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  • not able to Deserialize object

    - by Ravisha
    I am having following peice of code ,where in i am trying to serialize and deserailize object of StringResource class. Please note Resource1.stringXml = its coming from resource file.If i pass strelemet.outerXMl i get the object from Deserialize object ,but if i pass Resource1.stringXml i am getting following exception {"< STRING xmlns='' was not expected."} System.Exception {System.InvalidOperationException} class Program { static void Main(string[] args) { StringResource str = new StringResource(); str.DELETE = "CanDelete"; str.ID= "23342"; XmlElement strelemet = SerializeObjectToXmlNode (str); StringResource strResourceObject = DeSerializeXmlNodeToObject<StringResource>(Resource1.stringXml); Console.ReadLine(); } public static T DeSerializeXmlNodeToObject<T>(string objectNodeOuterXml) { try { TextReader objStringsTextReader = new StringReader(objectNodeOuterXml); XmlSerializer stringResourceSerializer = new XmlSerializer(typeof(T),string.Empty); return (T)stringResourceSerializer.Deserialize(objStringsTextReader); } catch (Exception excep) { return default(T); } } public static XmlElement SerializeObjectToXmlNode(object obj) { using (MemoryStream memoryStream = new MemoryStream()) { try { XmlSerializerNamespaces xmlNameSpace = new XmlSerializerNamespaces(); xmlNameSpace.Add(string.Empty, string.Empty); XmlWriterSettings writerSettings = new XmlWriterSettings(); writerSettings.CloseOutput = false; writerSettings.Encoding = System.Text.Encoding.UTF8; writerSettings.Indent = false; writerSettings.OmitXmlDeclaration = true; XmlWriter writer = XmlWriter.Create(memoryStream, writerSettings); XmlSerializer xmlserializer = new XmlSerializer(obj.GetType()); xmlserializer.Serialize(writer, obj, xmlNameSpace); writer.Close(); memoryStream.Position = 0; XmlDocument serializeObjectDoc = new XmlDocument(); serializeObjectDoc.Load(memoryStream); return serializeObjectDoc.DocumentElement; } catch (Exception excep) { return null; } } } } public class StringResource { [XmlAttribute] public string DELETE; [XmlAttribute] public string ID; }

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  • Is there a clean way to determine RSS version?

    - by Fenick
    Is it possible to determine the feed type and version in a way so that you can make sure that you have the correct version. At the lowest level so to speak. Namespaces is an obvious approach, but its not present for a lot of feeds. Any thoughts? (I'm trying to mashup varioius RSS Feeds). Thanks in advance for any help!

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  • How to add line breaks in RSS feeds?

    - by Candidasa
    I'm building my own custom RSS feed in PHP. I want the tag to contain line breaks to make the text more readable. However, I can't seem to figure out how to do it correctly. No matter what I try some RSS reader interprets it incorrectly. Is there some standard best way to add a line-break in and RSS 2.0 feed? I have tried "\n", which works in NetNewsWire on the Mac, but gets ignored by the built-in Safari browser's RSS reader. I have tried <br />, which works in the Safari RSS reader, but results in all the text after the being cut off in NetNewsWire.

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  • How to get a html elements with python lxml

    - by Damiano
    Hello! I have this html code: <table> <tr> <td class="test"><b><a href="">aaa</a></b></td> <td class="test">bbb</td> <td class="test">ccc</td> <td class="test"><small>ddd</small></td> </tr> <tr> <td class="test"><b><a href="">eee</a></b></td> <td class="test">fff</td> <td class="test">ggg</td> <td class="test"><small>hhh</small></td> </tr> </table> I use this Python code to extract all <td class="test"> with lxml module. import urllib2 import lxml.html code = urllib.urlopen("http://www.example.com/page.html").read() html = lxml.html.fromstring(code) result = html.xpath('//td[@class="test"][position() = 1 or position() = 4]') It works good! The result is: <td class="test"><b><a href="">aaa</a></b></td> <td class="test"><small>ddd</small></td> <td class="test"><b><a href="">eee</a></b></td> <td class="test"><small>hhh</small></td> (so the first and the fourth column of each <tr>) Now, I have to extract: aaa (the title of the link) ddd (text between <small> tag) eee (the title of the link) hhh (text between <small> tag) How could I extract these values? (the problem is that I have to remove <b> tag and get the title of the anchor on the first column and remove <small> tag on the forth column) Thank you!

