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Search found 19554 results on 783 pages for 'xml pull parser'.

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  • Using NSXMLParser to parse Vimeo XML on iPhone.

    - by Sonny Fazio
    Hello, I'm working on an iPhone app that will use Vimeo Simple API to give me a listing our videos by a certain user, in a convenient TableView format. I'm new to Parsing XML and have tried TouchXML, TinyXML, and now NSXMLParser with no luck. Most tutorials on parsing XML are for a blog, and not for an API XML sheet. I've tried modifying the blog parsers to search for the specific tags, but it doesn't seem to work. Right now I'm working with NSXMLParser and it seems to correctly find the value of an XML tag, but when it goes to append it to a NSMutableString, it writes a whole bunch of nulls in between it. I'm using a tutorial from theappleblog and modifying it to work with Vimeo API - (void)parser:(NSXMLParser *)parser foundCharacters:(NSString *)string{ if ([currentElement isEqualToString:@"video_title"]) { NSLog(@"String: %@",string); [currentTitle appendString:string]; } else if ([currentElement isEqualToString:@"video_url"]) { [currentLink appendString:string]; } else if ([currentElement isEqualToString:@"video_description"]) { [currentSummary appendString:string]; } else if ([currentElement isEqualToString:@"date"]) { [currentDate appendString:string]; } Here is the nulls it writes: http://grab.by/grabs/92d9cfc2df4fac3fe6579493b1a8e89f.png Then when it finishes, it has to add the NSMutableStrings into a NSMutableDictionary - (void)parser:(NSXMLParser *)parser didStartElement:(NSString *)elementName namespaceURI:(NSString *)namespaceURI qualifiedName:(NSString *)qName attributes:(NSDictionary *)attributeDict{ //NSLog(@"found this element: %@", elementName); currentElement = [elementName copy]; if ([elementName isEqualToString:@"item"]) { // clear out our story item caches... item = [[NSMutableDictionary alloc] init]; currentTitle = [[NSMutableString alloc] init]; currentDate = [[NSMutableString alloc] init]; currentSummary = [[NSMutableString alloc] init]; currentLink = [[NSMutableString alloc] init]; } } - (void)parser:(NSXMLParser *)parser didEndElement:(NSString *)elementName namespaceURI:(NSString *)namespaceURI qualifiedName:(NSString *)qName{ NSLog(@"Title: %@",currentTitle); if ([elementName isEqualToString:@"item"]) {// save values to an item, then store that item into the array... [item setObject:currentTitle forKey:@"video_title"]; NSLog(@"Current Title%@", currentTitle); [item setObject:currentLink forKey:@"video_url"]; [item setObject:currentSummary forKey:@"video_description"]; [item setObject:currentDate forKey:@"date"]; [stories addObject:[item copy]]; NSLog(@"adding story: %@", currentTitle); } } I would really appreciate it if someone has any advance

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  • Idiomatic way to build a custom structure from XML zipper in Clojure

    - by Checkers
    Say, I'm parsing an RSS feed and want to extract a subset of information from it. (def feed (-> "http://..." clojure.zip/xml-zip clojure.xml/parse)) I can get links and titles separately: (xml-> feed :channel :item :link text) (xml-> feed :channel :item :title text) However I can't figure out the way to extract them at the same time without traversing the zipper more than once, e.g. (let [feed (-> "http://..." clojure.zip/xml-zip clojure.xml/parse)] (zipmap (xml-> feed :channel :item :link text) (xml-> feed :channel :item :title text))) ...or a variation of thereof, involving mapping multiple sequences to a function that incrementally builds a map with, say, assoc. Not only I have to traverse the sequence multiple times, the sequences also have separate states, so elements must be "aligned", so to speak. That is, in a more complex case than RSS, a sub-element may be missing in particular element, making one of sequences shorter by one (there are no gaps). So the result may actually be incorrect. Is there a better way or is it, in fact, the way you do it in Clojure?

