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  • Getting Raw XML From SOAPMessage in Java

    - by Daniel Lew
    I've set up a SOAP WebServiceProvider in JAX-WS, but I'm having trouble figuring out how to get the raw XML from a SOAPMessage (or any Node) object. Here's a sample of the code I've got right now, and where I'm trying to grab the XML: @WebServiceProvider(wsdlLocation="SoapService.wsdl") @ServiceMode(value=Service.Mode.MESSAGE) public class SoapProvider implements Provider<SOAPMessage> { public SOAPMessage invoke(SOAPMessage msg) { // How do I get the raw XML here? } } Is there a simple way to get the XML of the original request? If there's a way to get the raw XML by setting up a different type of Provider (such as Source), I'd be willing to do that, too.

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  • Problem with compiler in Web.Config for generating xml doc

    - by asksuperuser
    I have several problems when putting code below in Web.Config to be able to generate xml doc with website (not webproject): <compiler language="c#;cs;csharp" extension=".cs" warningLevel="0" type="Microsoft.CSharp.CSharpCodeProvider, System, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089" compilerOptions="/doc:c:\doc\WebDocs.xml"> How do I put a directory with spaces instead of /doc:c:\doc\WebDocs.xml? How do I put a directory that is a subdirectory of current project? Why my xml file output is nearly empty? Is it because some properties, methods, ... have no xml comment?

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  • Read xml file and import only one table from multiple tables from xml file in the dataset at a time.

    - by Harikrishna
    I want to store the data in the xml file and retrieve the data from that. I have defined more than table in that xml file.Now to read the tables I am using dataset ds = new dataset(); ds.ReadXml(xmlfilepath); Now this dataset contains all the tables those are in xml file when we read the xml file into dataset. But I want only one specified table at a time in a dataset by condition. Like there are PersonalInfo,OtherInfo,PropertiesInfo tables in the xml file. But I want only OtherInfo table in dataset what I should do ?

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  • rails contoller defaults to respond with application/xml in production

    - by Dave Paroulek
    I have a standard contacts_controller.rb with index action that responds as follows: respond_to do |format| format.html format.xml { render :xml => @contacts } end In development, it works as intended: when I browse to http://localhost:3000/contacts, I get an html response. But, when I start the app using capistrano on a remote ubuntu server and browse to the same url, I get a xml response? If I go to http://remote_host:8000/contacts.html, then I see the html response. If I comment out the format.xml { render :xml => @contacts }, then I see the desired html response. Pretty sure I'm missing something subtle about difference between rails development and production modes? Any ideas about what I'm overlooking? Thanks, - Dave

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  • rails controller defaults to respond with application/xml in production

    - by Dave Paroulek
    I have a standard contacts_controller.rb with index action that responds as follows: respond_to do |format| format.html format.xml { render :xml => @contacts } end In development, it works as intended: when I browse to http://localhost:3000/contacts, I get an html response. But, when I start the app using capistrano on a remote Ubuntu server and browse to the same url, I get an xml response. If I go to http://remote_host:8000/contacts.html, then I see the html response. If I comment out the format.xml { render :xml => @contacts }, then I see the desired html response. Pretty sure I'm missing something subtle about difference between Rails development and production modes. Any ideas about what I'm overlooking? Thanks, - Dave

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  • How to convert jquery collection to xml string for an ajax request

    - by Jim
    I created a jquery collection that stores xml as follows: var rh_request = $('') .attr('user_id', user_id) .attr('company_id', company_id) .attr('action', 'x'); I want to post it to my server via an ajax request as follows: $.ajax({ type: "POST", url: mywebsiteURL, processData: false, dataType: "xml", data: rh_request.html(), success: mycallbackfunction }); My problem is that the "data" parameter of the ajax call needs a string version of the xml and it seems neither Jquery's .html() or .text() function yields this. I have older code that used straight javascript to form the outgoing xml and calling the DOM .xml() function yielded a string that worked. How is this done with a jquery collection???

