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• wmic output well formed xml on remote queries

  - by Mervin

  I want to use the WMI command line tool (wmic) to get information about windows computers on the network and output it as valid xml. However, I can't seem to find the right way to do this as the outputted xml currently contains invalid tokens for which I think I should use the /TRANSLATE:basicxml switch. The command: wmic /NODE:"tech-demo" /IMPLEVEL:Impersonate /USER:MyUser /PASSWORD:MyPassword /PRIVILEGES:DISABLE /AUTHLEVEL:Pkt /AUTHORITY:"ntlmdomain:companydomain.local" PATH Win32_LogicalDisk GET * /FORMAT:rawxml This command runs but returns invalid xml tokens ('<' and '' I think? edit: it appears to fail parsing at ‹) When I add the translate switch I get the message: Can not use credentials for local connections a bit strange that it tries to query the local pc when I add the switch.. Help would be greatly appreciated.

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• Consume an XML Feed with PowerPoint 2010

  - by Matt Schweers

  Hi there. I'm looking for a way to consume an XML feed from a web-service directly into PowerPoint 2010. I found the LiveWeb plugin (http://skp.mvps.org/liveweb.htm) for PowerPoint that, while pretty cool, really only pulls in actual web content in a way that feels more like an iframe. Ideally, I would like to consume raw XML web service/feed with PowerPoint, parse it, and stylize the results. Is this possible? Even reading from a static XML file would be a good start.

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• How to Read XML and Generate SQL Insert

  - by hackerkatt

  I am trying to write a VB Script to read a XML file (downloaded daily) and insert the information into a MSSQL DB. The content of the XML is a list if CDRs (Call Data Records). I need to parse the file and insert the cdr's into a table. I'm a Ruby,Perl,PHP,Javascript,SQL,... programer. But I've really never written any VB Script. I've done some googling and find a number of examples on how to generate XML from a SQL Query, but not the reverse. Any help/suggestions would be greatly appreciated. Thank you!

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• Django attribute error: 'module' object has no attribute 'is_usable'

  - by Robert A Henru

  Hi, I got the following error when calling the url in Django. It's working before, I guess it's related with some accidental changes I made, but I have no idea what they are. Thanks before for the help, Robert Environment: Request Method: GET Request URL: http://localhost:8000/time/ Django Version: 1.2 Python Version: 2.6.1 Installed Applications: ['django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'django.contrib.messages', 'django.contrib.admin', 'djlearn.books'] Installed Middleware: ('django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.middleware.csrf.CsrfViewMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'django.contrib.messages.middleware.MessageMiddleware') Traceback: File "/Library/Python/2.6/site-packages/django/core/handlers/base.py" in get_response 100. response = callback(request, *callback_args, **callback_kwargs) File "/Users/rhenru/Workspace/django/djlearn/src/djlearn/../djlearn/views.py" in current_datetime 16. return render_to_response('current_datetime.html',{'current_date':now,}) File "/Library/Python/2.6/site-packages/django/shortcuts/__init__.py" in render_to_response 20. return HttpResponse(loader.render_to_string(*args, **kwargs), **httpresponse_kwargs) File "/Library/Python/2.6/site-packages/django/template/loader.py" in render_to_string 181. t = get_template(template_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in get_template 157. template, origin = find_template(template_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in find_template 128. loader = find_template_loader(loader_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in find_template_loader 111. if not func.is_usable: Exception Type: AttributeError at /time/ Exception Value: 'module' object has no attribute 'is_usable'

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• Linq-to-XML explicit casting in a generic method