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  • How to find/extract data from xml with jQuery

    - by darryl
    I'm trying to extract the StateLongName and StateShortName values from the xml below. I know there has to be a simple elegant way to do this with jQuery. <NewDataSet> <Table> <StateLongName>Alabama</StateLongName> <StateShortName>AL</StateShortName> </Table> <Table> <StateLongName>Alaska</StateLongName> <StateShortName>AK</StateShortName> </Table> ...elments removed for brevity </NewDataSet> Here's what I've tried. Load the xml from above into a Javascript variable name xml. Try #1 $(xml).find("TABLE").each(function() { var stateName = $(this).find("StateLongName").innerText; var stateCode = $(this).find("StateShortName").innerText; }); Try #1 doesn't find anything and never goes inside to load the stateName and stateCode variables. Try #2 $(xml).find("StateLongName").each(function() { var stateName = $(this).find("StateLongName").innerText; var stateCode = $(this).find("StateShortName").innerText; }); Try #2 does find matches, however the stateName and stateCode are left undefined. Try #3 $(xml).find("StateLongName").each(function() { var stateName = $($(xml).find('StateLongName').parent()[0].innerHTML)[1].data; var stateCode = $($(xml).find('StateLongName').parent()[0].innerHTML)[5].data; }); Try #3 works but there has to be a better way. Please enlighten me. Thanks for you time!

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  • Python regex on list

    - by Peter Nielsen
    Hi there I am trying to build a parser and save the results as an xml file but i have problems.. For instance i get a TypeError: expected string or buffer when i try to run the code.. Would you experts please have a look at my code ? import urllib2, re from xml.dom.minidom import Document from BeautifulSoup import BeautifulSoup as bs osc = open('OSCTEST.html','r') oscread = osc.read() soup=bs(oscread) doc = Document() root = doc.createElement('root') doc.appendChild(root) countries = doc.createElement('countries') root.appendChild(countries) findtags1 = re.compile ('<h1 class="title metadata_title content_perceived_text(.*?)</h1>', re.DOTALL | re.IGNORECASE).findall(soup) findtags2 = re.compile ('<span class="content_text">(.*?)</span>', re.DOTALL | re.IGNORECASE).findall(soup) for header in findtags1: title_elem = doc.createElement('title') countries.appendChild(title_elem) header_elem = doc.createTextNode(header) title_elem.appendChild(header_elem) for item in findtags2: art_elem = doc.createElement('artikel') countries.appendChild(art_elem) s = item.replace('<P>','') t = s.replace('</P>','') text_elem = doc.createTextNode(t) art_elem.appendChild(text_elem) print doc.toprettyxml()

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  • ajax html vs xml/json responses - perfomance or other reasons

    - by pedalpete
    I've got a fairly ajax heavy site and some 3k html formatted pages are inserted into the DOM from ajax requests. What I have been doing is taking the html responses and just inserting the whole thing using jQuery. My other option is to output in xml (or possibly json) and then parse the document and insert it into the page. I've noticed it seems that most larger site do things the json/xml way. Google Mail returns xml rather than formatted html. Is this due to performance? or is there another reason to use xml/json vs just retrieving html? From a javascript standpoint, it would seem injecting direct html is simplest. In jQuery I just do this jQuery.ajax({ type: "POST", url: "getpage.php", data: requestData, success: function(response){ jQuery('div#putItHear').html(response); } with an xml/json response I would have to do jQuery.ajax({ type: "POST", url: "getpage.php", data: requestData, success: function(xml){ $("message",xml).each(function(id) { message = $("message",xml).get(id); $("#messagewindow").prepend(""+$("author",message).text()+ ": "+$("text",message).text()+ ""); }); } }); clearly not as efficient from a code standpoint, and I can't expect that it is better browser performance, so why do things the second way?

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  • Transform RSS-Feed into another "standard" XML-Format with PHP

    - by ChrisBenyamin
    Hey friends, quick question: I need to transform a default RSS Structure into another XML-format. The RSS File is like.... Name des RSS Feed Feed Beschreibung de http://xml-rss.de Sat, 1 Jan 2000 00:00:00 GMT Titel der Nachricht Die Nachricht an sich http://xml-rss.de/link-zur-nachricht.htm Sat, 1. Jan 2000 00:00:00 GMT 01012000-000000 Titel der Nachricht Die Nachricht an sich http://xml-rss.de/link-zur-nachricht.htm Sat, 1. Jan 2000 00:00:00 GMT 01012000-000000 Titel der Nachricht Die Nachricht an sich http://xml-rss.de/link-zur-nachricht.htm Sat, 1. Jan 2000 00:00:00 GMT 01012000-000000 ...and I want to extract only the item-elements (with childs and attributes) XML like: <?xml version="1.0" encoding="ISO-8859-1"?> <item> <title>Titel der Nachricht</title> <description>Die Nachricht an sich</description> <link>http://xml-rss.de/link-zur-nachricht.htm</link> <pubDate>Sat, 1. Jan 2000 00:00:00 GMT</pubDate> <guid>01012000-000000</guid> </item> ... It hasn't to be stored into a file. I need just the output. I tried different approaches with DOMNode, SimpleXML, XPath, ... but without success. Thanks chris