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  • write to xml file using objective c

    - by Mith
    Hi, ok, I managed to read from xml file using NSXMLParser but now i don't know how to write to xml file. I have a xml file , say <?xml version="1.0" encoding="UTF-8"?> <root> <user id="abcd" password="pass1"/> <user id="efg" password="pass2"/> </root> Now when a new user enters details, I want to store them in a new tag.. lets say like, the id is "hhhh" and password is"pass3" I want to add a new tag with attributes as such <user id="hhhh" password="pass3"/> to the xml file. How should I do this. Please explain in an elaborate way . I am a newbie here. Any links to tutorials or examples will be much helpful. Thanks

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  • reference to XML file is not a member of the R file

    - by yoavstr
    how can i had to class layout in R another xml file ? it should b autmatic as i had new resources to res but it's not someone knows what i did wrong ? i open an activity and now i want to open another activity that will work with another xml example i have menu and main.xml now i want to go for anther activity called gamescreen using this method : newGameButton.setOnClickListener(new OnClickListener() { public void onClick(View view) { Intent i = = new Intent(this, gameScreen.class); startActivity(i); } } i want to move to another "page" to another activity called gameScreen which should b associated to the xml called gameScreen.xml but in his onCreate : public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.gameScreen); } and gameScreen is not a member of the R file please help me i am sitting for the last 4 hours felling like an idiot ...

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  • Store multiple XML files in Android

    - by Federico Perez
    I am developing a survey android app so that: Each survey is downloaded from server as XML file. A survey can have multiple versions (new survey, new XML). When a new XML is downloaded it should overwrite its corresponding previous XML. I would like to store the files in a sqlite database. How can I insert my files into the db? Should I use BLOB or store the content of the XML as string? In any case how should I do it?

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  • post xml to Spring REST server returns Unsupported Media Type

    - by Mayra
    I'm trying to create a simple spring based webservice that supports a "post" with xml content. In spring, I define an AnnotationMethodHandler: <bean id="inboundMessageAdapter" class="org.springframework.web.servlet.mvc.annotation.AnnotationMethodHandlerAdapter"> <property name="messageConverters"> <util:list> <bean class="org.springframework.http.converter.xml.MarshallingHttpMessageConverter"> <property name="marshaller" ref="xmlMarshaller"/> <property name="unmarshaller" ref="xmlMarshaller"/> </bean> </util:list> </property> </bean> And a jaxb based xml marshaller: <bean id="xmlMarshaller" class="org.springframework.oxm.jaxb.Jaxb2Marshaller"> <property name="contextPaths"> <array> <value>com.company.schema</value> </array> </property> <property name="schemas"> <array> <value>classpath:core.xsd</value> </array> </property> </bean> My controller is annotated as follows, where "Resource" is a class autogenerated by jaxb: @RequestMapping(method = POST, value = "/resource") public Resource createResource(@RequestBody Resource resource) { // do work } The result of a webservice call is always "HTTP/1.1 415 Unsupported Media Type". Here is an example service call: HttpPost post = new HttpPost(uri); post.addHeader("Accept", "application/xml"); post.addHeader("Content-Type", "application/xml"); StringEntity entity = new StringEntity(request, "UTF-8"); entity.setContentType("application/xml"); post.setEntity(entity); It seems to me that I am setting the correct media type everywhere possible. Anyone have an ideas?

