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  • Large XML files in dataset (outofmemory)

    - by dklein
    Hi folks, I am currently trying to load a slightly large xml file into a dataset. The xml file is about 700 MB and every time I try to read the xml it needs plenty of time and after a while it throws an "out of memory" exception. DataSet ds = new DataSet(); ds.ReadXml(pathtofile); The main problem is, that it is necessary for me to use those datasets (I use it to import the data from xml file into a sybase database (foreach table, foreach row, foreach column)) and that I have no scheme file. I already googled a while, but I did only find solutions that won't be usable for me.

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  • Receive XML via POST with ASP.NET

    - by Mark Hurd
    I have to set up an XML "web service" that receives a POST where the 'Content-type header will specify “text/xml”.' What is the simplest way to get the XML into an XDocument for access by VB.NET's axis queries? I don't believe the web service is guaranteed to follow any protocol (e.g. SOAP, etc); just specific tags and sub-tags for various requests, and it will use Basic Authentication, so I will have to process the headers. (If it matters: * the live version will use HTTPS, and * the response will also be XML.)

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  • how to get mysql query as xml?

    - by mobibob
    I recall reading about XML support from MySql. Does anyone know how to get XML without writing code? My client-protocol expects XML and I have a data source that I can access from a web app (JSP using JDBC).

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  • implement INotifyCollectionChanged etc on xml file changed

    - by netmajor
    It's possible to implement INotifyCollectionChanged or other interface like IObservable to enable to bind filtered data from xml file on this file changed ? I see examples with properties or collection, but what with files changes ? I have that code to filter and bind xml data to list box: XmlDocument channelsDoc = new XmlDocument(); channelsDoc.Load("RssChannels.xml"); XmlNodeList channelsList = channelsDoc.GetElementsByTagName("channel"); this.RssChannelsListBox.DataContext = channelsList;

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  • How to digitally sign XML document using Oracle 9i PL/SQL

    - by Andris Krauze
    Let's say we have a simple XML document (doc.xml) like this: <?xml version="1.0" encoding="UTF-8"?> <Envelope xmlns="http://www.someexample.com/examples"> <Salutation Id="test"> Welcome! </Salutation> </Envelope> And a certificate file:test.p12 How to make a solution using Oracle 9i PL/SQL that digitally signs XML document according to http://www.w3.org/2000/09/xmldsig# Any Digital Signature form (e.g. Enveloped) and method (e.g. RSAwithSHA1) example would be great.

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  • How to display only necessary information from xml

    - by fakson
    I have xml table: <?xml version="1.0" encoding="UTF-8"?> <table> <user> <id>1</id> <info> <more> <nick>John</nick> <adress>blabla</adress> </more> <more> <nick>Mike</nick> <adress>blablabla</adress> <tel>blablabla</tel> </more> </info> </user> <user> <id>2</id> <info> <more> <nick>Fake</nick> <adress>blablabla</adress> <tel>blablabla</tel> </more> </info> </user> </table> And i need to read data from it. I made such parser. <?php $xml = simplexml_load_file("test.xml"); echo $xml->getName() . "<br /><br />"; echo '<hr />'; foreach ($xml->children() as $child1){ //echo $child1->getName() . ": " . $child1 . "<br />"; //el foreach($child1->children() as $child2){ if ((string)$child2 == '2') { echo "<strong>" .$child2 . "</strong><br />"; foreach($child2->children() as $child3){ echo "<strong>".$child3->getName()."</strong>:" . $child3 . "<br />"; foreach($child3->children() as $child4){ echo $child4->getName() . ": " . $child4 . "<br />"; } } } } echo '<hr />'; echo '<br />'; } ?> I want to make search in xml file, for example get all information about user with id 2, i try to realize it with if function but i get only id. What is wrong?. Any suggestions??

