title, _category, _pubDate, _link); rssItems.add(rssItem); conn.disconnect(); }

    Read the article

  • Querying XML using node numbers

    - by CP
    Okay, so I'm writing a utility that compares 2 XML documents using Microsoft's XML diff patch tool. The result looks something like this: <?xml version="1.0" encoding="utf-16"?><xd:xmldiff version="1.0" srcDocHash="10728157883908851288" options="IgnoreChildOrder IgnoreComments IgnoreWhitespace " fragments="yes" xmlns:xd="http://schemas.microsoft.com/xmltools/2002/xmldiff"><xd:node match="1"><xd:node match="1"><xd:node match="1"><xd:node match="2"><xd:node match="1"><xd:node match="1"><xd:node match="2"><xd:change match="1">testi22n2123</xd:change></xd:node></xd:node><xd:add match="/1/1/1/2/1/8" opid="1" /><xd:node match="7"><xd:node match="1"><xd:change match="1">31</xd:change></xd:node><xd:node match="2"><xd:change match="1">test2ing</xd:change></xd:node></xd:node><xd:remove match="8" opid="1" /></xd:node></xd:node></xd:node></xd:node></xd:node><xd:descriptor opid="1" type="move" /></xd:xmldiff> What I'm trying to do is go back into the source document and get the source data that represents the difference. I initially tried creating an Xpath query, but as I understand it now this XmlDiff thing works off the DOM... which seems like the dinosaur of XML objects these days. What's the best way to get at the node in the source XML by using the numbers provided in the diff result?

    Read the article

  • How to ignore the validation of Unknown tags ?

    - by infant programmer
    One more challenge to the XSD capability,I have been sending XML files by my clients, which will be having 0 or more undefined or [call] unexpected tags (May appear in hierarchy). Well they are redundant tags for me .. so I have got to ignore their presence, but along with them there are some set of tags which are required to be validated. This is a sample XML: <root> <undefined_1>one</undefined_1> <undefined_2>two</undefined_2> <node>to_be_validated</node> <undefined_3>two</undefined_3> <undefined_4>two</undefined_4> </root> And the XSD I tried with: <xs:element name="root" type="root"></xs:element> <xs:complexType name="root"> <xs:sequence> <xs:any maxOccurs="2" minOccurs="0"/> <xs:element name="node" type="xs:string"/> <xs:any maxOccurs="2" minOccurs="0"/> </xs:sequence> </xs:complexType XSD doesn't allow this, due to certain reasons. The above mentioned example is just a sample. The practical XML comes with the complex hierarchy of XML tags .. Kindly let me know if you can get a hack of it. By the way, The alternative solution is to insert XSL-transformation, before validation process. Well, I am avoiding it because I need to change the .Net code which triggers validation process, which is supported at the least by my company.

    Read the article

  • Problems using Custom Layout in XML

    - by Kevin
    I've created a new GridLayout class that I want to use in an XML File. The class is in another project. I've created the attrs.xml file in the other project with my properties but when the constructor gets called with the AttributeSet, none of the values are set. In my xml for my screen layout, I refer to the layout with the following: xmlns:gridLayout="@com.mastertechsoftware.AndroidUtil:http://schemas.android.com/apk/res/com.mastertechsoftware.AndroidUtil" Not sure if that is right. I have the attrs.xml file being compiled to R.java in com/mastertechsoftware/AndroidUtil. All the right styleable values are there. In my constructor, I use: TypedArray a = c .obtainStyledAttributes(attrs, R.styleable.GridLayout); this.numRows = a.getInt(R.styleable.GridLayout_numRows, -1); Nothing comes back (-1 does but that means nothing is there) If I use int n = a.getIndexCount(); I get 0.

    Read the article

  • Write Mysql tables to XML : Security Issue

    - by jasmine
    I want to make a news portal(php) with minimum mysql force. :create a cron, fetch data from mysql and write to a php file . (I dont know is it right way) But Can I use xml instead of php file? Write mysql data to xml. Is this a secure way? What is the best way? XML or php file? Thanks in advance

    Read the article

  • Nokogiri extract data from xml

    - by Awea
    Hi guys, i try to extract data from a xml in rails application with the Nokogiri gem, the xml : <item> <description> <img src="something" title="anothething"> <p>text, bla bla...</p> </description> </item> Actually i do something like this to extract data from the xml : def test_content @return = Array.new site = 'http://www.les-encens.com/modules/feeder/rss.php?id_category=0' @doc = Nokogiri::XML(open(site, "UserAgent" => "Ruby-OpenURI")) @doc.xpath("//item").each do |n| @return << [ n.xpath('description') ] end end Could you show me how extract just the src attribute from the img tag ?

    Read the article

  • Help me choose between XML or SQL Lite on android

    - by Ngetha
    I have an android app that periodically, say once a week downloads content from a server in XML. The content is used by the app, different Acitivities use different parts of the content. My question is a design one, should I save the data in SQlite or just keep it as an XML file, which one would be faster to read? The app can only use one content piece at a time, which means subsequent XML content downloads replace the old one.

    Read the article

  • Serializing a class containing a custom class

    - by Netfangled
    I want to serialize an object as xml that contains other custom classes. From what I understand (I've been reading MSDN and SO mostly), the XmlSerializer doesn't take this into account. This is the line that's confusing me: XML serialization serializes only the public fields and property values of an object into an XML stream. XML serialization does not include type information. For example, if you have a Book object that exists in the Library namespace, there is no guarantee that it will be deserialized into an object of the same type. Taken from MSDN, here For example, I want to serialize an object of type Order, but it contains a list of Products, and each one contains an object of type Category: class Order { List<Product> products; } class Product { Category type; } class Category { string name; string description; } And I want my Order object to be serialized like so: <Order> <Product> <Category Name=""> <Description></Description> </Category> </Product> <Product> <Category Name=""> <Description></Description> </Category> </Product> <Order> Does the XmlSerializer already do this? If not, is there another class that does or do I have to define the serialization process myself?

    Read the article

< Previous Page | 33 34 35 36 37 38 39 40 41 42 43 44  | Next Page >