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  • Parse Exception: At line 1, column 0: no element found

    - by Jeffrey
    Hi everyone, I have a weird issue. I receive the following error that causes a force-close: org.apache.harmony.xml.ExpatParser$ParseException: At line 1, column 0: no element found at org.apache.harmony.xml.ExpatParser.parseFragment(ExpatParser.java:508) at org.apache.harmony.xml.ExpatParser.parseDocument(ExpatParser.java:467) at org.apache.harmony.xml.ExpatReader.parse(ExpatReader.java:329) at org.apache.harmony.xml.ExpatReader.parse(ExpatReader.java:286) After clicking the Force Close button, the Activity is recreated and the parsing completes without a hitch. I'm using the following code snippet inside doInBackground of an AsyncTask: URL serverAddress = new URL(url[0]); HTTPURLConnection connection = (HttpURLConnection) serverAddress.openConnection(); connection.setRequestMethod("GET"); connection.setDoOutput(true); connection.setReadTimeout(10000); connection.connect(); InputStream stream = connection.getInputStream(); SAXParserFactory spf = SAXParserFactory.newInstance(); SAXParser sp = spf.newSAXParser(); XMLReader xr = sp.getXMLReader(); xr.parse(new InputSource(stream)); // The line that throws the exception Why would the Activity force-close and then run without any problems immediately after? Would a BufferedInputStream be any different? I'm baffled. :( Thanks for your time everyone.

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  • StAX - Setting the version and encoding using XMLStreamWriter

    - by Anurag
    Hi, I am using StAX for creating XML files and then validating the file with and XSD. I am getting an error while creating the XML file: javax.xml.stream.XMLStreamException: Underlying stream encoding 'Cp1252' and input paramter for writeStartDocument() method 'UTF-8' do not match. at com.sun.xml.internal.stream.writers.XMLStreamWriterImpl.writeStartDocument(XMLStreamWriterImpl.java:1182) Here is the code snippet: XMLOutputFactory xof = XMLOutputFactory.newInstance(); try{ XMLStreamWriter xtw = xof.createXMLStreamWriter(new FileWriter(fileName)); xtw.writeStartDocument("UTF-8","1.0");} catch(XMLStreamException e) { e.printStackTrace(); } catch(IOException ie) { ie.printStackTrace(); } I am running this code on unix. Does anybody know how to set the version and encoding style.

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  • How do I stop XElement.Save from escaping characters?

    - by Daniel I-S
    I'm populating an XElement with information and writing it to an xml file using the XElement.Save(path) method. At some point, certain characters in the resulting file are being escaped - for example, > becomes &gt;. This behaviour is unacceptable, since I need to store information in the XML that includes the > character as part of a password. How can I write the 'raw' content of my XElement object to XML without having these escaped?

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  • Messing with Encoding and XslCompiledTransform

    - by Josemalive
    Hello, im messing with the encodings. For one hand i have a url that is responding me in UTF-8 (im pretty sure thanks to firebug plugin). Im opening the url reading the content in UTF-8 using the following code: StreamReader reader = new StreamReader(response.GetResponseStream(),System.Text.Encoding.UTF8); For other hand i have a transformation xslt sheet with the following code: <?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="xml" indent="yes"/> <xsl:template match="@* | node()"> <xsl:copy> <xsl:apply-templates select="@* | node()"/> <br/> hello </xsl:copy> </xsl:template> </xsl:stylesheet> This xslt sheet is saved in UTF-8 format too. I use the following code to mix the xml with the xslt: StringWriter writer = new StringWriter(); XslCompiledTransform transformer = new XslCompiledTransform(); transformer.Load(HttpContext.Current.Server.MapPath("xslt\\xsltsheet.xslt"); XmlWriterSettings xmlsettings = new XmlWriterSettings(); xmlsettings.Encoding = System.Text.Encoding.UTF8; transformer.Transform(xmlreader, null, writer); return writer; Then after all im render in the webbrowser the content of this writer and im getting the following error: The XML page cannot be displayed Cannot view XML input using style sheet. Please correct the error and then click the Refresh button, or try again later. Switch from current encoding to specified encoding not supported. Error processing resource 'http://localhost:2541/Results.... <?xml version="1.0" encoding="utf-16"?> Im wondering where is finding the UTF-16 encoding take in count that: All my files are saved as UTF-8 The response from the server is in UTF-8 The way of read the xslt sheet is configured as UTF-8. Any help would be appreciated. Thanks in advance. Best Regards. Jose.

