Search Results

Search found 840 results on 34 pages for 'shekar xsl'.

Page 18/34 | < Previous Page | 14 15 16 17 18 19 20 21 22 23 24 25  | Next Page >

  • MS Word to Stylesheet

    - by Chris Johnson
    Is there an easy way to automatically convert a bunch of MS Word documents to xslt stylesheets that can be displayed in the browser? What I have is a large collection of forms in Word format that have to be displayed in the browser, or sent to the user, with known fields populated from a data source, edited by a user and, finally, printed (including the original headers and footers). The data entered by the user will not need to be saved. I'm not sure if converting the documents to stylesheets is even feasible. Maybe someone has a better idea of how to achieve this? Installing Office on the server is not an option in my case.

    Read the article

  • Sorting XML file through multiple attributes

    - by jbugeja
    I want to sort a 'free-form' XML file through multiple attributes (first by T and then by L). The XML is a bit complex and it is structured as shown below: <?xml version="1.0" encoding="utf-8"?> <wb xmlns:cf="http://www.macromedia.com/2004/cfform" xmlns:a="urn:dummy"> <a:form name="chart"> <a:fieldset FIELD="a" FIELDNAME="FieldSet1"> <a:select1 FIELDNUMBER="01" L="1" T="2" /> <a:input FIELDNUMBER="02" INDEX="4" L="200" T="1" /> </a:fieldset> <a:fieldset FIELD="b" FIELDNAME="FieldSet1"> <a:select1 FIELDNUMBER="03" T="3" L="1" /> <a:input FIELDNUMBER="04" INDEX="7" T="4" L="200" /> <a:fieldset FIELD="c" FIELDNAME="FieldSet1"> <a:input FIELDNUMBER="05" T="10" INDEX="6" L="400" /> <a:input FIELDNUMBER="06" T="8" INDEX="8" L="200" /> </a:fieldset> </a:fieldset> <a:input FIELDNUMBER="08" INDEX="3" L="3" T="5" /> <a:input FIELDNUMBER="09" INDEX="2" L="2" T="4" /> </a:form> </wb> PS: The root element is wb and this is always followed by a:form The L and T are always found in elements that have a tag in the namespace a, the only exception being a:fieldset which does not have L and T a:fieldset could have multiple children of the namespace a including another a:fieldset We can also assume that L denotes Left and T denotes Top. So, the idea of this is that when I view the transformed XML I can immediately note which elements precede what. What's your take on this?

    Read the article

  • Is it possible to come over the time out issue for a function call in C#?

    - by infant programmer
    In my program I call a method xslTransform.Load(strXmlQueryTransformPath, xslSettings, new XmlUrlResolver()); The problem I am facing is: sometimes this function doesn't execute well within the time. Sometimes compiler raises the time out issue after a long time of trial.. which inturn causes this part of application to shut. That is what I want to avoid. So if it exceeds certain time say 10 seconds I need to recall the method. Is it possible to add some code lines adjacent to this, which can meet the requirement?

    Read the article

  • Selecting unique elements using XSLT.

    - by AJ
    I have the following XML: <option> <title>ABC</title> <desc>123</desc> </option> <option> <title>ABC</title> <desc>12345</desc> </option> <option> <title>ABC</title> <desc>123</desc> </option> <option> <title>EFG</title> <desc>123</desc> </option> <option> <title>EFG</title> <desc>456</desc> </option> Using XSLT, I want to transform it into: <choice> <title>ABC</title> <desc>123</desc> <desc>12345</desc> </choice> <choice> <title>EFG</title> <desc>123</desc> <desc>456</desc> </choice>

    Read the article

  • Using XSLT, how can I produce a table with elements at the position of the the node's attributes?

    - by Dr. Sbaitso
    Given the following XML: <items> <item> <name>A</name> <address>0</address> <start>0</start> <size>2</size> </item> <item> <name>B</name> <address>1</address> <start>2</start> <size>4</size> </item> <item> <name>C</name> <address>2</address> <start>5</start> <size>2</size> </item> </items> I want to generate the following output including colspan's +---------+------+------+------+------+------+------+------+------+ | Address | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 | +---------+------+------+------+------+------+------+------+------+ | 0 | | | | | | | A | +---------+------+------+------+------+------+------+------+------+ | 1 | | | B | | | +---------+------+------+------+------+------+------+------+------+ | 2 | | C | | | | | | +---------+------+------+------+------+------+------+------+------+ | 3 | | | | | | | | | +---------+------+------+------+------+------+------+------+------+ I think I would be able to accomplish this with a mutable xslt variable, but alas, there's no such thing. Is it even possible? How?