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  • Why doesn't XmlSerializer support Dictionary?

    - by theburningmonk
    Just curious as to why Dictionary is not supported by XmlSerializer? You can get around it easily enough by using DataContractSerializer and writing the object to a XmlTextWriter, but what are the characteristics of a Dictionary that makes it difficult for a XmlSerializer to deal with considering it's really an array of KeyValuePairs. In fact, you can pass an IDictionary<TKey, TItem> to a method expecting an IEnumerable<KeyValuePairs<TKey, ITem>>.

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  • How to register XHTML namespaces??

    - by John Tyler
    My page use to validate for XHTML but I added the addthis buttons to page and it gives a new namespace: E.G.: <a addthis:url="http://domain.tld/path/to/stuff" addthis:title="Teh Title here"> I tried: <html xmlns="http://www.w3.org/1999/xhtml" xmlns:addthis="http://addthis.com/" lang="en"> Doesn't work. Can we register namespaces for the validator??? I AM EMOTIONALLY ATTACHED TO VALID XHTML. I like addthis, I think its the best button of its kind (yes, I need to use the namespace properties, I know you dont have to but I do for what im doing PHP project) :(

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  • iPad/iPhone: Form filling application pointers

    - by raj.tiwari
    Folks, I am starting work on an iPad/iPhone application that is essentially a form-filing UI. The requirement is to present a (rather large) form to the user. The form is composed of sections and questions, like so: Form Question 0.1 Question 0.2 Section 1 Question 1.1 etc. The user can take various paths down the form based on answers to questions. I would like to architect this by defining a declarative markup that can be used to author the form questionnaire including traversal rules. My questions are: Can anyone recommend a markup/language that would satisfy the declaration requirement? Is there any existing library that would ease the implementation as described above? Thanks for your time.

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  • Implicitly including optional dependencies in Maven

    - by Jon Todd
    I have a project A which has a dependency X. Dependency X has an optional dependency Y which doens't get included in A by default. Is there a way to include Y in my POM without explicitly including it? In Ivy they have a way to essentailly say include all optional dependencies of X, does Maven have a way to do this?

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  • Processing RSS/RDF via xml.dom.minidom

    - by Bill
    I'm trying to process a delicious rss feed via python. Here's a sample: ... <item rdf:about="http://weblist.me/"> <title>WebList - The Place To Find The Best List On The Web</title> <dc:date>2009-12-24T17:46:14Z</dc:date> <link>http://weblist.me/</link> ... </item> <item rdf:about="http://thumboo.com/"> <title>Thumboo! Free Website Thumbnails and PHP Script to Generate Web Screenshots</title> <dc:date>2006-10-24T18:11:32Z</dc:date> <link>http://thumboo.com/</link> ... The relevant code is: def getText(nodelist): rc = "" for node in nodelist: if node.nodeType == node.TEXT_NODE: rc = rc + node.data return rc dom = xml.dom.minidom.parse(file) items = dom.getElementsByTagName("item") for i in items: title = i.getElementsByTagName("title") print getText(title) I would think this would print out each title, but instead I get basically get blank output. I'm sure I'm doing something stupid wrong, but no idea what?

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  • How to get parameter values from an XmlNode in C#

    - by Brian
    How do I get the values for parameters in a XmlNode tag. For example: <weather time-layout="k-p24h-n7-1"> <name>Weather Type, Coverage, and Intensity</name> <weather-conditions weather-summary="Mostly Sunny"/> </weather> I want to get the value for the parameter 'weather-summary' in the node 'weather-conditions'.

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  • button in XSL for AJAX usage

    - by phingko
    Hi guys, I wonder if its is possible to do AJAX when I put a button inside the xsl file; <input type = "button" id="laptop" value = "Add to Cart" onclick="sendCartRequest('Add');" /> That's what I do in my xsl file then in my js file I pass the id to the DOM and try to alert it make sure it is passed. And the alert appear to be empty. Is it a mistake to put the button in the xsl? or that's something else that cause it's empty? May be my DOMpath? Please point me a right direction. Thanks in advanced.