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  • Easiest Way to Parse data from twitter with Cocoa

    - by happyCoding25
    Hello, I've followed the tutorial from here: Twitter Client Tutorial to make a little twitter app. Now I need to find out how to parse the XML twitter gives you when you make a request. I've looked at tons of tutorials on phrasing xml on the iPhone but none have made much sense because Im still new to cocoa. Twitter stores the text of the tweet in something like this <text> Some tweet here </text>. From reading the tutorials I think this would involve nsxmlparser but I'm not sure. If anyone could share some code that could parse the <text> Some tweet here </text> things into an array that would be really great. Thanks in advance

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  • C# XML export for Excel table using XmlDocument

    - by Mark
    I am trying to write the following into an XML document: <head> <xml> <x:ExcelWorkbook> <x:ExcelWorksheets> <x:ExcelWorksheet> other code here </x:ExcelWorksheet> </x:ExcelWorksheets> </x:ExcelWorkbook> </xml> </head> However, if I use the following code, it strips out the 'x:'. System.Xml.XmlDocument document = new System.Xml.XmlDocument(); System.Xml.XmlElement htmlNode = document.CreateElement("html"); htmlNode.SetAttribute("xmlns:o", "urn:schemas-microsoft-com:office:office"); htmlNode.SetAttribute("xmlns:x", "urn:schemas-microsoft-com:office:excel"); htmlNode.SetAttribute("xmlns", "http://www.w3.org/TR/REC-html40"); document.AppendChild(htmlNode); System.Xml.XmlElement headNode = document.CreateElement("head"); htmlNode.AppendChild(headNode); headNode.AppendChild( document.CreateElement("xml")).AppendChild( document.CreateElement("x:ExcelWorkbook"))).AppendChild( document.CreateElement("x:ExcelWorksheets")).AppendChild( document.CreateElement("x:ExcelWorksheet")).InnerText="other code here"; How can I stop this from happening? Thanks!

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  • How to check if a node's value in one XML is present in another XML with a specific attribute?

    - by Manish
    The question seems to be little confusing. So let me describe my situation through an example: Suppose I have an XML: A.xml <Cakes> <Cake>Egg</Cake> <Cake>Banana</Cake> </Cakes> Another XML: B.xml <Cakes> <Cake show="true">Egg</Cake> <Cake show="true">Strawberry</Cake> <Cake show="false">Banana</Cake> </Cakes> Now I want to show some text say "TRUE" if all the Cake in A.xml have show="true" in B.xml else "FALSE". In the above case, it will print FALSE. I need to develop an XSL for that. I can loop through all the Cake in A.xml and check if that cake has show="true" in B.xml but I don't know how to break in between (and set a variable) if a show="false" is found. Is it possible? Any help/comments appreciated.

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  • Having trouble parsing XML with jQuery

    - by Jack
    Hi Guys, I'm trying to parse some XML data using jQuery, and as it stands I have extracted the 'ID' attribute of the required nodes and stored them in an array, and now I want to run a loop for each array member and eventually grab more attributes from the notes specific to each ID. The problem currently is that once I get to the 'for' loop, it isn't looping, and I think I may have written the xml path data incorrectly. It runs once and I recieve the 'alert(arrayIds.length);' only once, and it only loops the correct amount of times if I remove the subsequent xml path code. Here is my function: var arrayIds = new Array(); $(document).ready(function(){ $.ajax({ type: "GET", url: "question.xml", dataType: "xml", success: function(xml) { $(xml).find("C").each(function(){ $("#attr2").append($(this).attr('ID') + "<br />"); arrayIds.push($(this).attr('ID')); }); for (i=0; i<arrayIds.length; i++) { alert(arrayIds.length); $(xml).find("C[ID='arrayIds[i]']").(function(){ // pass values alert('test'); }); } } }); }); Any ideas?