  - by vlad

  I've looked for a similar question, but the only one that was close didn't help me in the end. I have an XML file that looks like this: <Fields> <Field name="abc" value="2011-01-01" /> <Field name="xyz" value="" /> <Field name="tuv" value="123.456" /> </Fields> I'm trying to use Linq-to-XML to get the values from these fields. The values can be of type Decimal, DateTime, String and Int32. I was able to get the fields one by one using a relatively simple query. For example, I'm getting the 'value' from the field with the name 'abc' using the following: private DateTime GetValueFromAttribute(IEnumerable<XElement> fields, String attName) { return (from field in fields where field.Attribute("name").Value == "abc" select (DateTime)field.Attribute("value")).FirstOrDefault() } this is placed in a separate function that simply returns this value, and everything works fine (since I know that there is only one element with the name attribute set to 'abc'). however, since I have to do this for decimals and integers and dates, I was wondering if I can make a generic function that works in all cases. this is where I got stuck. here's what I have so far: private T GetValueFromAttribute<T>(IEnumerable<XElement> fields, String attName) { return (from field in fields where field.Attribute("name").Value == attName select (T)field.Attribute("value").Value).FirstOrDefault(); } this doesn't compile because it doesn't know how to convert from String to T. I tried boxing and unboxing (i.e. select (T) (Object) field.Attribute("value").Value but that throws a runtime Specified cast is not valid exception as it's trying to convert the String to a DateTime, for instance. Is this possible in a generic function? can I put a constraint on the generic function to make it work? or do I have to have separate functions to take advantage of Linq-to-XML's explicit cast operators?

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• Android app crashes when I change the default xml layout file to another

  - by mib1413456

  I am currently just starting to learn android development and have created a basic "Hello world" app that uses "activity_main.xml" for the default layout. I tried to create a new layout xml file called "new_layout.xml" with a text view, a text field and a button and did the following changes in the MainActivity.java file: setContentView(R.layout.new_layout); I did nothing else expect for adding a new_layout.xml in the res/layout folder, I have tried restarting and cleaning the project but nothing. Below is my activity_main.xml file, new_layout.xml file and MainActivity.java activity_main.xml: <FrameLayout xmlns:android="http://schemas.android.com/apk/res/android" xmlns:tools="http://schemas.android.com/tools" android:id="@+id/container" android:layout_width="match_parent" android:layout_height="match_parent" tools:context="org.example.androidsdk.demo.MainActivity" tools:ignore="MergeRootFrame" /> new_layout.xml: <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="match_parent" android:layout_height="match_parent" android:orientation="horizontal" > <TextView android:id="@+id/textView1" android:layout_width="wrap_content" android:layout_height="wrap_content" android:text="TextView" /> <EditText android:id="@+id/editText1" android:layout_width="wrap_content" android:layout_height="wrap_content" android:layout_weight="1" android:ems="10" > <requestFocus /> </EditText> <Button android:id="@+id/button1" android:layout_width="wrap_content" android:layout_height="wrap_content" android:text="Button" /> MainActivity.java file package org.example.androidsdk.demo; import android.app.Activity; import android.app.ActionBar; import android.app.Fragment; import android.os.Bundle; import android.view.LayoutInflater; import android.view.Menu; import android.view.MenuItem; import android.view.View; import android.view.ViewGroup; import android.os.Build; public class MainActivity extends Activity { @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.new_layout); if (savedInstanceState == null) { getFragmentManager().beginTransaction() .add(R.id.container, new PlaceholderFragment()) .commit(); } } @Override public boolean onCreateOptionsMenu(Menu menu) { // Inflate the menu; this adds items to the action bar if it is present. getMenuInflater().inflate(R.menu.main, menu); return true; } @Override public boolean onOptionsItemSelected(MenuItem item) { // Handle action bar item clicks here. The action bar will // automatically handle clicks on the Home/Up button, so long // as you specify a parent activity in AndroidManifest.xml. int id = item.getItemId(); if (id == R.id.action_settings) { return true; } return super.onOptionsItemSelected(item); } /** * A placeholder fragment containing a simple view. */ public static class PlaceholderFragment extends Fragment { public PlaceholderFragment() { } @Override public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) { View rootView = inflater.inflate(R.layout.fragment_main, container, false); return rootView; } } }

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• Is it possible to use .properties files in web.xml in conjunction with contextConfigLocation paramet

  - by Vladimir

  Here is part of my web.xml: <context-param> <param-name>contextConfigLocation</param-name> <param-value> classpath:application-config.xml </param-value> </context-param> application-config.xml uses property placeholder: <context:property-placeholder location="classpath:properties/db.properties"/> Is it possible somehow to define which properties file to use in web.xml rather than in application-config.xml?