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  • get property from XML using PHP

    - by Adnan
    Hello, I am using PHP's SimpleXML to get some values out of the following XML; - <entry> <id>http://www.google.com/m8/feeds/contacts/email_address%40gmail.com/base/0</id> <updated>2010-01-14T22:06:26.565Z</updated> <category scheme="http://schemas.google.com/g/2005#kind" term="http://schemas.google.com/contact/2008#contact" /> <title type="text">Customer Name</title> <link rel="http://schemas.google.com/contacts/2008/rel#edit-photo" type="image/*" href="http://www.google.com/m8/feeds/photos/media/email_address%40gmail.com/0/34h5jh34j5kj3444" /> <link rel="self" type="application/atom+xml" href="http://www.google.com/m8/feeds/contacts/email_address%40gmail.com/full/0" /> <link rel="edit" type="application/atom+xml" href="http://www.google.com/m8/feeds/contacts/email_address%40gmail.com/full/0/5555" /> <gd:email rel="http://schemas.google.com/g/2005#other" address="[email protected]" primary="true" /> </entry> I can get the title with: $xml = new SimpleXMLElement($response_h1); foreach ($xml->entry as $entry) { echo $entry->title, '<br />'; } But how to get the address="[email protected]" property?

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  • Proper use of HTTP status codes in a "validation" server

    - by Romulo A. Ceccon
    Among the data my application sends to a third-party SOA server are complex XMLs. The server owner does provide the XML schemas (.xsd) and, since the server rejects invalid XMLs with a meaningless message, I need to validate them locally before sending. I could use a stand-alone XML schema validator but they are slow, mainly because of the time required to parse the schema files. So I wrote my own schema validator (in Java, if that matters) in the form of an HTTP Server which caches the already parsed schemas. The problem is: many things can go wrong in the course of the validation process. Other than unexpected exceptions and successful validation: the server may not find the schema file specified the file specified may not be a valid schema file the XML is invalid against the schema file Since it's an HTTP Server I'd like to provide the client with meaningful status codes. Should the server answer with a 400 error (Bad request) for all the above cases? Or they have nothing to do with HTTP and it should answer 200 with a message in the body? Any other suggestion? Update: the main application is written in Ruby, which doesn't have a good xml schema validation library, so a separate validation server is not over-engineering.

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  • SQL Server insert with XML parameter - empty string not converting to null for numeric

    - by Mayo
    I have a stored procedure that takes an XML parameter and inserts the "Entity" nodes as records into a table. This works fine unless one of the numeric fields has a value of empty string in the XML. Then it throws an "error converting data type nvarchar to numeric" error. Is there a way for me to tell SQL to convert empty string to null for those numeric fields in the code below? -- @importData XML <- stored procedure param DECLARE @l_index INT EXECUTE sp_xml_preparedocument @l_index OUTPUT, @importData INSERT INTO dbo.myTable ( [field1] ,[field2] ,[field3] ) SELECT [field1] ,[field2] ,[field3] FROM OPENXML(@l_index, 'Entities/Entity', 1) WITH ( field1 int 'field1' ,field2 varchar(40) 'field2' ,field3 decimal(15, 2) 'field3' ) EXECUTE sp_xml_removedocument @l_index EDIT: And if it helps, sample XML. Error is thrown unless I comment out field3 in the code above or provide a value in field3 below. <?xml version="1.0" encoding="utf-16"?> <Entities> <Entity> <field1>2435</field1> <field2>843257-3242</field2> <field3 /> </Entity> </Entities>

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  • Display part of an XML file while parsing it

    - by Andy M
    Hey, Consider the following XML file : <cookbook> <recipe xml:id="MushroomSoup"> <title>Quick and Easy Mushroom Soup</title> <ingredient name="Fresh mushrooms" quantity="7" unit="pieces"/> <ingredient name="Garlic" quantity="1" unit="cloves"/> </recipe> <recipe xml:id="AnotherRecipe"> <title>XXXXXXX</title> <ingredient name="Tomatoes" quantity="8" unit="pieces"/> <ingredient name="PineApples" quantity="2" unit="cloves"/> </recipe> </cookbook> Let's say I want to parse this file and gather each recipe as XML, each one as a separated QString. For example, I would like to have a QString that contains : <recipe xml:id="MushroomSoup"> <title>Quick and Easy Mushroom Soup</title> <ingredient name="Fresh mushrooms" quantity="7" unit="pieces"/> <ingredient name="Garlic" quantity="1" unit="cloves"/> </recipe> How could I do this ? Do you guys know a quick and clean method to perform this ? Thanks in advance for your help !

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  • Best way XML File handling bullet point •

    - by dbengals
    For an XML file I am creating I have data that contains a bullet • what is the best method for handling this in xml data? It opens in an XML editor and reads fine, but I cannot import the file via SSIS, I get an error regarding this point. <xmldata>• Bullet</xmldata> Renders fine, but cannot import with SSIS.