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  • Cross domain AJAX requests using YQL

    - by nav
    Hi , I need to query a locations WOEID and grab the WOEID value from the xml returned. So the user would type e.g. London, UK and I need to load the query as below: http://query.yahooapis.com/v1/public/yql?q=select%20woeid%20from%20geo.places%20where%20text%20%3D%20%22London%2C%20UK%2C%20UK%22&format=xml After which I need to grab the WOEID value from the XML content returned. I know this can be done when HTML content is returned as this link shows - http://ajaxian.com/archives/using-yql-as-a-proxy-for-cross-domain-ajax Is there a way to use similar code to query the XML data returned? Thanks alot

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  • Using PHP/MySQL with Google Maps

    - by Anders Kitson
    Hiya, I followed this tutorial below http://code.google.com/apis/maps/articles/phpsqlajax_v3.html#outputxml I ran into trouble, near then end, I am hoping someone else here has got this working and can help me discover my problem. Simply there are 4 steps to this tutorial Creating the Table Populating the Table Outputting XML with PHP Creating the Map I successfully have completed all the steps, however the outputted xml isn't read by the google map I created. The files are all on the same directory, and I didn't change any of the file names from the tutorial. The tutorial has a step to test if the php file called phpsqlajax_genxml.php is outputting the xml and I successfully tested it and it was. The problem is that the map isn't rendering the items I have in the database, that should be converted to xml for the map to read. Any help, or pointing me in the right direction would be much appreciated.

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  • PHP - XML Feed get print values