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• How to select visibility of specific worksheet at the opening of Excel using Spreadsheet XML

  - by user211607

  Hi, I am getting spreadsheet xml from a code logic (Flex Grids to spreadsheet xml). I have 3 worksheets (A, B, C) in that spreadsheet xml. I am opening this spreadsheet xml in Excel. I want to view worksheet B when I am opening the spreadhseet xml in Excel. Is there any tag/code, I need to add so that worksheet B will be visible at initial? I can add that code in code logic Thanks ... Atul

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• Print XML file and download it

  - by Pankaj

  I am create xml using serialization and trying to print them using this code string xmlDate = xml.GetXML(); string name = string.Format("{0}_HighEST", ProjectName); Response.AddHeader("Content-disposition", "attachment; filename=\"" + name + "_HighEST.xml\""); Response.ContentType = string.Format("application/.xml", name); how can assign my xmldata to Response so that data will write on downloaded xml?

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• Parsing secure entries XML file with jquery

  - by user573131

  Apologies if this is elementary. I'm primarily a front end designer/dev. I have webform through a form service called wufoo. Wufoo generates a lovely XML (or json) file that can be grabed and parsed. I'm trying to grab the entries xml feed that is associated with the form and parse it via jquery to show who has entered. Im using the following code (which works with a local xml file). http://bostonwebsitemakeover.com/2/test <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.js"></script> <script> $(document).ready(function () { $.ajax({ type: "GET", url: "people.xml", dataType: "xml", success: xmlParser }); }); function xmlParser(xml) { $('#load').fadeOut(); $(xml).find("Entry").each(function () { $(".main").append('<div class="entry">' + $(this).find("Field1").text() + ' ' + $(this).find("Field2").text() + ' http://twitter.com/' + $(this).find("Field17").text() + '</div>'); $(".entry").fadeIn(1000); }); } </script> My XML file contains the following: <?xml version="1.0"?> <Entries> <Entry> <EntryId>1</EntryId> <Field1>Meaghan</Field1> <Field2>Severson</Field2> <Field17/> </Entry> <Entry> <EntryId>2</EntryId> <Field1>Michael</Field1> <Field2>Flint</Field2> <Field17>michaelflint</Field17> </Entry> <Entry> <EntryId>3</EntryId> <Field1>Niki</Field1> <Field2>Brown</Field2> <Field17>nikibrown</Field17> </Entry> <Entry> <EntryId>4</EntryId> <Field1>Niki</Field1> <Field2>Brown</Field2> <Field17>nikibrown</Field17> </Entry> </Entries> I'm wondering how I would do this with the xml file hosted on the wufoo (which is https) So I guess Im asking how do I authenticate the feed via jquery? Or do i need to do this via json? Could someone explain how?

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• Which XML library for what purposes?

  - by John Mee

  A search for "python" and "xml" returns a variety of libraries for combining the two. This list probably faulty: xml.dom xml.etree xml.sax xml.parsers.expat PyXML beautifulsoup? HTMLParser htmllib sgmllib Be nice if someone can offer a quick summary of when to use which, and why.

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• How to replace invalid characters in XML using Javascript or PhP

  - by Raind

  Hi, Need help here for the following: Running PhP, javascript, MySQL, XML. 1) Retrieving file from MySQL and stored it onto XML file. 2) Use javascript function to load XML file (that stored those data). 3) It produces invalid characters in XML file. STEP 1 : Sample of the code in PhP - Loading MySQL DB to store data onto XML file $file= fopen("MapDeals2.xml", "w"); $_xml ="\n"; $_xml .="\n"; while($row1_ThisWeek = mysql_fetch_array($result1_ThisWeek)) { $rRName = $row1_ThisWeek['Retailer_Name']; $rRAddress = $row1_ThisWeek['Retailer_Address1']; $rRAddressPostCode = $row1_ThisWeek['Retailer_AddressPostCode1']; } $_xml .= "<DEAL>\n"; $_xml .= "<DealDescription>" . $d_Description . "</DealDescription>\n"; $_xml .= "<DealURL>" . $d_URL . "</DealURL>\n"; $_xml .= "<DealRName>" . $rRName . "</DealRName>\n"; $_xml .= "<DealRAddress>" . $rRAddress . "</DealRAddress>\n"; $_xml .= "<DealRPostCode>" . $rRAddressPostCode . "</DealRPostCode>\n"; $_xml .= "</DEAL>\n"; } } $_xml .="\n"; fwrite($file, $_xml); fclose($file); STEP 2 : Sample of the code in Javscript - Loading XML file xhttp.open("GET","Test2.xml", false); xhttp.send(""); xmlDoc=xhttp.responseXML; var x=xmlDoc.getElementsByTagName("Employee"); parser = new DOMParser(); xmlDoc = parser.parseFromString("MapDeals2.xml", "text/xml"); for (i=0;i"; . . . } Is there a solution for the above? Looking forward to hear from you soon. Cheers