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  • How to draw some lines in a view element defined in the xml layout

    - by Nils
    Hello, I have problems drawing some simple lines in a view object (Android programming). First I created the layout with the view element(kind of painting area) in it (XML file). [...] < View android:id="@+id/viewmap" android:layout_width="572px" android:layout_height="359px" android:layout_x="26px" android:layout_y="27px" [...] ... and tried then to access it to draw some lines. Unfortunately the program is running and other UI elements like buttons are displayed, but I can't see the drawings. What's wrong ? [...] viewmap = (View) findViewById(R.id.viewmap); Canvas canvas = new Canvas(); viewmap.draw(canvas); Paint p = new Paint(); p.setColor(Color.BLUE); p.setStyle(Paint.Style.STROKE); canvas.drawColor(Color.WHITE); p.setColor(Color.BLUE); canvas.drawLine(4, 4, 29, 5, p); p.setColor(Color.RED); viewmap.draw(canvas); [...] Thanks for help :) !

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  • Null Exception RelativeLayout

    - by theblixguy
    I am trying to remove objects from my relative layout and replace the background with another image but I get a java.lang.NullPointerException on this line: RelativeLayout ths = (RelativeLayout)findViewById(R.layout.activity_main); Below is my code: package com.ssrij.qrmag; import android.app.Activity; import android.content.Intent; import android.net.Uri; import android.os.Bundle; import android.view.View; import android.view.animation.Animation; import android.view.animation.TranslateAnimation; import android.view.animation.Animation.AnimationListener; import android.widget.Button; import android.widget.RelativeLayout; import android.widget.TextView; public class MainActivity extends Activity { @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.activity_main); // Initialize animations Animation a = new TranslateAnimation(1000,0,0,0); Animation a1 = new TranslateAnimation(1000,0,0,0); Animation a2 = new TranslateAnimation(1000,0,0,0); Animation a3 = new TranslateAnimation(1000,0,0,0); // Set animation durations (ms) a.setDuration(1200); a1.setDuration(1400); a2.setDuration(1600); a3.setDuration(1800); // Get a reference to the objects we want to apply the animation to final TextView v = (TextView)findViewById(R.id.textView1); final TextView v1 = (TextView)findViewById(R.id.textView2); final TextView v2 = (TextView)findViewById(R.id.TextView3); final Button v3 = (Button)findViewById(R.id.tap_scan); // Clear existing animations, just in case... v.clearAnimation(); v1.clearAnimation(); v2.clearAnimation(); v3.clearAnimation(); // Start animating v.startAnimation(a); v1.startAnimation(a1); v2.startAnimation(a2); v3.startAnimation(a3); a1.setAnimationListener(new AnimationListener() { @Override public void onAnimationStart(Animation animation) { // TODO Auto-generated method stub } @Override public void onAnimationRepeat(Animation animation) { // TODO Auto-generated method stub } @Override public void onAnimationEnd(Animation animation) { v.setVisibility(View.INVISIBLE); v1.setVisibility(View.INVISIBLE); v2.setVisibility(View.INVISIBLE); v3.setVisibility(View.INVISIBLE); RelativeLayout ths = (RelativeLayout)findViewById(R.layout.activity_main); ths.setBackgroundResource(R.drawable.blurbg); } }); } public void ScanQr(View v) { // Open the QR Scan page Intent a = new Intent(MainActivity.this, ScanActivity.class); startActivity(a); } } Is there anything that I am doing wrong?

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  • Xpath how to select an element based on an order and dependent of its existence

    - by hokkos
    Hi, how to select an element based on an order and dependent of its existence in XPath ? For example how to select the best quality video if it exist. <VIDEOS> <LOW_RES>video_L.flv</LOW_RES> <HI_RES>video_H.flv</HI_RES> <HD/> </VIDEOS> this should return video_H.flv because the hd version doesn't exist this case can exist (the videos names can be random): <VIDEOS> <LOW_RES>video_L.flv</LOW_RES> <HI_RES>video_H.flv</HI_RES> <HD>video_hd.mp4</HD> </VIDEOS> this should return video_hd.mp4 because the hd version exist. Many thanks.