    Read the article

  • Use DataSource in DataFormWebPart

    - by Bryan Shen
    I'm writing a custom web part that extends DataFormWebPart. public class MyCustomWebPart : DataFormWebPart{ // other methods public override void DataBind() { XmlDataSource source = new XmlDataSource() { Data = @" <Person> <name cap='true'>Bryan</name> <occupation>student</occupation> </Person> " }; DataSources.Add(source); base.DataBind(); } } The only noticeable thing I do is overriding the DataBind() method, where I use xml as the data source. After I deploy the web part, I set the following XSL to it: <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/> <xsl:template match="/"> <xmp> <xsl:copy-of select="*"/> </xmp> </xsl:template> </xsl:stylesheet> This xsl will surround the input xml with a tag . So I expected the web part to display the original xml data as I wrote in C# code behind. But what shows up in the web part is this: <Person> <name cap="true" /> <occupation /> </Person> All the values within the inner-most tags disappear. What's going on? Can anybody help me? Thanks.

    Read the article

  • How to Use DataSource Property in DataFormWebPart

    - by Bryan Shen
    I'm writing a custom web part that extends DataFormWebPart. public class MyCustomWebPart : DataFormWebPart{ // other methods public override void DataBind() { XmlDataSource source = new XmlDataSource() { Data = @" <Person> <name cap='true'>Bryan</name> <occupation>student</occupation> </Person> " }; DataSources.Add(source); base.DataBind(); } } The only noticeable thing I do is overriding the DataBind() method, where I use xml as the data source. After I deploy the web part, I set the following XSL to it: <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/> <xsl:template match="/"> <xmp> <xsl:copy-of select="*"/> </xmp> </xsl:template> </xsl:stylesheet> This xsl will surround the input xml with a tag . So I expected the web part to display the original xml data as I wrote in C# code behind. But what shows up in the web part is this: <Person> <name cap="true" /> <occupation /> </Person> All the values within the inner-most tags disappear. What's going on? Can anybody help me? Thanks.

    Read the article

  • xslt broken: pattern does not match

    - by krisvandenbergh
    I'm trying to query an xml file using the following xslt: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#" xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#" xmlns:bpmn="http://dkm.fbk.eu/index.php/BPMN_Ontology"> <!-- Participants --> <xsl:template match="/"> <html> <body> <table> <xsl:for-each select="Package/Participants/Participant"> <tr> <td><xsl:value-of select="ParticipantType" /></td> <td><xsl:value-of select="Description" /></td> </tr> </xsl:for-each> </table> </body> </html> </xsl:template> </xsl:stylesheet> Here's the contents of the xml file: <?xml version="1.0" encoding="utf-8"?> <?xml-stylesheet type="text/xsl" href="xpdl2bpmn.xsl"?> <Package xmlns="http://www.wfmc.org/2008/XPDL2.1" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" Id="25ffcb89-a9bf-40bc-8f50-e5afe58abda0" Name="1 price setting" OnlyOneProcess="false"> <PackageHeader> <XPDLVersion>2.1</XPDLVersion> <Vendor>BizAgi Process Modeler.</Vendor> <Created>2010-04-24T10:49:45.3442528+02:00</Created> <Description>1 price setting</Description> <Documentation /> </PackageHeader> <RedefinableHeader> <Author /> <Version /> <Countrykey>CO</Countrykey> </RedefinableHeader> <ExternalPackages /> <Participants> <Participant Id="008af9a6-fdc0-45e6-af3f-984c3e220e03" Name="customer"> <ParticipantType Type="RESOURCE" /> <Description /> </Participant> <Participant Id="1d2fd8b4-eb88-479b-9c1d-7fe6c45b910e" Name="clerk"> <ParticipantType Type="ROLE" /> <Description /> </Participant> </Participants> </Package> Despite, the simple pattern, the foreach doesn't work. What is wrong with Package/Participants/Participant ? What do I miss here? Thanks a lot!

    Read the article

  • How can I transform XML to invalid XML using XSLT?