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• Set attribute to all child elements via xsl:choose

  - by Camal

  Hi, assuming I got following XML file : <?xml version="1.0" encoding="ISO-8859-1" ?> <MyCarShop> <Car gender="Boy"> <Door>Lamborghini</Door> <Key>Skull</Key> </Car> <Car gender="Girl"> <Door>Normal</Door> <Key>Princess</Key> </Car> </MyCarShop> I want to perform a transformation so the xml looks like this : <?xml version="1.0" encoding="ISO-8859-1" ?> <MyCarShop> <Car gender="Boy"> <Door color="blue">Lamborghini</Door> <Key color="blue">Skull</Key> </Car> <Car gender="Girl"> <Door color="red">Normal</Door> <Key color="red">Princess</Key> </Car> </MyCarShop> So I want to add a color attribut to each subelement of Car depending on the gender information. I came up with this XSLT but it doesnt work : <?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()"/> </xsl:copy> </xsl:template>--> <xsl:template match="/"> <xsl:element name="MyCarShop"> <xsl:attribute name="version">1.0</xsl:attribute> <xsl:apply-templates/> </xsl:element> </xsl:template> <xsl:template match="Car"> <xsl:element name="Car"> <xsl:apply-templates/> </xsl:element> </xsl:template> <xsl:template match="Door"> <xsl:element name="Door"> <xsl:attribute name="ViewSideIndicator"> <xsl:choose> <xsl:when test="gender = 'Boy' ">Front</xsl:when> <xsl:when test="gender = 'Girl ">Front</xsl:when> </xsl:choose> </xsl:attribute> </xsl:element> </xsl:template> <xsl:template match="Key"> <xsl:element name="Key"> <xsl:apply-templates/> </xsl:element> </xsl:template> </xsl:stylesheet> Does anybody know what might be wrong ? Thanks again!