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  • [Android] XML Parser

    - by lemon
    I'm trying to extract n0Y7ezLlIYA8R0K54rEmHaTOraBQVSPDjQaGlQxlGso4jdVN1kRxtcfskEs= using w3c dom <html> <div id='token' style='display:none;'> n0Y7ezLlIYA8R0K54rEmHaTOraBQVSPDjQaGlQxlGso4jdVN1kRxtcfskEs= </div> </html> but I seem to be stuck DocumentBuilder builder = DocumentBuilderFactory.newInstance().newDocumentBuilder(); Document doc = builder.parse(con.getInputStream()); NodeList list = doc.getElementsByTagName("div"); Can someone please point me to some basic tutorials that would help me solve my dilemma. Thanks.

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  • Multiple XML/XSLT files in PHP, transform one with XSLT and add others but process it first with PHP

    - by ipalaus
    I am processing XML files transformations with XSLT in PHP correctly. Actually I use this code: $xml = new DOMDocument; $xml->LoadXML($xml_contents); $xsl = new DOMDocument; $xsl->load($xsl_file); $proc = new XSLTProcesoor; $proc->importStyleSheet($xsl); echo $proc->transformToXml($xml); $xml_contents is the XML processed with PHP, this is done by including the XML file first and then assigning $xml_contents = ob_get_contents(); ob_end_clean();. This forces to process the PHP code on the XML, and it works perfectly. My problem is that I use more than one XML file and this XML files has PHP code on it that need to be processed AND have a XSLT file associated to process the data. Actually I'm including this files in XSLT with the next code: <!-- First I add the XML file --> <xsl:param name="menu" select="document('menu.xml')" /> <!-- Next I add the transformations for menu.xml file --> <xsl:include href="menu.xsl" /> <!-- Finally, I process it on the actual ("parent") XML --> <xsl:apply-templates select="$menu/menu" /> My questiion is how I can handle this. I need to add mutiple XML(+XSLT) files to my first XML file that will containt PHP so it needs to be processed. Thank you in advance!

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  • How to make restrictions on XML Schema Complex type?

    - by chobo2
    Hi I am reading the tutorials on w3cschools ( http://www.w3schools.com/schema/schema_complex.asp ) but they don't seem to mention how you could add restrictions on complex types. Like for instance I have this schema. <xs:element name="employee"> <xs:complexType> <xs:sequence> <xs:element name="firstname" type="xs:string"/> <xs:element name="lastname" type="xs:string"/> </xs:sequence> </xs:complexType> </xs:element> now I want to make sure the firstname is no more then 10 characters long. How do I do this? I tried to put in the simple type for the firstname but it says I can't do that since I am using a complex type. So how do I put restrictions like that on the file so the people who I give the schema to don't try to make the firstname 100 characters.

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  • Webserver directory index: index.xml?

    - by Marius
    Hello there, I am making my first RSS-Feed, and I want to host it like this: www.example.com/rss/ I tried to name the xml-file "index.xml" and place it inside the directory, however, when I type http://www.example.com/rss/ i arrive at "Index of /rss" where the file is listed as being part of the directory, but it is not loaded automatically. What can be done about this? Thank you for your time. Kind regards, Marius

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  • Displaying XML in Chrome Browser

    - by Josh
    I love the Chrome browser but I use XML quite a lot in my development work and when I view it in Chrome I just get the rendered text. I know that the source view is slightly better but I'd really like to see the layout and functionality that IE adds to XML namely: Highlighting Open/close nodes Any ideas how I can get this on Chrome? Thanks, Josh UPDATE: The XMLTree Extension is available on Google Chrome Extension Beta Site. Thanks again for your help.

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  • Retrieve MS SQL database or table struture in XML

    - by clutch
    Is there a way to export the database schema in well formed XML of a MS 2000 SQL Server. I'm looking for just the structure not the data and the more detailed the better. The XML may be used in a migration processes. I'm more familiar with MySQL then with SQL Server so please be detailed if you hav time. Thanks

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  • Search in multiple xml files

    - by Ram
    I have a windows Xp Sp2 system where the windows explorer search is not able to find the text in xml files. Is there some setting that enable the search in xml files? It finds in text in text / doc files in the same folder.

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  • Notepad++ Reindent XML breaks lines.

    - by C. Ross
    I'm attempting to use Notepad++ TextFX HTMLTidy - Tidy: Reindent XML. This works well with the following exception. Lines longer than 70 character are wrapped, breaking the validity of my xml. This: <RevieweeDepartmentName>BB AAAAAAAAAA AAAAAAAAAA AAAA</RevieweeDepartmentName> Becomes this: <RevieweeDepartmentName>BB AAAAAAAAAA AAAAAAAAAA AAAA</RevieweeDepartmentName> How can I get it to stop this behavior?

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