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  • Go XML Unmarshal example doesn't compile

    - by marketer
    The Xml example in the go docs is broken. Does anyone know how to make it work? When I compile it, the result is: xmlexample.go:34: cannot use "name" (type string) as type xml.Name in field value xmlexample.go:34: cannot use nil as type string in field value xmlexample.go:34: too few values in struct initializer Here is the relevant code: package main import ( "bytes" "xml" ) type Email struct { Where string "attr"; Addr string; } type Result struct { XMLName xml.Name "result"; Name string; Phone string; Email []Email; } var buf = bytes.NewBufferString ( ` <result> <email where="home"> <addr>[email protected]</addr> </email> <email where='work'> <addr>[email protected]</addr> </email> <name>Grace R. Emlin</name> <address>123 Main Street</address> </result>`) func main() { var result = Result{ "name", "phone", nil } xml.Unmarshal ( buf , &result ) println ( result.Name ) }

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  • Perl - Encoding String for XML

    - by Sho Minamimoto
    I'm not too fluent with the perl XML libraries (actually, I really suck at understanding encoding in general), all I'm doing is taking a string that possibly has characters such as "à" and putting it in an XML file, but when I open the file, I get an encoding error at the line containing such a character. So I just need a lightweight way to take a string and encode it for XML.

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  • How to return plain XML from ADO.NET data service

    - by KHALIL
    Hi, I was wondering how to return plain XML from ADO.net data services I have exposed an ADO.net data service to different DEPARTMENTS in our company who are not so technical. The data returned is ATOM FEED which is kind a hard to read / interpret with its format, too much information is returned people from various departments would execute different queries ( HTTP Request) and i wanted them to display simple XML or atleast something more user friendly like HTML I have tried ACCEPT attribute of the request to be plain XML and it still returns ATOM Thanks -- Khalil

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  • Windows RPC vs XML-RPC

    - by Y.Z
    Is there any benchmark about encoding/decoding certain common typed data in Microsoft RPC NDR engine (DCE 1.1) in comparison with that in XML-RPC-C/C++ in the de-facto C/C++ implementation in XML-RPC? Actually I have to choose between Windows RPC and XML-RPC-C/C++ to implement my own common object infrastructure for High Performance Computing on Windows. Any recommandation about which with regard to their performance? Thank you. Best Regards, Yang

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  • Getting values from DataGridView back to XDocument (using LINQ-to-XML)

    - by Pretzel
    Learning LINQ has been a lot of fun so far, but despite reading a couple books and a bunch of online resources on the topic, I still feel like a total n00b. Recently, I just learned that if my query returns an Anonymous type, the DataGridView I'm populating will be ReadOnly (because, apparently Anonymous types are ReadOnly.) Right now, I'm trying to figure out the easiest way to: Get a subset of data from an XML file into a DataGridView, Allow the user to edit said data, Stick the changed data back into the XML file. So far I have Steps 1 and 2 figured out: public class Container { public string Id { get; set; } public string Barcode { get; set; } public float Quantity { get; set; } } // For use with the Distinct() operator public class ContainerComparer : IEqualityComparer<Container> { public bool Equals(Container x, Container y) { return x.Id == y.Id; } public int GetHashCode(Container obj) { return obj.Id.GetHashCode(); } } var barcodes = (from src in xmldoc.Descendants("Container") where src.Descendants().Count() > 0 select new Container { Id = (string)src.Element("Id"), Barcode = (string)src.Element("Barcode"), Quantity = float.Parse((string)src.Element("Quantity").Attribute("value")) }).Distinct(new ContainerComparer()); dataGridView1.DataSource = barcodes.ToList(); This works great at getting the data I want from the XML into the DataGridView so that the user has a way to manipulate the values. Upon doing a Step-thru trace of my code, I'm finding that the changes to the values made in DataGridView are not bound to the XDocument object and as such, do not propagate back. How do we take care of Step 3? (getting the data back to the XML) Is it possible to Bind the XML directly to the DataGridView? Or do I have to write another LINQ statement to get the data from the DGV back to the XDocument? Suggstions?