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• Create attribute in XML

  - by user560411

  Hello. I have the following php code that adds data into XML and works correctly. However, in my second step I will create a form that deletes some of the elements. The problem is that I want to add an ID number and then the PHP file will search for it and delete entire node. My question is how can i add an ID into CD for this to work ? For example ( <cd id="xxxx"> ) Also, any ideas or examples of a code that deletes CD having the ID would be appreciate. insert.php ( my index file with the form ) <h1>Playlist</h1> <form action="insert2.php" method="post"> <fieldset> <label for="TITLE">TITLE:</label><input type="text" id="title" name="title" /><br /> <label for="title">BAND:</label> <input type="text" id="band" name="band"/><br /> <label for="path">YEAR:</label> <input type="text" id="year" name="year" /> <br /> <input type="submit" /> </fieldset> </form> <h2>Current entries:</h2> <p>TITLE - BAND - YEAR</p> <?php $doc = new DOMDocument(); $doc->load( 'insert.xml' ); $CATEGORIES = $doc->getElementsByTagName( "CD" ); foreach( $CATEGORIES as $CD ) { $TITLES = $CD->getElementsByTagName( "TITLE" ); $TITLE = $TITLES->item(0)->nodeValue; $BANDS= $CD->getElementsByTagName( "BAND" ); $BAND= $BANDS->item(0)->nodeValue; $YEARS = $CD->getElementsByTagName( "YEAR" ); $YEAR = $YEARS->item(0)->nodeValue; echo "<b>$TITLE - $BAND - $YEAR\n</b><br>"; } ?> inser2.php ( the main code ) <?php $CD = array( 'TITLE' => $_POST['title'], 'BAND' => $_POST['band'], 'YEAR' => $_POST['year'], ); $doc = new DOMDocument(); $doc->load( 'insert.xml' ); $doc->formatOutput = true; $r = $doc->getElementsByTagName("CATEGORIES")->item(0); $b = $doc->createElement("CD"); $TITLE = $doc->createElement("TITLE"); $TITLE->appendChild( $doc->createTextNode( $CD["TITLE"] ) ); $b->appendChild( $TITLE ); $BAND = $doc->createElement("BAND"); $BAND->appendChild( $doc->createTextNode( $CD["BAND"] ) ); $b->appendChild( $BAND ); $YEAR = $doc->createElement("YEAR"); $YEAR->appendChild( $doc->createTextNode( $CD["YEAR"] ) ); $b->appendChild( $YEAR ); $r->appendChild( $b ); $doc->save("insert.xml"); ?> the XML file <?xml version="1.0" encoding="utf-8"?> <MY_CD> <CATEGORIES> <CD> <TITLE>NEVER MIND THE BOLLOCKS</TITLE> <BAND>SEX PISTOLS</BAND> <YEAR>1977</YEAR> </CD> <CD> <TITLE>NEVERMIND</TITLE> <BAND>NIRVANA</BAND> <YEAR>1991</YEAR> </CD> </CATEGORIES> </MY_CD>

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• How to get the content of two xml tags even when they have another xml tags inside them (PHP-XML)

  - by Harry

  In simple terms, let's suppose you have the following xml structure: <TEXT>Well, I need some help as you <CUSTOMTAG>can</CUSTOMTAG> see.</TEXT> When extracting the text of this node in PHP with strip_tags(), I am not getting the content of tags. First step: What I want to do, is to extract and thus have the following string: "Well, I need some help as you can see." Second step: I would also like to convert the <CUSTOMTAG> and </CUSTOMTAG> to something else, like <e> and </e> for example, and finally have the following string: "Well, I need some help as you <e>can</e> see." I would appreciate only tested and working code. Thanks in advance! Kind Regards

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• How to add XmlInclude attribute dynamically

  - by Anindya Chatterjee

  I have the following classes [XmlRoot] public class AList { public List<B> ListOfBs {get; set;} } public class B { public string BaseProperty {get; set;} } public class C : B { public string SomeProperty {get; set;} } public class Main { public static void Main(string[] args) { var aList = new AList(); aList.ListOfBs = new List<B>(); var c = new C { BaseProperty = "Base", SomeProperty = "Some" }; aList.ListOfBs.Add(c); var type = typeof (AList); var serializer = new XmlSerializer(type); TextWriter w = new StringWriter(); serializer.Serialize(w, aList); } } Now when I try to run the code I got an InvalidOperationException at last line saying that The type XmlTest.C was not expected. Use the XmlInclude or SoapInclude attribute to specify types that are not known statically. I know that adding a [XmlInclude(typeof(C))] attribute with [XmlRoot] would solve the problem. But I want to achieve it dynamically. Because in my project class C is not known prior to loading. Class C is being loaded as a plugin, so it is not possible for me to add XmlInclude attribute there. I tried also with TypeDescriptor.AddAttributes(typeof(AList), new[] { new XmlIncludeAttribute(c.GetType()) }); before var type = typeof (AList); but no use. It is still giving the same exception. Does any one have any idea on how to achieve it?

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• Loading attribute in XPATH, problem

  - by Nguyen Quoc Hung

  I've got a question about loading attribute in XPATH. I write short XML code to test: <?xml version="1.0" encoding="iso-8859-1"?> <?xml-stylesheet type="text/xsl" href="testDate.xsl"?> <element attribute="1/1/2100"> Hung </element> My XSL code: <?xml version="1.0" encoding="iso-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <!--Handle the document: set up HTML page--> <xsl:template match="/"> <html> <head> </head> <body> This is a test <xsl:value-of select="element@attribute"/> </body> </html> </xsl:template> </xsl:stylesheet> Why it produces an error when loading the stylesheet? Would you please help me explain this? Thank you

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• How to spread XML Sitemaps over several webservers behind AWS loadbalancer?