    - by Damovisa
    I need to transform a valid XML document to the OFX v1.0.2 format. This format is more or less XML, but it's technically invalid and therefore cannot be parsed as XML. I'm having trouble getting my Xml transformation working because the .Net XslCompiledTransform object insists on interpreting the XSL as an XML document (which is fair enough). If I escape the xml-ish tags using &lt; and &gt, they get removed when I download the file. Here's the start of my XSLT: <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output method="text"></xsl:output> <xsl:param name="currentdate"></xsl:param> <xsl:template match="Transactions"> OFXHEADER:100 DATA:OFXSGML VERSION:102 SECURITY:NONE ENCODING:USASCII CHARSET:1252 COMPRESSION:NONE OLDFILEUID:NONE NEWFILEUID:NONE <OFX> <SIGNONMSGSRSV1> <SONRS> <STATUS> <CODE>0 <SEVERITY>INFO </STATUS> <DTSERVER><xsl:value-of select="$currentdate" /> <LANGUAGE>ENG Any suggestions?

    Read the article

  • XSLT Transformation of XML File

    - by Russ Clark
    I've written a simple XML Document that I am trying to transform with an XSLT file, but I get no results when I run the code. Here is my XML document: <?xml version="1.0" encoding="utf-8" ?> <Employee xmlns:i="http://www.w3.org/2001/XMLSchema-instance" xmlns="XSLT_MVC.Controllers"> <ID>42</ID> <Name>Russ</Name> </Employee> And here is the XSLT file: <?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" xmlns:ex="XSLT_MVC.Controllers" > <xsl:output method="xml" indent="yes"/> <xsl:template match="/"> <xsl:copy> <!--<xsl:apply-templates select="@* | node()"/>--> <xsl:value-of select="ex:Employee/Name"/> </xsl:copy> </xsl:template> </xsl:stylesheet> Here is the code (from a C# console app) I am trying to run to perform the transform: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Xml; using System.Xml.Xsl; using System.Xml.XPath; namespace XSLT { class Program { static void Main(string[] args) { Transform(); } public static void Transform() { XPathDocument myXPathDoc = new XPathDocument(@"docs\sampledoc.xml"); XslTransform myXslTrans = new XslTransform(); myXslTrans.Load(@"docs\new.xslt"); XmlTextWriter myWriter = new XmlTextWriter( "results.html", null); myXslTrans.Transform(myXPathDoc, null, myWriter); myWriter.Close(); } } } When I run the code I get a blank html file. I think I may have problems with the namespaces, but am not sure. Can anyone help with this?

    Read the article

  • How to add some complexe structure in multiple places in an XML file

    - by Guillaume
    I have an XML file which has many section like the one below : <Operations> <Action [some attibutes ...]> [some complexe content ...] </Action> <Action [some attibutes ...]> [some complexe content ...] </Action> </Operations> I have to add an to every . It seems that an XSLT should be a good solution to this problem : <xsl:template match="Operations/Action[last()]"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> <Action>[some complexe content ...]</Action> </xsl:template> <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> My problem is that the content of my contains some xPath expressions. For exemple : <Action code="p_histo01"> <customScript languageCode="gel"> <gel:script xmlns:core="jelly:core" xmlns:gel="jelly:com.niku.union.gel.GELTagLibrary" xmlns:soap="jelly:com.niku.union.gel.SOAPTagLibrary" xmlns:soap-env="http://schemas.xmlsoap.org/soap/envelope/" xmlns:sql="jelly:sql" xmlns:x="jelly:xml" xmlns:xog="http://www.niku.com/xog" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <sql:param value="${gel_stepInstanceId}"/> </gel:script> </customScript> </Action> The '${gel_stepInstanceId}' is interpreted by my XSLT but I would like it to be copied as-is. Is that posible ? How ?

    Read the article

  • How to Create Element in XSLT When Copying Using Templates

    - by John Dumont
    Hello, I'm trying to create an element in an XML where the basic content is copied and modified. My XML is something like <root> <node> <child>value</child> <child2>value2</child2> </node> <node2>bla</node2> </root> The number of children of node may change as well as the children of root. The XSLT should copy the whole content, modify some values and add some new. The copying and modifying is no problem: <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"> <xsl:output method="xml" encoding="UTF-8"/> <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> </xsl:stylesheet> (+ further templates for modifications). But how do I add a new element in this structure on some path, e.g. I want to add an element as the last element of the "node" node. The "node" element itself always exists.

    Read the article

  • XSLT: Disable output escaping in an entire document.

