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  • In php, Prepare string and create XML/RSS Feed

    - by Bill
    I want to create my own RSS/XML feed. I fetch data from the database to display, but keep getting invalid character errors. If the string has an ampersand or other strange characters in it, the XML will be invalid. I tried using urlencode and htmlentities, but these don't capture all possible characters which need to be escaped. Does anyone know of a PHP function which will prepare a string for XML output?

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  • Simple XML parsing with Google Picasa API and JQuery

    - by zeedog
    I'm starting to look into Google's Picasa API for photo data, which provides you with a big XML file with info about albums and photos. I'm doing some quick and dirty tests to parse the XML file (which is saved locally to my hard drive for now) with JQuery and pull out the album id's, which are stored as "gphoto:id" tags, and display them in a div: $(document).ready(function() { $.get( 'albums.xml', function(data) { $(data).find('entry').each(function() { var albumId = $(this).children('gphoto:id').text(); $('#photos').append(albumId + '<br />'); }) }) }) I'm getting the following error in the console: jquery.js:3321 - Uncaught Syntax error, unrecognized expression: Syntax error, unrecognized expression: id This will work with other tags in the XML file (such as title, author, updated, etc.), but I'm trying to understand what's going on here. Does it have to do with the colon in "gphoto:id", somehow?

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  • How to write two-dimensional array to xml in Scala 2.8.0

    - by Shadowlands
    The following code (copied from a question from about a year ago) works fine under Scala 2.7.7, but does not behave correctly under Scala 2.8.0 (Beta 1, RC8). import scala.xml class Person(name : String, age : Int) { def toXml(): xml.Elem = <person><name>{ name }</name><age>{ age }</age></person> } def peopleToXml(people: Array[Person]): xml.Elem = { <people>{ for {person <- people} yield person.toXml }</people> } val data = Array(new Person("joe",40), new Person("mary", 35)) println(peopleToXml(data)) The output (according to 2.7.7) should be: <people><person><name>joe</name><age>40</age></person><person><name>mary</name><age>35</age></person></people> but instead comes out as: <people>\[Lscala.xml.Elem;@17821782</people> How do I get this to behave as it did in 2.7.x?

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  • How to display information contained in XML file from another website

    - by Tristan
    Hello, I have an XML file ( XML file I produce ) which contains information about my parteners. I want them to display on their website information relative to them by picking them into the XML file. I have no idea to do that, ecxept that i need to write a 'parser' in javascript to display information. could you please provide me examples to do that ? (how to write a parser, how to display only information for one partener ?) Thank you, Regards

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  • Extract some data from a lot of xml files

    - by LifeH2O
    I have cricket player profiles saved in the form of .xml files in a folder. each file has these tags in it <playerid>547</playerid> <majorteam>England</majorteam> <playername>Don</playername> the playerid is same as in .xml (each file is of different size,1kb to 5kb). These are about 500 files. What i need is to extract the playername, majorteam, and playerid from all these files to a list. I will convert that list to XML later. If you know how can i do it directly to XML i will be very thankful.

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  • Convert XML to .plist

    - by Dror Sabbag
    Hey, I have an XML exported from Oracle DB, which will be downloaded into my application main bundle. i would like to convert this XML file into .plist file so i can assign the values into NSDictionary and NSArrays.. Is there a way to get this to work? or is there a better way to work with an external XML file? note that one of the fields in the XML is a full HTML content example: <main> <DATA_RECORD> <ID>ID1</ID> <NO>1234512</NO> <TYPE>NEW</TYPE> <TYPE_NO>0</TYPE_NO> <TEXT_ID>TEXT1</TEXT_ID> <TEXT><HTML>some html goes here</HTML></TEXT> </DATA_RECORD> </main>

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  • XML - python prints extra lines

    - by horse
    `from xml import xpath from xml.dom import minidom xmldata = minidom.parse('model.xml').documentElement for maks in xpath.Evaluate('/cacti/results/maks/text()', xmldata): print maks.nodeValue ` and I get result: 85603399.14 398673062.66 95785523.81 But I needed to be: 85603399.14 NO SPACE 398673062.66 NO SPACE 95785523.81 Can somebody help me, i new at programing :( ?