  - by Jurik

  We have a web portal with almost a million products and way more other urls. I wrote a script that checks database. If there is a new url needed or an old one update, this script will update/create the XML Sitemaps. But we have several servers behind the load balancer at our rented AWS space. Further this script checks database for each url if there was an update so that it updates the appropriate xml file too. My question is how to spread those XML Sitemaps over all webservers behind this AWS load balancer? Our approaches/ideas: we could just generate them on one server with a cron job and copy them to the other servers, but this could be difficult because of automatic raising numbers of servers and so on. we put them on our S3 - but this one is not avaible thru our domain, so I guess google will have a problem with it I let my script run on every webserver but change it in a way that it will generate each time all xml files if they do not exist. But then I would have conflicts with updated URLs in my database, where I saved timestamp of last changed value of every url Is there another - better - solution that I do not know? Are there any special services by amazon for such cases?

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• Create attribute in existing XML

  - by user560411

  Hello. I have the following php code that inserts data into XML and works correctly. However, I want to add an ID number like below <CD id="xxxx"> My question is how can i add an ID into CD for this to work ? I use form to parse the id. ** the main code ** <?php $CD = array( 'TITLE' => $_POST['title'], 'BAND' => $_POST['band'], 'YEAR' => $_POST['year'], ); $doc = new DOMDocument(); $doc->load( 'insert.xml' ); $doc->formatOutput = true; $r = $doc->getElementsByTagName("CATEGORIES")->item(0); $b = $doc->createElement("CD"); $TITLE = $doc->createElement("TITLE"); $TITLE->appendChild( $doc->createTextNode( $CD["TITLE"] ) ); $b->appendChild( $TITLE ); $BAND = $doc->createElement("BAND"); $BAND->appendChild( $doc->createTextNode( $CD["BAND"] ) ); $b->appendChild( $BAND ); $YEAR = $doc->createElement("YEAR"); $YEAR->appendChild( $doc->createTextNode( $CD["YEAR"] ) ); $b->appendChild( $YEAR ); $r->appendChild( $b ); $doc->save("insert.xml"); ?> the XML file <?xml version="1.0" encoding="utf-8"?> <MY_CD> <CATEGORIES> <CD> <TITLE>NEVER MIND THE BOLLOCKS</TITLE> <BAND>SEX PISTOLS</BAND> <YEAR>1977</YEAR> </CD> <CD> <TITLE>NEVERMIND</TITLE> <BAND>NIRVANA</BAND> <YEAR>1991</YEAR> </CD> </CATEGORIES> </MY_CD>

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• Avoid extra xmlns:xsi while adding an attribute to XML root