    - by Kragen
    I'm trying to generate some C# code using xslt - its working great until I get to generics and need to output some text like this: MyClass<Type> In this case I've found that the only way to emit this is to do the following: MyClass<xsl:text disable-output-escaping="yes">&lt;</xsl:text>Type<xsl:text disable-output-escaping="yes">&gt;</xsl:text> Where: Often it all needs to go on one line, otherwise you end up with line breaks in the generated code In the above example I technically could have used only 1 <xsl:text />, however usually the type Type is given by some other template, e.g: <xsl:value-of select="@type" /> I don't mind having to write &lt; a lot, but I would like to avoid writing <xsl:text disable-output-escaping="yes">&lt;</xsl:text> for just a single character! Is there any way of doing disable-output-escaping="yes" for the entire document?

    Read the article

  • XSLT creating a table with varying amount of columns

    - by H4mm3rHead
    Hi, I have a RSS feed i need to display in a table (its clothes from a webshop system). The images in the RSS vary in width and height and I want to make a table that shows them all. First off i would be happy just to show all of the items in 3 columns, but further down the road i need to be able to specify through a parameter the amount of columns in my table. I run into a problem showing the tr tag, and making it right, this is my Code so far: <xsl:template match="item"> <xsl:choose> <xsl:when test="position() mod 3 = 0 or position()=1"> <tr> <td> <xsl:value-of select="title"/> </td> </tr> </xsl:when> <xsl:otherwise> <td> <xsl:value-of select="title"/> </td> </xsl:otherwise> </xsl:choose> </xsl:template> In RSS all "item" tags are on the same level in the xml, and thus far i need only the title shows. Thr problem seems to be that i need to specify the start tag as well as the end tag of the tr element, and cant get all 3 elements into it, anyone know how to do this?

    Read the article

  • Navigating sorted XML data (XSLT)

    - by Andrew Parisi
    I have an XML file with company data in it, for 30 companies across 8 industries, on a portfolio page. The user has the option to sort this data by Industry, and this XML file will be added to constantly. This sorting is done in my XSL file using <xsl:choose>. Example: <xsl:when test="(invest[@investid='con'])"> <xsl:for-each select="$invest-port/portfolio/company[@industry='Communications']"> <xsl:sort select="name" /> <div class="invest-port-thumb"> <a> <xsl:attribute name="href"> <xsl:value-of select="link" /> </xsl:attribute> </a> </div> </xsl:for-each> </xsl:when> When navigating to an individual company's page, there are "previous" and "next" buttons at the bottom of the window. My issue is that I need these to dynamically link to the previous and next elements from within the XML data that has been sorted. Is this possible? Or is there an easier way to do this? (such as place each company in industry-divided XML files instead of one) Any insight would be greatly appreciated!

    Read the article

  • XSLT Exclude Specific Tags

    - by MMKD
    I have a problem i am trying to resolve in an XSLT but i an unable to find a solution. The example below is related to a payment system it is adding items to a basket then removing them. The out XML provides a audit trail of actions conducted on a basket. Senario: Add Item (Id 1) Add Item (Id 1) With a price change Void Item (Id 1) Void Item (Id 1) With a price change Add Item (Id 1) Add Item (Id 1) Expected Outcome Remove: Add Item (Id 1) Add Item (Id 1) With a price change Output XML Contains Void Item (Id 1) Void Item (Id 1) With a price change Add Item (Id 1) Add Item (Id 1) Input XML: <xml> <product void="false"> <sequence_number>1</sequence_number> <item_id>11111111</item_id> <price>12</price> </product> <product void="false"> <sequence_number>2</sequence_number> <item_id>11111111</item_id> <price>12</price> <price_change> <price>10</price> </price_change> </product> <product void="true"> <sequence_number>3</sequence_number> <item_id>11111111</item_id> <price>12</price> <price_change> <price>10</price> </price_change> </product> <product void="true"> <sequence_number>4</sequence_number> <item_id>11111111</item_id> <price>12</price> </product> <product void="false"> <sequence_number>5</sequence_number> <item_id>11111111</item_id> <price>12</price> </product> <product void="false"> <sequence_number>6</sequence_number> <item_id>11111111</item_id> <price>12</price> </product> </xml> Expected outcome: <xml> <product void="true"> <sequence_number>3</sequence_number> <item_id>11111111</item_id> <price>12</price> <price_change> <price>10</price> </price_change> </product> <product void="true"> <sequence_number>4</sequence_number> <item_id>11111111</item_id> <price>12</price> </product> <product void="false"> <sequence_number>5</sequence_number> <item_id>11111111</item_id> <price>12</price> </product> <product void="false"> <sequence_number>6</sequence_number> <item_id>11111111</item_id> <price>12</price> </product> </xml> XSLT <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> <xsl:template match="//product[@void='false']"> <xsl:if test="item_id != //product[@void='true']/item_id"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:if> </xsl:template> </xsl:stylesheet> The problem with this is that it is deleting all products that are not voided and have the same id and not taking into account the number of void items vs the number of none void items. If you have 1 void item it should only delete one product that is not voided but has exactly the same tags as itself

    Read the article

  • XSLT - Adding a class to something with a class?

