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  • Parsing mutiple xml files in android

    - by cppdev
    Hi, I ma writing an application where I have to parse multiple xml files as a response from a server. Till now, I have written different xml parser classes for each xml depepending on the tags present in different xmls ? Can I combine all xml parser classes and write a single xml parser that handles all different tags in different xmls ? Will it work ? Will combining xml parser classes add an overhead because xml parser will check for every tag irrespective of whether that is part of file or not ? Please help.

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  • XML testing in Rails - Fixed attributes order in Builder::XmlMarkup in ruby -

    - by Daniel Cukier
    I have the following test in my Rails Application: it "should validate xml" do builder = Builder::XmlMarkup.new builder.server(:name => "myServer", :ip => "192.168.1.1").should == "<server name=\"myServer\" ip=\"192.168.1.1\"/>" end The problem is that this test passes sometimes, because the order of the xml tag attributes is unpredictable. Is there a way to force this order? Is there any other easy way to build xml? This example is simplified, I have a big XML. My problem is that I want to do an integration test, which compares a WebService call with a fixed XML file. Otherwise, I would have to parse the xml and verify element by element in the XML.

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  • Browser Not Reading Entire XML File

    - by Chris
    I have an XML file that is written by a PHP script. The data for the XML file is gathered from several different RSS feeds. The PHP script is invoked every 5 minutes by a Cron Job. The PHP Script takes maybe 5-10 seconds to write the XML File. Here's the problem: After the XML file is written, I can open it through DreamWeaver and read everything just fine - but when I enter the XML File's URL into my Web Browser (IE or Firefox), I get a "XML Parsing Error: not well-formed" Error in the Browser. When I do View Source in the Browser, the XML file appears incomplete - but when I open the file directly off the server, it is complete. Anyone know what's going on here?

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  • XML and .NET: How to replace specific node with many other ones loaded from raw xml data

    - by Entrase
    Let's suppose we have an element like this one in the main xml file: <group name="gr1" filename="groups/gr1.xml"/>. The second file gr1.xml contains something like this: <item name="i1">Item one</item> <item name="i2">Item two</item> <item name="i3">Item three</item> Note that there is no XML declaration in gr1.xml, just plain items without single parent node. So… Which is the best way to replace <group/> with its <item/>s? I have already tried some things like manual enclosing of gr1.xml content into a single node with XML declaration and loading it into XmlDocument, but it doesn't look like a good solution.

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  • XSLT: How to get XML

    - by Jyotsna Sonawane
    I have a XSL that transforms one format of XML into another. In input XML I have a node with following value - which is actually a XML string if we replace &lt; with < (less than) for e.g. &lt;Paragraph&gt;&lt;Title&gt;&lt;!CDATA[Pour les nuits du 2012-10-01 - 2012-10-30]]&gt;&lt;/Title&gt;&lt;Text&gt;&lt;![CDATA[TAXES INCLUSES.]]&gt;&lt;/Text&gt;&lt;/Paragraph&gt; I want to have the content of otherInfo as a XML node in output XML. if I do , I do not get it as a XML node - it is output just as text. How can I make XSL output the content of otherInfo as XML node ?

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  • bash script to find/grep particular string in xml files in particular folders

    - by user3702188
    i have a problem at work, where i need to simplify the process how i searrh for logs. I would like to ask for help from experts here. We have different services for every channel. The structure is following: - root/channel_1/service_1/2014-05-21/file_54544654541.xml - root/channel_1/server2_2/2014-05-20/file_74272172.xml - root/channel_1/service_3/2014-05-22/file_45456546.xml - root/channel_2/service_4/2014-05-23/file_78754456.xml - root/channel_2/service_5/2014-05-24/file_546546546.xml my main problem is to find particular string in these xml files. Lets say, i know the channel name but i dont know the service name under which my particular string should be present. Also i know the date. So in search i want to enter the channel name the date and string. The search would be going via all service folders and looking for string only in all xml files under particular date folder and particular channel. any ideas for quickest and easiest solution to achieve this? Either by bash or perl? Any help would be appreciated thanks