    - by danit
    Here is my feed: <entry> <id>http://api.visitmix.com/OData.svc/Sessions(guid'816995df-b09a-447a-9391-019512f643a0')</id> <title type="text">Building Web Applications with Microsoft SQL Azure</title> <summary type="text">SQL Azure provides a highly available and scalable relational database engine in the cloud. In this demo-intensive and interactive session, learn how to quickly build web applications with SQL Azure Databases and familiar web technologies. We demonstrate how you can quickly provision, build and populate a new SQL Azure database directly from your web browser. Also, see firsthand several new enhancements we are adding to SQL Azure based on the feedback we&#x2019;ve received from the community since launching the service earlier this year.</summary> <published>2010-01-25T00:00:00-05:00</published> <updated>2010-03-05T01:07:05-05:00</updated> <author> <name /> </author> <link rel="edit" title="Session" href="Sessions(guid'816995df-b09a-447a-9391-019512f643a0')" /> <link rel="http://schemas.microsoft.com/ado/2007/08/dataservices/related/Speakers" type="application/atom+xml;type=feed" title="Speakers" href="Sessions(guid'816995df-b09a-447a-9391-019512f643a0')/Speakers"> <m:inline> <feed> <title type="text">Speakers</title> <id>http://api.visitmix.com/OData.svc/Sessions(guid'816995df-b09a-447a-9391-019512f643a0')/Speakers</id> <updated>2010-03-25T11:56:06Z</updated> <link rel="self" title="Speakers" href="Sessions(guid'816995df-b09a-447a-9391-019512f643a0')/Speakers" /> <entry> <id>http://api.visitmix.com/OData.svc/Speakers(guid'3395ee85-d994-423c-a726-76b60a896d2a')</id> <title type="text">David-Robinson</title> <summary type="text"></summary> <updated>2010-03-25T11:56:06Z</updated> <author> <name>David Robinson</name> </author> <link rel="edit-media" title="Speaker" href="Speakers(guid'3395ee85-d994-423c-a726-76b60a896d2a')/$value" /> <link rel="edit" title="Speaker" href="Speakers(guid'3395ee85-d994-423c-a726-76b60a896d2a')" /> <link rel="http://schemas.microsoft.com/ado/2007/08/dataservices/related/Sessions" type="application/atom+xml;type=feed" title="Sessions" href="Speakers(guid'3395ee85-d994-423c-a726-76b60a896d2a')/Sessions" /> <category term="EventModel.Speaker" scheme="http://schemas.microsoft.com/ado/2007/08/dataservices/scheme" /> <content type="image/jpeg" src="http://live.visitmix.com/Content/images/speakers/lrg/default.jpg" /> <m:properties xmlns:m="http://schemas.microsoft.com/ado/2007/08/dataservices/metadata" xmlns:d="http://schemas.microsoft.com/ado/2007/08/dataservices"> <d:SpeakerID m:type="Edm.Guid">3395ee85-d994-423c-a726-76b60a896d2a</d:SpeakerID> <d:SpeakerFirstName>David</d:SpeakerFirstName> <d:SpeakerLastName>Robinson</d:SpeakerLastName> <d:LargeImage m:null="true"></d:LargeImage> <d:SmallImage m:null="true"></d:SmallImage> <d:Twitter m:null="true"></d:Twitter> </m:properties> </entry> </feed> </m:inline> </link> <link rel="http://schemas.microsoft.com/ado/2007/08/dataservices/related/Tags" type="application/atom+xml;type=feed" title="Tags" href="Sessions(guid'816995df-b09a-447a-9391-019512f643a0')/Tags" /> <link rel="http://schemas.microsoft.com/ado/2007/08/dataservices/related/Files" type="application/atom+xml;type=feed" title="Files" href="Sessions(guid'816995df-b09a-447a-9391-019512f643a0')/Files" /> <category term="EventModel.Session" scheme="http://schemas.microsoft.com/ado/2007/08/dataservices/scheme" /> <content type="application/xml"> <m:properties> <d:SessionID m:type="Edm.Guid">816995df-b09a-447a-9391-019512f643a0</d:SessionID> <d:Location>Breakers L</d:Location> <d:Type>Seminar</d:Type> <d:Code>SVC07</d:Code> <d:StartTime m:type="Edm.DateTime">2010-03-17T12:00:00</d:StartTime> <d:EndTime m:type="Edm.DateTime">2010-03-17T13:00:00</d:EndTime> <d:Slug>SVC07</d:Slug> <d:CreatedDate m:type="Edm.DateTime">2010-01-26T18:14:24.687</d:CreatedDate> <d:SourceID m:type="Edm.Guid">cddca9b7-6830-4d06-af93-5fd87afb67b0</d:SourceID> </m:properties> </content> </entry> I want to print the: Session Title (Building Web Applications with Microsoft SQL Azure) The Author (David Robinson) The Location (Breakers L) And display the speakers image (http://live.visitmix.com/Content/images/speakers/lrg/default.jpg) I presume I can use filegetcontents and then transform to simplexmlstring, but I dont know how to get the deeper items in I want, like Author, and image. Any chance of a bit of coding genius here?

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  • help with grouping and sorting for TreeView in xaml

    - by danhotb
    I am having problems getting my head around grouping and sorting in xaml and hope someone can get me straightened out! I have creaed an xml file from a tree of files and folders (just like windows explorer) that can be serveral levels deep. I have bound a TreeView control to an xml datasource and it works great! It sorts everything alphabetically but ... I would like it to sort all folders first then all files, rather than folders listed with files, as it does now. the xml : if you load this to a treeviw it will display the two files before the folder because they are first in alpha-order. here is my code: <!-- This will contain the XML-data. --> <XmlDataProvider x:Key="xmlDP" XPath="*"> <x:XData> <Select_Project /> </x:XData> </XmlDataProvider> <!-- This HierarchicalDataTemplate will visualize all XML-nodes --> <HierarchicalDataTemplate DataType="project" ItemsSource ="{Binding}"> <TextBlock Text="{Binding XPath=@name}" /> </HierarchicalDataTemplate> <HierarchicalDataTemplate DataType="folder" ItemsSource ="{Binding}"> <TextBlock Text="{Binding XPath=@name}" /> </HierarchicalDataTemplate> <HierarchicalDataTemplate DataType="file" ItemsSource ="{Binding}"> <TextBlock Text="{Binding XPath=@name}" /> </HierarchicalDataTemplate> <CollectionViewSource x:Key="projectView" Source="{StaticResource xmlDP}"> <CollectionViewSource.SortDescriptions> <!-- ADD SORT DESCRIPTION HERE --> </CollectionViewSource.SortDescriptions> </CollectionViewSource> <TreeView Margin="11,79.992,18,19.089" Name="tvProject" BorderThickness="1" FontSize="12" FontFamily="Verdana"> <TreeViewItem ItemsSource="{Binding Source={StaticResource xmlDP}, XPath=*}" Header="Project"/> </TreeView>