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  • Row as XML in SP

    - by user171523
    From SP i need to get a row as as XML (Includeing all fileds.) Is there any way we can get like below. Declare @xmlMsg varchar(4000) select * into #tempTable from dbo.order for xml raw select @xmlMsg = 1 from #tempTable print '@xmlMsg' + @xmlMsg Row i would like to get it as XML output.

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  • Preventing referenced assembly PDB and XML files copied to output

    - by Jason Morse
    I have a Visual Studio 2008 C#/.NET 3.5 project with a post build task to ZIP the contents. However I'm finding that I'm also getting the referenced assemblies' .pdb (debug) and .xml (documentation) files in my output directory (and ZIP). For example, if MyProject.csproj references YourAssembly.dll and there are YourAssembly.xml and YourAssembly.pdb files in the same directory as the DLL they will show up in my output directory (and ZIP). I can exclude *.pdb when ZIP'ing but I cannot blanket exclude the *.xml files as I have deployment files with the same extension. Is there a way to prevent the project from copying referenced assembly PDB and XML files?

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  • Middel East XML Currency Conversion

    - by Tim
    Hi, Using the following script to do currency conversion which relies on an xml feed. http://www.white-hat-web-design.co.uk/articles/php-currency-conversion.php It grabs the data from the following feed... var $xml_file = "www.ecb.int/stats/eurofxref/eurofxref-daily.xml"; However this xml feed has limited currencies and i require currencies for the middle east. Does anyone know where i can find an xml file with middle east currencies or have any better suggestions? Any help would be appreciated.

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  • XML document being parsed as single element instead of sequence of nodes

    - by Rob Carr
    Given xml that looks like this: <Store> <foo> <book> <isbn>123456</isbn> </book> <title>XYZ</title> <checkout>no</checkout> </foo> <bar> <book> <isbn>7890</isbn> </book> <title>XYZ2</title> <checkout>yes</checkout> </bar> </Store> I am getting this as my parsed xmldoc: >>> from xml.dom import minidom >>> xmldoc = minidom.parse('bar.xml') >>> xmldoc.toxml() u'<?xml version="1.0" ?><Store>\n<foo>\n<book>\n<isbn>123456</isbn>\n</book>\n<t itle>XYZ</title>\n<checkout>no</checkout>\n</foo>\n<bar>\n<book>\n<isbn>7890</is bn>\n</book>\n<title>XYZ2</title>\n<checkout>yes</checkout>\n</bar>\n</Store>' Is there an easy way to pre-process this document so that when it is parsed, it isn't parsed as a single xml element?

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  • sql server 2000 and for xml explicit

    - by Marcin
    Hi everyone, I've got a problem with using for xml explicit in SQL Server 2000 (so I can't use the new path() stuff from sql 2005/8) Essentially I have two tables and the XML structure I want to have is <xml> <table_1 field1="foo" field2="foobar2" field3="foobar3"> <a_row_from_table_2 field1="goo" field2="goobar2" field3="goobar3" /> <a_row_from_table_2 field1="hoo" field2="hoobar2" field3="hoobar3" /> </table_1> </xml> That is, table_1 has a one-to-many relationship with table_2, and I want to make a hierarchy of it. So far I can't seem to get it, the closest I've managed to get is all the records from table1, with all the records from table2 appended to the very last element of table1 Any help with setting up this kind of relationship would be greatly appreciated. -Marcin

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  • Databinding question: DataGridView <=> XDocument (using LINQ-to-XML)

    - by Pretzel
    Learning LINQ has been a lot of fun so far, but despite reading a couple books and a bunch of online resources on the topic, I still feel like a total n00b. Recently, I just learned that if my query returns an Anonymous type, the DataGridView I'm populating will be ReadOnly (because, apparently Anonymous types are ReadOnly.) Right now, I'm trying to figure out the easiest way to: Get a subset of data from an XML file into a DataGridView, Allow the user to edit said data, Stick the changed data back into the XML file. So far I have Steps 1 and 2 figured out: public class Container { public string Id { get; set; } public string Barcode { get; set; } public float Quantity { get; set; } } // For use with the Distinct() operator public class ContainerComparer : IEqualityComparer<Container> { public bool Equals(Container x, Container y) { return x.Id == y.Id; } public int GetHashCode(Container obj) { return obj.Id.GetHashCode(); } } var barcodes = (from src in xmldoc.Descendants("Container") where src.Descendants().Count() > 0 select new Container { Id = (string)src.Element("Id"), Barcode = (string)src.Element("Barcode"), Quantity = float.Parse((string)src.Element("Quantity").Attribute("value")) }).Distinct(new ContainerComparer()); dataGridView1.DataSource = barcodes.ToList(); This works great at getting the data I want from the XML into the DataGridView so that the user has a way to manipulate the values. Upon doing a Step-thru trace of my code, I'm finding that the changes to the values made in DataGridView are not bound to the XDocument object and as such, do not propagate back. How do we take care of Step 3? (getting the data back to the XML) Is it possible to Bind the XML directly to the DataGridView? Or do I have to write another LINQ statement to get the data from the DGV back to the XDocument? Suggstions?