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  • Confused as to use a class or a function: Writing XML files using lxml and Python

    - by PulpFiction
    Hi. I need to write XML files using lxml and Python. However, I can't figure out whether to use a class to do this or a function. The point being, this is the first time I am developing a proper software and deciding where and why to use a class still seems mysterious. I will illustrate my point. For example, consider the following function based code I wrote for adding a subelement to a etree root. from lxml import etree root = etree.Element('document') def createSubElement(text, tagText = ""): etree.SubElement(root, text) # How do I do this: element.text = tagText createSubElement('firstChild') createSubElement('SecondChild') As expected, the output of this is: <document> <firstChild/> <SecondChild/> </document> However as you can notice the comment, I have no idea how to do set the text variable using this approach. Is using a class the only way to solve this? And if yes, can you give me some pointers on how to achieve this?

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  • Java Node.cloneNode()

    - by Tom Brito
    Talking about the org.w3c.dom package; When I call Node.cloneNode() method from a Element(extends Node) object, which Document is used to create the new cloned Element? Example: import org.w3c.dom; class MyClass { public static void main(String[] args) throws Exception { DocumentBuilder builder = DocumentBuilderFactory.newInstance().newDocumentBuilder(); Document doc = builder.newDocument(); Element element = doc.createElement("myElement"); Element cloneElement = (Element) element.cloneNode(true); } } Which Document was used to create cloneElement?

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  • How to get node without children in xQuery?

    - by mbrevoort
    So I have two nodes of elements that I'm essentially trying to join. I want the top level node to stay the same but the child nodes to be replaced by those cross referenced. Given: <stuff> <item foo="foo" boo="1"/> <item foo="bar" boo="2" /> <item foo="baz" boo="3"/> <item foo="blah boo="4""/> </stuff> <list a="1" b="2"> <foo>bar</foo> <foo>baz</foo> </list> I want to loop through "list" and cross reference elements in "stuff" for this result: <list a="1" b="2"> <item foo="bar" boo="2" /> <item foo="baz" boo="3"/> </list> I want to do this without having to know about what attributes might be on "list". In other words I don't want to have to explicitly call them out like attribute a { $list/@a }, attribute b { $list/@b }

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  • Sorting based on existing elements in xslt

    - by Teelo
    Hi , I want to sort in xslt based on existing set of pattern . Let me explain with the code: <Types> <Type> <Names> <Name>Ryan</Name> </Names> <Address>2344</Address> </Type> <Type> <Names> </Name>Timber</Name> </Names> <Address>1234</Address> </Type> <Type> <Names> </Name>Bryan</Name> </Names> <Address>34</Address> </Type> </Types> Right now I m just calling it and getting it like (all hyperlinks) Ryan Timber Bryan Now I don't want sorting on name but I have existing pattern how I want it to get displayed.Like Timber Bryan Ryan (Also I don't want to lose the url attached to my names earlier while doing this) I was thinking of putting earlier value in some array and sort based on the other array where I will store my existing pattern. But I am not sure how to achieve that.. My xslt looks like this now(there can be duplicate names also) <xsl:for-each select="/Types/Type/Names/Name/text()[generate-id()=generate-id(key('Name',.)[1])]"> <xsl:call-template name="typename"> </xsl:call-template> </xsl:for-each> <xsl:template name="typename"> <li> <a href="somelogicforurl"> <xsl:value-of select="."/> </a> </li> </xsl:template> I am using xsl 1.0

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  • How to switch a view after i parse data?

    - by charles Graffeo
    OK my problem is this, i parse a document and after the document is parsed then i want to load to the next view sounds simple but ive been here for like 4 hours playing with code and id appreciete any help u can give me atm. k heres my parserDidEndDocumentCode - (void)parserDidEndDocument:(NSXMLParser *)parser{ IpadSlideShowViewController temp=(IpadSlideShowViewController)[[IpadSlideShowViewController alloc]init]; [temp performSelectorOnMainThread:@selector(switchViews) withObject:nil waitUntilDone:TRUE]; [temp release]; } now what i think should happen is my code will switch the view after it parses the documentation, n its not doing it idk why

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