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  • There is an error in XML document... When calling to web service

    - by Sigurjón Guðbergsson
    I have created a web service and a function in it that should return a list of 11thousand records retreived from a pervasive database Here is my function in the web service. [WebService(Namespace = "http://tempuri.org/")] [WebServiceBinding(ConformsTo = WsiProfiles.BasicProfile1_1)] [System.ComponentModel.ToolboxItem(false)] public class BBI : System.Web.Services.WebService { [WebMethod] public List<myObject> getAll() { List<myObject> result = new List<myObject>(); PsqlConnection conn = new PsqlConnection("Host=soemthing;Port=something;Database=something;Encoding=IBM861"); conn.Open(); string strSql = "select 0, 1, 2, 3, 4, 5 from something"; PsqlCommand DBCmd = new PsqlCommand(strSql, conn); PsqlDataReader myDataReader; myDataReader = DBCmd.ExecuteReader(); while (myDataReader.Read()) { myObject b = new myObject(); b.0 = Convert.ToInt32(myDataReader[0].ToString()); b.1 = myDataReader[1].ToString(); b.2 = myDataReader[2].ToString(); b.3 = myDataReader[3].ToString(); b.4 = myDataReader[4].ToString(); b.5 = myDataReader[5].ToString(); result.Add(b); } conn.Close(); myDataReader.Close(); return result; } } Then i add web reference to this web service in my client program and call the reference BBI. Then i call to the getAll function and get the error : There is an error in XML document (1, 63432). public List<BBI.myObject> getAll() { BBI.BBI bbi = new BBI.BBI(); List<BBI.myObject> allBooks = bbi.getAll().OfType<BBI.myObject>().ToList(); return allBooks; } Here is the total exception detail System.InvalidOperationException was unhandled by user code Message=There is an error in XML document (1, 71897). Source=System.Xml StackTrace: at System.Xml.Serialization.XmlSerializer.Deserialize(XmlReader xmlReader, String encodingStyle, XmlDeserializationEvents events) at System.Xml.Serialization.XmlSerializer.Deserialize(XmlReader xmlReader, String encodingStyle) at System.Web.Services.Protocols.SoapHttpClientProtocol.ReadResponse(SoapClientMessage message, WebResponse response, Stream responseStream, Boolean asyncCall) at System.Web.Services.Protocols.SoapHttpClientProtocol.Invoke(String methodName, Object[] parameters) at BBI.BBI.getAllBooks() in c:\WINDOWS\Microsoft.NET\Framework\v4.0.30319\Temporary ASP.NET Files\vefur\73db60db\a4ee31dd\App_WebReferences.jl1r8jv6.0.cs:line 252 at webServiceFuncions.getAllBooks() in c:\Documents and Settings\forritari\Desktop\Vefur - Nýr\BBI\trunk\Vefur\App_Code\webServiceFuncions.cs:line 59 InnerException: System.Xml.XmlException Message='', hexadecimal value 0x01, is an invalid character. Line 1, position 71897. Source=System.Xml LineNumber=1 LinePosition=71897 SourceUri="" StackTrace: at System.Xml.XmlTextReaderImpl.Throw(Exception e) at System.Xml.XmlTextReaderImpl.Throw(String res, String[] args) at System.Xml.XmlTextReaderImpl.Throw(Int32 pos, String res, String[] args) at System.Xml.XmlTextReaderImpl.ParseNumericCharRefInline(Int32 startPos, Boolean expand, StringBuilder internalSubsetBuilder, Int32& charCount, EntityType& entityType) at System.Xml.XmlTextReaderImpl.ParseCharRefInline(Int32 startPos, Int32& charCount, EntityType& entityType) at System.Xml.XmlTextReaderImpl.ParseText(Int32& startPos, Int32& endPos, Int32& outOrChars) at System.Xml.XmlTextReaderImpl.ParseText() at System.Xml.XmlTextReaderImpl.ParseElementContent() at System.Xml.XmlTextReaderImpl.Read() at System.Xml.XmlTextReader.Read() at System.Xml.XmlReader.ReadElementString() at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationReaderBBI.Read2_Book(Boolean isNullable, Boolean checkType) at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationReaderBBI.Read20_getAllBooksResponse() at Microsoft.Xml.Serialization.GeneratedAssembly.ArrayOfObjectSerializer35.Deserialize(XmlSerializationReader reader) at System.Xml.Serialization.XmlSerializer.Deserialize(XmlReader xmlReader, String encodingStyle, XmlDeserializationEvents events) InnerException: The database records are containing all kind of strange symbols, for example ¤rmann Kr. Einarsson and Tv” ‘fint˜ri Can someone see what im doing wrong here?