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  • Parsing a UTF-16 encoded xml file in ruby with REXML

    - by Matthew Toohey
    Hello, I'm trying to parse the following UTF-16 encoded xml file in REXML: http://www.abc.net.au/triplej/feeds/playout/triplejsydneyplayout.xml?_523525 REXML encounters an error after the following: >> require 'rexml/document' => true >> include REXML => Object >> require 'net/http' => true >> triplejString = Net::HTTP.get('www.abc.net.au', '/triplej/feeds/playout/triplejsydneyplayout.xml?_523525') => "\377\376<\000?\000x\000m\000l\000 \000v\000e\000r\000s\000i\000o\000n\000=\000\"\0001\000.\0000\000\"\000 \000e\000n\000c\000o\000d\000i\000n\000g\000=\000\"\000u\000t\000f\000-\0001\0006\000\"\000?\000>\000<\000a\000b\000c\000m\000u\000s\000i\000c\000_\000p\000l\000a\000y\000o\000u\000t\000>\000<\000c\000h\000a\000n\000n\000e\000l\000>\000J\000J\000J\000<\000/\000c\000h\000a\000n\000n\000e\000l\000>\000<\000p\000u\000b\000l\000i\000s\000h\000t\000i\000m\000e\000>\000F\000r\000i\000,\000 \0003\0000\000 \000A\000p\000r\000 \0002\0000\0001\0000\000 \0001\0001\000:\0005\0007\000:\0001\0007\000 \000G\000M\000T\000<\000/\000p\000u\000b\000l\000i\000s\000h\000t\000i\000m\000e\000>\000<\000i\000t\000e\000m\000s\000>\000<\000i\000t\000e\000m\000>\000<\000p\000l\000a\000y\000i\000n\000g\000>\000n\000o\000w\000<\000/\000p\000l\000a\000y\000i\000n\000g\000>\000<\000t\000i\000t\000l\000e\000>\000D\000o\000c\000t\000o\000r\000,\000 \000D\000o\000c\000t\000o\000r\000<\000/\000t\000i\000t\000l\000e\000>\000<\000t\000r\000a\000c\000k\000i\000d\000>\000<\000/\000t\000r\000a\000c\000k\000i\000d\000>\000<\000p\000l\000a\000y\000e\000d\000t\000i\000m\000e\000>\000F\000r\000i\000,\000 \0003\0000\000 \000A\000p\000r\000 \0002\0000\0001\0000\000 \0001\0001\000:\0005\0007\000:\0001\0007\000 \000G\000M\000T\000<\000/\000p\000l\000a\000y\000e\000d\000t\000i\000m\000e\000>\000<\000p\000u\000b\000l\000i\000s\000h\000e\000r\000>\000<\000/\000p\000u\000b\000l\000i\000s\000h\000e\000r\000>\000<\000d\000a\000t\000e\000c\000o\000p\000y\000r\000i\000g\000h\000t\000e\000d\000>\0002\0000\0000\0003\000<\000/\000d\000a\000t\000e\000c\000o\000p\000y\000r\000i\000g\000h\000t\000e\000d\000>\000<\000d\000u\000r\000a\000t\000i\000o\000n\000>\0001\0006\0003\000<\000/\000d\000u\000r\000a\000t\000i\000o\000n\000>\000<\000a\000u\000s\000t\000>\000N\000o\000<\000/\000a\000u\000s\000t\000>\000<\000t\000r\000a\000c\000k\000n\000o\000t\000e\000>\000<\000/\000t\000r\000a\000c\000k\000n\000o\000t\000e\000>\000<\000t\000r\000a\000c\000k\000l\000i\000n\000k\000>\000<\000/\000t\000r\000a\000c\000k\000l\000i\000n\000k\000>\000<\000s\000h\000o\000w\000>\000<\000/\000s\000h\000o\000w\000>\000<\000t\000a\000l\000e\000n\000t\000>\000<\000/\000t\000a\000l\000e\000n\000t\000>\000<\000a\000l\000b\000u\000m\000>\000<\000a\000l\000b\000u\000m\000n\000a\000m\000e\000>\000D\000r\000i\000v\000i\000n\000g\000 \000F\000o\000r\000 \000T\000h\000e\000 \000S\000t\000o\000r\000m\000/\000D\000o\000c\000t\000o\000r\000 \000D\000o\000c\000t\000o\000r\000<\000/\000a\000l\000b\000u\000m\000n\000a\000m\000e\000>\000<\000a\000l\000b\000u\000m\000i\000d\000>\0008\0003\000-\0004\0002\0002\0006\0009\000<\000/\000a\000l\000b\000u\000m\000i\000d\000>\000<\000a\000l\000b\000u\000m\000i\000m\000a\000g\000e\000>\000h\000t\000t\000p\000:\000/\000/\000w\000w\000w\000.