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  • MSSQL 2005 FOR XML

    - by Lima
    Hi, I am wanting to export data from a table to a specifically formated XML file. I am fairly new to XML files, so what I am after may be quite obvious but I just cant find what I am looking for on the net. The format of the XML results I need are: <data> <event start="May 28 2006 09:00:00 GMT" end="Jun 15 2006 09:00:00 GMT" isDuration="true" title="Writing Timeline documentation" image="http://simile.mit.edu/images/csail-logo.gif"> A few days to write some documentation </event> </data> My table structure is: name VARCHAR(50), description VARCHAR(255), startDate DATETIME, endDate DATETIME (I am not too interested in the XML fields image or isDuration at this point in time). I have tried: SELECT [name] ,[description] ,[startDate] ,[endTime] FROM [testing].[dbo].[time_timeline] FOR XML RAW('event'), ROOT('data'), type Which gives me: <data> <event name="Test1" description="Test 1 Description...." startDate="1900-01-01T00:00:00" endTime="1900-01-01T00:00:00" /> <event name="Test2" description="Test 2 Description...." startDate="1900-01-01T00:00:00" endTime="1900-01-01T00:00:00" /> </data> What I am missing, is the description needs to be outside of the event attributes, and there needs to be a tag. Is anyone able to point me in the correct direction, or point me to a tutorial or similar on how to accomplish this? Thanks, Matt

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  • Working with Database using LINQ to XML

    - by M.H
    Hi,I am Working on a project (C#) in the university and they said that we can't use a DBMS like SQL Server so we decide to use Linq and XML...we learned some basics in Linq to Xml But really we don't know how we can create tables and fields and work with them in Xml.any suggestions ?

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  • Parsing XML with jQuery

    - by Jamie
    I've used this: $(document).ready(function () { $.ajax({ type: "GET", url: "http://domain.com/languages.php", dataType: "xml", success: xmlParser }); }); function xmlParser(xml) { $('#load').fadeOut(); $(xml).find("result").each(function () { $(".main").append('' + $(this).find("language").text() + ''); $(".lang").fadeIn(1000); }); } I used a local XML file on my computer, it works fine, but when I change the URL to an website, it just keeps loading all the time... How do I get this to work?

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  • How to XML escaping with Apache Velocity?

    - by Jan Algermissen
    I am generating XML using Apache Velocity. What is the best (most straight-forward) way to XML-escape the output? (I saw there is an escape tool, but could not figure out it's dev state. I also think that XML escaping is something that is very likely supported by Velocity directly.)

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  • Embedding XSL Stylesheet into XML

    - by user700996
    I have the following XML: <?xml version="1.0" encoding="ISO-8859-1"?> <?xml-stylesheet type="text/xsl" href="http://www.fakedomain.com/sally.xsl"?> And the following content in sally.xsl: <?xml version="1.0" encoding="ISO-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="/"> <html> <body> <xsl:for-each select="documentcollection/document"> <p> <xsl:for-each select="rss/channel/item"> <xsl:value-of select="title"/><br /> <xsl:value-of select="description"/><br /> <xsl:value-of select="link"/><br /> </xsl:for-each> </p> </xsl:for-each> </body> </html> </xsl:template> </xsl:stylesheet> However, the browser displays the XML as though the XSL line is not present. Do you know why the browser is ignoring the XSL stylesheet? Is the style sheet wrong? Thanks

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  • MSSQL "for xml", multiple tables, multiple nodes

    - by Nelson
    Is it possible to select multiple tables at once? For example, I can do: SELECT ( SELECT * FROM Articles FOR XML PATH('article'), TYPE ) FOR XML PATH('articles'), ROOT('data') and SELECT ( SELECT * FROM ArticleTypes FOR XML PATH('articleType'), TYPE ) FOR XML PATH('articleTypes'), ROOT('data') Can I join both so that I get the following output? I can't use UNION because the table structures don't match. <data> <articles> <article>...</article> ... </articles> <articleTypes> <articleType>...</articleType> ... </articleTypes> </data>

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  • C# Retrieving Data From XML File

    - by Christopher MacKinnon
    I seem to be having a problem with retrieving XML values with C#, which I know it is due to my very limited knowledge of C# and .XML. I was given the following XML file <PowerBuilderRunTimes> <PowerBuilderRunTime> <Version>12</Version> <Files> <File>EasySoap110.dll</File> <File>exPat110.dll</File> <File>pbacc110.dll</File> </File> </PowerBuilderRunTime> </PowerBuilderRunTimes> I am to process the XML file and make sure that each of the files in the exist in the folder (that's the easy part). It's the processing of the XML file that I have having a hard time with. Here is what I have done thus far: var runtimeXml = File.ReadAllText(string.Format("{0}\\{1}", configPath, Resource.PBRuntimes)); var doc = XDocument.Parse(runtimeXml); var topElement = doc.Element("PowerBuilderRunTimes"); var elements = topElement.Elements("PowerBuilderRunTime"); foreach (XElement section in elements) { //pbVersion is grabbed earlier. It is the version of PowerBuilder if( section.Element("Version").Value.Equals(string.Format("{0}", pbVersion ) ) ) { var files = section.Elements("Files"); var fileList = new List<string>(); foreach (XElement area in files) { fileList.Add(area.Element("File").Value); } } } My issue is that the String List is only ever populated with one value, "EasySoap110.dll", and everything else is ignored. Can someone please help me, as I am at a loss.