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  • C# Windows Service XML

    - by Goober
    Scenario I have a windows service written in C# that performs some processing based on parsing an XML file and use that data to carry out various tasks. The service also does various bits of logging - which uses settings from an APP.Config file. The Problem When the service is compiled, installed and run, the XML file seems to disappear. I'm getting the impression that it is just ignored or something like that. So far I've tried using TWO App.Config files, one named App.Config that contains settings for the service, and the other called MyService.exe.config that contains all of the data that was used in the XML file (the idea being that I can parse the XML from a config file that actually gets compiled and appears in my installation directory. However When I do this, all that happens is that ONE config file appears (with the name MyService.exe.config), but it contains the contents of the App.Config file and not the XML data that I want to parse. What I need All I want is to have a config file for my settings, and an XML file for my data. Question Is this possible? I know the application works as it was originally built as a console application that ran fine. Other The application has to be designed this way (as in, I need my data stored as XML, and my settings stored in a config file). Thoughts If I could somehow combine the contents of the two files into ONE config file, that would be one way of solving the problem. However, I have tried this and of course I get a "Type Initialisation Exception", as the config file cannot interprate the XML data (probably because the tags are custom and do not form any part of the config schema - or something like that). Ideas Please could someone explain to me if it is possible for me to have an XML file AND a config file that will actually be compiled and stored in my installation directory for the service when it is run? CODE Custom XML/Data Config File <?xml version="1.0" encoding="utf-8" ?> <configuration> <servers> <SV066930> <add name="Name" value = "SV066930" /> <processes> <SimonTest1> <add name="ProcessName" value="notepad.exe" /> <add name="CommandLine" value="C:\\WINDOWS\\system32\\notepad.exe C:\\WINDOWS\\Profiles\\TA2TOF1\\Desktop\\SimonTest1.txt" /> </SimonTest1> </processes> </SV066930> </servers> </configuration> APP.Config Settings File <?xml version="1.0" encoding="utf-8" ?> <configuration> <configSections> <section name="dataConfiguration" type="Microsoft.Practices.EnterpriseLibrary.Data.Configuration.DatabaseSettings, Microsoft.Practices.EnterpriseLibrary.Data, Version=4.0.0.0, Culture=neutral, PublicKeyToken=xxxxxxxxxxx" /> </configSections> <connectionStrings> <add name="DB" connectionString="Data Source=etc......" /> </connectionStrings> </configuration> Help greatly appreciated.