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  • JAXB: @XmlTransient on third-party or external super class

    - by Phil
    Hi, I need some help regarding the following issue with JAXB 2.1. Sample: I've created a SpecialPerson class that extends a abstract class Person. Now I want to transform my object structure into a XML schema using JAXB. Thereby I don't want the Person XML type to appear in my XML schema to keep the schema simple. Instead I want the fields of the Person class to appear in the SpecialPerson XML type. Normally I would add the annotation @XmlTransient on class level into the Person code. The problem is that Person is a third-party class and I have no possibility to add @XmlTransient here. How can I tell JAXB that it should ignore the Person class without annotating the class. Is it possible to configure this externally somehow? Have you had the same problem before? Any ideas what the best solution for this problem would be?

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  • Dependent on CVS tagging for automated builds

    - by OMG Ponies
    My current work relies on using tags in CVS for an automated build process (ANT currently) to build for respective environments (development, QA, production). From our research, neither Git or Subversion support tagging in the same manner. If we use Subversion or Git, they don't support tags (in the same manner - please correct me?). So how would ANT or Maven know what to pick up for the respective build? Example: For a webapp, when viewing our repository say for the web.xml file -- the history would look like: web.xml v1 ... web.xml v1.2.3 Tag: Prod web.xml v1.2.4 web.xml v1.2.5 Tag: QA web.xml v1.2.6 web.xml v1.2.7 Head The ANT build scripts are run as CRON jobs, at different times & intervals for different environments. The environment build is based on the repository checkout, based on the tag. Development continues, and eventually the respective tags are moved: web.xml v1 ... web.xml v1.2.3 web.xml v1.2.4 web.xml v1.2.5 web.xml v1.2.6 Tag: Prod web.xml v1.2.7 Tag: QA web.xml v1.2.8 Head

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  • How do I specify a crossdomain policy file to allow Flash to grab a bitmap from an RTMP (Wowza) vide

    - by Ken Smith
    I'm trying to get a bitmap/snapshot of a Wowza video stream playing on my client, like so: var bitmapData:BitmapData = new BitmapData(view.videoPlayerComponent.width, view.videoPlayerComponent.height); bitmapData.draw(view.videoPlayerComponent); When I do this, I get this error message: SecurityError: Error #2123: Security sandbox violation: BitmapData.draw: http://localhost:51150/Resources/WRemoteWebCam.swf cannot access rtmp://localhost/videochat/smithkl42._default/. No policy files granted access. I presume the error comes from not being able to locate the appropriate crossdomain.xml file. I'm not quite sure where it's looking for it, and a wireshark sniff was inconclusive, so I've tried placing one in each of the following places: http://localhost/crossdomain.xml http://localhost:1935/crossdomain.xml http://localhost:51150/crossdomain.xml I can retrieve the file successfully from each of those three locations. (I'm pretty sure that the last one wouldn't have any effect, since it's just the location of the web site which hosts the page that hosts the .swf file, but on the off chance...) These are the contents of the file that it's grabbing in each instance: <cross-domain-policy> <allow-access-from domain="*" to-ports="*" /> </cross-domain-policy> And it's still throwing that same error message. I've also followed the instructions on the Wowza forums, to turn on StreamVideoSampleAccess in the [install]\conf[appname]\Application.xml, with no joy: <Client> <IdleFrequency>-1</IdleFrequency> <Access> <StreamReadAccess>*</StreamReadAccess> <StreamWriteAccess>*</StreamWriteAccess> <StreamAudioSampleAccess>*</StreamAudioSampleAccess> <StreamVideoSampleAccess>*</StreamVideoSampleAccess> <SharedObjectReadAccess>*</SharedObjectReadAccess> <SharedObjectWriteAccess>*</SharedObjectWriteAccess> </Access> </Client> Any thoughts?