\000a\000b\000c\000.\000n\000e\000t\000.\000a\000u\000/\000t\000r\000i\000p\000l\000e\000j\000/\000c\000o\000v\000e\000r\000s\000/\000G\000y\000r\000o\000s\000c\000o\000p\000e\000 \000-\000 \000D\000r\000i\000v\000i\000n\000g\000 \000F\000o\000r\000 \000T\000h\000e\000 \000S\000t\000o\000r\000m\000/\000D\000o\000c\000t\000o\000r\000 \000D\000o\000c\000t\000o\000r\000 \000(\0002\0000\0000\0003\000)\000.\000j\000p\000g\000<\000/\000a\000l\000b\000u\000m\000i\000m\000a\000g\000e\000>\000<\000/\000a\000l\000b\000u\000m\000>\000<\000a\000r\000t\000i\000s\000t\000>\000<\000a\000r\000t\000i\000s\000t\000n\000a\000m\000e\000>\000G\000y\000r\000o\000s\000c\000o\000p\000e\000<\000/\000a\000r\000t\000i\000s\000t\000n\000a\000m\000e\000>\000<\000a\000r\000t\000i\000s\000t\000i\000d\000>\000<\000/\000a\000r\000t\000i\000s\000t\000i\000d\000>\000<\000a\000r\000t\000i\000s\000t\000n\000o\000t\000e\000>\000<\000/\000a\000r\000t\000i\000s\000t\000n\000o\000t\000e\000>\000<\000a\000r\000t\000i\000s\000t\000l\000i\000n\000k\000>\000<\000/\000a\000r\000t\000i\000s\000t\000l\000i\000n\000k\000>\000<\000/\000a\000r\000t\000i\000s\000t\000>\000<\000/\000i\000t\000e\000m\000>\000<\000/\000i\000t\000e\000m\000s\000>\000<\000/\000a\000b\000c\000m\000u\000s\000i\000c\000_\000p\000l\000a\000y\000o\000u\000t\000>\000" >> xmlDoc = REXML::Document.new(triplejString) REXML::ParseException: #<REXML::ParseException: malformed XML: missing tag start Line: Position: Last 80 unconsumed characters: <?xml version="1.0" encoding="utf-16"?><a> /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/rexml/parsers/baseparser.rb:356:in `pull' /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/rexml/parsers/treeparser.rb:22:in `parse' /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/rexml/document.rb:227:in `build' /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/rexml/document.rb:43:in `initialize' (irb):19:in `new' (irb):19:in `irb_binding' /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/irb/workspace.rb:52:in `irb_binding' /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/irb/workspace.rb:52 ... malformed XML: missing tag start Line: Position: Last 80 unconsumed characters: <?xml version="1.0" encoding="utf-16"?><a Line: Position: Last 80 unconsumed characters: <?xml version="1.0" encoding="utf-16"?><a from /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/rexml/parsers/treeparser.rb:92:in `parse' from /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/rexml/document.rb:227:in `build' from /System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/1.8/rexml/document.rb:43:in `initialize' from (irb):19:in `new' from (irb):19 Any ideas?

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