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  • Leave entity intact in XML + XSLT

    - by Kuroki Kaze
    I transform XML to (sort of) HTML with XSL stylesheets (using Apache Xalan). In XML there can be entities like &mdash;, which must be left as is. In beginning of XML file I have a doctype which references these entities. What should I do for entity to be left unchanged? <!DOCTYPE article [ <!ENTITY mdash "&mdash;"><!-- em dash --> ]> gives me SAXParseException: Recursive entity expansion, 'mdash' when encountering &mdash in XML text.

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  • How to extract byte-array from one xml and store it in another in Java

    - by grobartn
    So I am using DocumentBuilderFactory and DocumentBuilder to parse an xml. So it is DOM parser. But what I am trying to do is extract byte-array data (its an image encoded in base64) Store it in one object and later in code write it out to another xml encoded in base64. What is the best way to store this in btw. Store it as string? or as ByteArray? How can I extract byte array data in best way and write it out. I am not experienced with this so wanted to get opinion from the group. UPDATE: I am given XML I do not have control of incoming XML that comes in binary64 encoded < byte-array > ... base64 encoded image ... < /byte-array > Using parser I have I need to store this node and question is should that be byte or string and then writing it out to another node in new xml. again in base64 encoding. thanks

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  • loading xml slow down my site

    - by Ayrton
    Hi is it possible that loading an xml is slowing down my site? I've written this little function in php to iterate over an array of Strings to calculate the total amount of my followers function getFeedCount() { foreach ($array as $value) { $xml = simplexml_load_file("http://api.feedburner.com/awareness/1.0/GetFeedData?uri=$value") or die ("Unable to load XML file!"); $circulation += $xml->feed->entry['circulation']; } return $circulation; } the array is about 10 items big and since I started using it, it really slowed down my site. What could I do fix this issue.

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  • Perl, deleting an .xml file created and open with IO::File and XML::Writer?

    - by Sho Minamimoto
    So I'm running through a list of things and have code that creates an .xml file with IO::File called $doc, then I make a new writer with XML::Writer(OUTPUT = $doc). More code runs and I build a big xml file with XML::Writer. Then, near the end of the file, I find out if I need this file at all. If I do need it, I just $writer-end(); $doc-close(); but if I don't need it, what should I enter to just delete all data I've stored/saved and move onto the next file? I tried unlink($docpath) (before and after $doc-close()), the file was not deleted.

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  • Convert an XML object back into a string in ColdFusion

    - by jpmyob
    In ColdFusion, I can parse a string of XML formatted data into an XML Object using xmlParse(). How can I convert it back into a string? When I tried using toString() it throws an error that "it can't convert complex object to simple objects....", which is ironic because that's what it's supposed to do. I need to use XMLTransform() which requires the first argument to be an xml string. But I also need to use xmlSearch() to get a node to pass into my transform, and xmlSearch returns an xmlObject. So now I need to put that object back into xml string format to pass into xmlTransform.

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  • what persistence layer (xml or mysql) should i use for this xml data?

    - by fayer
    i wonder how i could store a xml structure in a persistence layer. cause the relational data looks like: <entity id="1000070"> <name>apple</name> <entities> <entity id="7002870"> <name>mac</name> <entities> <entity id="7002907"> <name>leopard</name> <entities> <entity id="7024080"> <name>safari</name> </entity> <entity id="7024701"> <name>finder</name> </entity> </entities> </entity> </entities> </entity> <entity id="7024080"> <name>iphone</name> <entities> <entity id="7024080"> <name>3g</name> </entity> <entity id="7024701"> <name>3gs</name> </entity> </entities> </entity> <entity id="7024080"> <name>ipad</name> </entity> </entities> </entity> as you can see, it has no static structure but a dynamical one. mac got 2 descendant levels while iphone got 1 and ipad got 0. i wonder how i could store this data the best way? what are my options. cause it seems impossible to store it in a mysql database due to this dynamical structure. is the only way to store it as a xml file then? is the speed of getting information (xpath/xquery/simplexml) from a xml file worse or greater than from mysql? what are the pros and cons? do i have other options? is storing information in xml files, suited for a lot of users accessing it at the same time? would be great with feedbacks!! thanks! EDIT: now i noticed that i could use something called xml database to store xml data. could someone shed a light on this issue? cause apparently its not as simple as just store data in a xml file?

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