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  • Base64 en/decoding between xstream and iPhone SDK

    - by Matt McMinn
    I am passing a byte array from a java server to an iPad client in XML. The server is using xstream to convert the byte array to XML with the EncodedByteArrayConverter, which should convert the array to Base 64. Using xstream, I can decode the xml back to the proper byte array in a java client, but in the iPad client, I'm getting an invalid length error. To do my decoding, I'm using the code at the bottom of this page. The length of the string is indeed not a multiple of 4, so there must be something strange with my string - although since xstream can decode it just fine, I'm guessing there's just something I need to to on the iPad side to get it to decode. I've tried cutting off padding at the end of the string to get it down to the right size, and that does allow the decoder to work, but I end up with JPG's that have invalid headers, and are not displayable. On the server side, I'm using the following code: Object rtrn = getByteArray(); XStream xstream = new XStream(); String xml = xstream.toXML(rtrn); On the client side, I'm calling the above decoder from the XML parsing callback like this: -(void)parser:(NSXMLParser *)parser foundCharacters:(NSString *)string { NSLog(@"Converting data; string length: %d", [string length]); //NSLog(@"%@", string); NSData *data = [Base64 decode:string]; NSLog(@"converted data length: %d", [data length]); } Any ideas what could be going wrong?

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  • Search XDocument with LINQ with out knowing the Namespace

    - by BarDev
    Is there a way to search a XDocument without knowing the Namespace. I have a process that logs all soap requests and encrypts the sensitive data. I want to find any elements based on name. Something like, give me all elements where the name is CreditCard. I don't care what the namespace is. My problem seems to be with LINQ and requiring a xml namespace. I have other processes that retrieve values from XML, but I know the namespace for these other process. XDocument xDocument = XDocument.Load(@"C:\temp\Packet.xml"); XNamespace xNamespace = "http://CompanyName.AppName.Service.Contracts"; var elements = xDocument.Root.DescendantsAndSelf().Elements().Where(d = d.Name == xNamespace + "CreditCardNumber"); But what I really want, is to have the ability to search xml without knowing about namespaces, something like this: XDocument xDocument = XDocument.Load(@"C:\temp\Packet.xml"); var elements = xDocument.Root.DescendantsAndSelf().Elements().Where(d = d.Name == "CreditCardNumber") But of course this will not work be cause I do no have a namespace. BarDev

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  • PHP simplexml and xpath

    - by FFish
    I have an XML file like this: <?xml version="1.0" encoding="UTF-8"?> <gallery> <album tnPath="tn/" lgPath="imm/" fsPath="iml/" > <img src="001.jpg" /> <img src="002.jpg" /> </album> </gallery> I am reading the file with: $xmlFile = "xml.xml"; $xmlStr = file_get_contents($xmlFile . "?" . time()); $xmlObj = simplexml_load_string($xmlStr); Now I am rebuilding the XML file and would like to save the album node with it's attributes in a var I was thinking with xpath: // these all return arrays with the images... // echo $xmlObj->xpath('/gallery/album@tnPath'); // echo $xmlObj->xpath('//album[@tnPath]'); // echo $xmlObj->xpath('//@tnPath'); But that doesn't seem to work? Any help?

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