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  • Converting XML to RSS

    - by hkw
    I've got an xml feed that will be published in one location of our website, and would like to repurpose that for an RSS feed. A few different pages on the website will also be referencing the same xml - all of those transformations are set up and working. The base xml file (XMLTEST.xml) is using this structure: <POST> <item> <POST_ID>80000852</POST_ID> <POST_TITLE>title</POST_TITLE> <POST_CHANNEL>I</POST_CHANNEL> <POST_DESC>description</POST_DESC> <LINK>http://www...</LINK> <STOC>N</STOC> </item> </POST> I'm trying to transform the xml into an RSS feed using the following setup (feed.xml + rss.xsl) feed.xml: <?xml-stylesheet href="rss.xsl" type="text/xsl"?> <RSSChannels> <RSSChannel src="XMLTEST.xml"/> </RSSChannels> rss.xsl: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" omit-xml-declaration="yes" /> <xsl:template match="RSSChannels"> <rss version="2.0"> <channel> <title>site title</title> <link></link> <description>Site description...</description> <xsl:apply-templates /> </channel> </rss> </xsl:template> <xsl:template match="RSSChannel"> <xsl:apply-templates select="document(@src)" /> </xsl:template> <xsl:template match="item"> <xsl:choose> <xsl:when test="STOC = 'Y'"></xsl:when> <xsl:when test="POST_CHANNEL = 'I'"></xsl:when> <xsl:otherwise> <item> <title> <xsl:value-of select="*[local-name()='POST_TITLE']" /> </title> <link> <xsl:value-of select="*[local-name()='LINK']" /> </link> <description> <xsl:value-of select="*[local-name()='POST_DESC']" /> </description> </item> </xsl:otherwise> </xsl:choose> </xsl:template> </xsl:stylesheet> When I try to view the output of feed.xml in Firefox, all of the filtering is applied correctly (sorting out items that shouldn't be published in that channel), but the page outputs as plain text, rather than the usual feed-detection that takes place in Firefox. Any ideas on what I'm missing? Thanks for any suggestions you can provide.

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  • Replacing innertext of XML node using PHP DOMDocument

    - by Rohan Kumar
    I want to replace innertext of a XML node my XML file named test.xml is <?xml version="1.0" encoding="utf-8"?> <ads> <loop>no</loop> <item> <description>Description 1</description> </item> <item> <description>Text in item2</description> </item> <item> <description>Let play with this XML</description> </item> </ads> I want to change the value of loop and description tag both, and it should be saved in test.xml like: <?xml version="1.0" encoding="utf-8"?> <ads> <loop>yes</loop> <item> <description>Description Changing Here</description> </item> <item> <description>Changing text in item2</description> </item> <item> <description>We will play later</description> </item> </ads> I tried code in PHP: <? $file = "test.xml"; $fp = fopen($file, "rb") or die("cannot open file"); $str = fread($fp, filesize($file)); $dom=new DOMDocument(); $dom->formatOutput = true; $dom->preserveWhiteSpace = false; $dom->loadXML($str) or die("Error"); //$dom->load("items.xml"); $root=$dom->documentElement; // This can differ (I am not sure, it can be only documentElement or documentElement->firstChild or only firstChild) $loop=$root->getElementsByTagName('loop')->item(0);//->textContent; //echo $loop; if(trim($loop->textContent)=='no') { echo 'ok'; $root->getElementsByTagName('loop')->item(0)->nodeValue ='yes'; } echo "<xmp>NEW:\n". $dom->saveXML() ."</xmp>"; ?> I tried only for loop tag.I don't know how to replace nodevalue in description tag. When I run this page it shows output like: ok NEW: <?xml version="1.0" encoding="utf-8"?> <ads> <loop>yes</loop> <item> <description>Description 1</description> </item> <item> <description>Changing text in item2</description> </item> <item> <description>Let play with this XML</description> </item> </ads> It gives the value yes in browser but don't save it in test.xml any reason?

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  • Why is the XML DTD not found by the browser

    - by hyperuser
    When I load my XML file in a browser, it complains there is 'no style information': "This XML file does not appear to have any style information associated with it. The document tree is shown below." So I wrote an external DTD, then an internal DTD, but keep getting the same 'no style information' error. It doesn't even show the DTD! What am I doing wrong? <?xml version="1.0"?> <!DOCTYPE fotos [ <!ELEMENT fotos (titel,auteur)> <!ELEMENT titel (#PCDATA)> <!ELEMENT auteur (#PCDATA)> ]> <fotos> <titel>titel1</titel> <auteur>jan</auteur> </fotos>

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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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  • 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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  • 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 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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  • 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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