  - by Rakesh kumar

  I am creating an xml file using this code: XmlDocument d = new XmlDocument(); XmlDeclaration xmlDeclaration = d.CreateXmlDeclaration("1.0", "utf-8", null); d.InsertBefore(xmlDeclaration,d.DocumentElement); XmlElement root = d.CreateElement("ITRETURN","ITR","http://incometaxindiaefiling.gov.in/ITR1"); XmlAttribute xsiNs = d.CreateAttribute("xsi:schemaLocation", "http://www.w3.org/2001/XMLSchema-instance"); xsiNs.Value = "http://incometaxindiaefiling.gov.in/main ITRMain10.xsd"; root.SetAttributeNode(xsiNs); //root.SetAttribute("xsi:schemaLocation", "http://www.w3.org/2001/XMLSchema-instance"); root.SetAttribute("xmlns:ITR1FORM", "http://incometaxindiaefiling.gov.in/ITR1"); root.SetAttribute("xmlns:ITR2FORM", "http://incometaxindiaefiling.gov.in/ITR2"); root.SetAttribute("xmlns:ITR3FORM", "http://incometaxindiaefiling.gov.in/ITR3"); root.SetAttribute("xmlns:ITR4FORM", "http://incometaxindiaefiling.gov.in/ITR4"); d.AppendChild(root); d.Save("c://myxml.xml"); and I am getting an output like this <?xml version="1.0" encoding="utf-8" ?> - <!-- Sample XML file generated by XMLSpy v2007 sp2 (http://www.altova.com) --> - <ITRETURN:ITR xsi:schemaLocation="http://incometaxindiaefiling.gov.in/main ITRMain10.xsd" xmlns:ITR1FORM="http://incometaxindiaefiling.gov.in/ITR1" xmlns:ITR2FORM="http://incometaxindiaefiling.gov.in/ITR2" xmlns:ITR3FORM="http://incometaxindiaefiling.gov.in/ITR3" xmlns:ITR4FORM="http://incometaxindiaefiling.gov.in/ITR4" xmlns:xsi="http://incometaxindiaefiling.gov.in/main ITRMain10.xsd" xmlns:ITRETURN="http://incometaxindiaefiling.gov.in/ITR1"> <ITR1FORM:ITR1>root node</ITR1FORM:ITR1> </ITRETURN:ITR> But my requirement is like <ITRETURN:ITR xsi:schemaLocation="http://incometaxindiaefiling.gov.in/main ITRMain10.xsd" xmlns:ITR1FORM="http://incometaxindiaefiling.gov.in/ITR1" xmlns:ITR2FORM="http://incometaxindiaefiling.gov.in/ITR2" xmlns:ITR3FORM="http://incometaxindiaefiling.gov.in/ITR3" xmlns:ITR4FORM="http://incometaxindiaefiling.gov.in/ITR4" xmlns:ITRETURN="http://incometaxindiaefiling.gov.in/main" xmlns:ITRForm="http://incometaxindiaefiling.gov.in/master" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> + <ITR1FORM:ITR1>any value </ITR1FORM:ITR1> </ITRETURN:ITR> What do I need to change in my code to get the desired output?

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• XMLOutputStream, repairing namespaces, and attributes without namespaces

  - by comment_bot

  A simple task: write an element with an unnamespaced attribute: String nsURI = "http://example.com/"; XMLOutputFactory outF = XMLOutputFactory.newFactory(); outF.setProperty(XMLOutputFactory.IS_REPAIRING_NAMESPACES, true); XMLStreamWriter out = outF.createXMLStreamWriter(System.out); out.writeStartElement(XMLConstants.DEFAULT_NS_PREFIX, "element", nsURI); out.writeAttribute(XMLConstants.DEFAULT_NS_PREFIX, XMLConstants.NULL_NS_URI, "attribute", "value"); out.writeEndElement(); out.close(); Woodstox's answer: <element xmlns="http://example.com/" attribute="value"></element> JDK 6 answer: <zdef-159241566:element xmlns="" xmlns:zdef-159241566="http://example.com/" attribute="value"></zdef-159241566:element> What?! Further, if we add a prefix to the element: out.writeStartElement(ns, "element", nsURI); JDK 6 no longer attempts to emit xmlns="": <ns:element xmlns:ns="http://example.com/" attribute="value"></ns:element> I'm fairly sure this is a bug in JDK 6. Am I right? And could anyone suggest a work around that will keep both libraries (and any others) happy? I don't want to require woodstox if I can help it.

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• XMl to XHTML using XSL with namespace

  - by user256007

  How to handle XML Elements with namespace in an XSL. Here goes my XML Document <?xml version="1.0" encoding="UTF-8"?> <?xml-stylesheet type="text/xml" href="wpanel.xsl" ?> <html xmlns="http://www.w3.org/1999/xhtml" xmlns:bong="http://bong/glob/wpanel" xml:lang="en" lang="en"> <head> <title> Transitional DTD XHTML Example </title> <link rel="stylesheet" type="text/css" href="wpanel.css" /> </head> <body> <bong:QButton text="Submit"></bong:QButton> <bong:QButton text="Reset"></bong:QButton> </body> </html> and My XSL goes here <?xml version="1.0" ?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:bong="http://bong/glob/wpanel" xmlns:fn="http://www.w3.org/2005/xpath-functions" > <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" /> <xsl:template match="*|@*"> <xsl:copy> <xsl:apply-templates select="@*"/> <xsl:apply-templates/> </xsl:copy> </xsl:template> <xsl:template match="processing-instruction()|comment()"> <xsl:copy>.</xsl:copy> </xsl:template> I want to copy the elements without bong namespace as it is But transform those using the bong namespace. I think there will be an xsl element or attribute for that.I've serched the net. but haven't found yet. Thank You.

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• Parsing XML via jQuery, nested loops

  - by Coughlin

  I am using jQuery to parse XML on my page using $.ajax(). My code block is below and I can get this working to display say each result on the XML file, but I am having trouble because each section can have MORE THAN ONE and im trying to print ALL grades that belong to ONE STUDENT. Here is an example of the XML. <student num="505"> <name gender="male">Al Einstein</name> <course cid="1">60</course> <course cid="2">60</course> <course cid="3">40</course> <course cid="4">55</course> <comments>Lucky if he makes it to lab, hopeless.</comments> </student> Where you see the I am trying to get the results to print the grades for EACH student in each course. Any ideas on what I would do? Thanks, Ryan $.ajax({ type: "GET", url: "final_exam.xml", dataType: "xml", success: function(xml) { var student_list = $('#student-list'); $(xml).find('student').each(function(){ $(xml).find('course').each(function(){ gradeArray = $(this).text(); console.log(gradeArray); }); var name = $(this).find("name").text(); var grade = $(this).find("course").text(); var cid = $(this).find("course").attr("cid"); //console.log(cid); student_list.append("<tr><td>"+name+"</td><td>"+cid+"</td><td>"+grade+"</td></tr>"); }); } });

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• XML::LibXML::XPathContext - Question

  - by sid_com

  This script works with and without XPathContext. Why should I use it with XPathContext? #!/usr/bin/env perl use warnings; use strict; use XML::LibXML; use 5.012; my $parser = XML::LibXML->new; my $doc = $parser->parse_string(<<EOT); <?xml version="1.0"?> <xml> Text im Dokument <element id="myID" name="myname" style="old" /> <object objid="001" objname="Object1" /> <element id="002" name="myname" /> </xml> EOT #/ # without XPathContext my $nodes = $doc->findnodes( '/xml/element[@id=002]' ); # with XPathContext #my $root = $doc->documentElement; #my $xc = XML::LibXML::XPathContext->new( $root ); #my $nodes = $xc->findnodes( '/xml/element[@id=002]' ); for my $node ( $nodes->get_nodelist ) { say "Node: ", $node->nodeName; print "Attribute: "; print $_->getName, '=', $_->getValue, ' ' for $node->attributes; say ""; }

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• Update element values using xml.dom.minidom

  - by amnesia-55

  Hello, I have an XML structure which looks similar to: <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> Using xml.dom.minidom only (restrictions) i would like to 1)traverse through the XML file 2)Search/Get for particular element, depending on its parent Example: checkout element for author1, isbn for author2 3)Change/Set that element's value 4)Write the new XML structure to a file Can anyone help here? Thank you! UPDATE: This is what i have done till now import xml.dom.minidom checkout = "yes" def getLoneChild(node, tagname): assert ((node is not None) and (tagname is not None)) elem = node.getElementsByTagName(tagname) if ((elem is None) or (len(elem) != 1)): return None return elem def getLoneLeaf(node, tagname): assert ((node is not None) and (tagname is not None)) elem = node.getElementsByTagName(tagname) if ((elem is None) or (len(elem) != 1)): return None leaf = elem[0].firstChild if (leaf is None): return None return leaf.data def setcheckout(node, tagname): assert ((node is not None) and (tagname is not None)) child = getLoneChild(node, 'foo') Check = getLoneLeaf(child[0],'checkout') Check = tagname return Check doc = xml.dom.minidom.parse('test.xml') root = doc.getElementsByTagName('Store')[0] output = setcheckout(root, checkout) tmp_config = '/tmp/tmp_config.xml' fw = open(tmp_config, 'w') fw.write(doc.toxml()) fw.close()

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