    - by Probocop
    When using XSLT how do I apply a class to an element which already has a class? The way I'm doing it it replaces the class that is already present? How would I add the class in addition to the existing class? My code is as follows: <xsl:if test="data[@alias = 'off'] = 1"> <xsl:attribute name="class">off</xsl:attribute> </xsl:if> <xsl:if test="$currentPage/ancestor-or-self::node/@id = current()/@id"> <xsl:attribute name="class">active</xsl:attribute> </xsl:if> Thanks.

    Read the article

  • Is there an effective way to test XSL transforms/BizTalk maps?

    - by nlawalker
    Creating repeatable tests for BizTalk maps is frustrating. I can't find a way to handle testing them like I'd do unit testing, because I can't find ways to break them into logical chunks. They tend to be one big monolithic unit, and any change has the potential to ripple through the map and break a lot of unit tests. Even if I could break it up, creating XML test inputs is painful and error prone. Is there any effective way of testing these? I'd settle for recommendations for testing XSL transforms in general, but I specifically mention BizTalk maps primarily for the reason that when using the mapper, there really isn't any way to break your XSLT into templates (which I'd imagine you could use to break up your logic into testable chunks, but I've honestly never gotten that far with XSLT).

    Read the article

  • Translate report data export from RUEI into HTML for import into OpenOffice Calc Spreadsheets

    - by [email protected]
    A common question of users is, How to import the data from the automated data export of Real User Experience Insight (RUEI) into tools for archiving, dashboarding or combination with other sets of data.XML is well-suited for such a translation via the companion Extensible Stylesheet Language Transformations (XSLT). Basically XSLT utilizes XSL, a template on what to read from your input XML data file and where to place it into the target document. The target document can be anything you like, i.e. XHTML, CSV, or even a OpenOffice Spreadsheet, etc. as long as it is a plain text format.XML 2 OpenOffice.org SpreadsheetFor the XSLT to work as an OpenOffice.org Calc Import Filter:How to add an XML Import Filter to OpenOffice CalcStart OpenOffice.org Calc andselect Tools > XML Filter SettingsNew...Fill in the details as follows:Filter name: RUEI Import filterApplication: OpenOffice.org Calc (.ods)Name of file type: Oracle Real User Experience InsightFile extension: xmlSwitch to the transformation tab and enter/select the following leaving the rest untouchedXSLT for import: ruei_report_data_import_filter.xslPlease see at the end of this blog post for a download of the referenced file.Select RUEI Import filter from list and Test XSLTClick on Browse to selectTransform file: export.php.xmlOpenOffice.org Calc will transform and load the XML file you retrieved from RUEI in a human-readable format.You can now select File > Open... and change the filetype to open your RUEI exports directly in OpenOffice.org Calc, just like any other a native Spreadsheet format.Files of type: Oracle Real User Experience Insight (*.xml)File name: export.php.xml XML 2 XHTMLMost XML-powered browsers provides for inherent XSL Transformation capabilities, you only have to reference the XSLT Stylesheet in the head of your XML file. Then open the file in your favourite Web Browser, Firefox, Opera, Safari or Internet Explorer alike.<?xml version="1.0" encoding="ISO-8859-1"?><!-- inserted line below --> <?xml-stylesheet type="text/xsl" href="ruei_report_data_export_2_xhtml.xsl"?><!-- inserted line above --><report>You can find a patched example export from RUEI plus the above referenced XSL-Stylesheets here: export.php.xml - Example report data export from RUEI ruei_report_data_export_2_xhtml.xsl - RUEI to XHTML XSL Transformation Stylesheetruei_report_data_import_filter.xsl - OpenOffice.org XML import filter for RUEI report export data If you would like to do things like this on the command line you can use either Xalan or xsltproc.The basic command syntax for xsltproc is very simple:xsltproc -o output.file stylesheet.xslt inputfile.xmlYou can use this with the above two stylesheets to translate RUEI Data Exports into XHTML and/or OpenOffice.org Calc ODS-Format. Or you could write your own XSLT to transform into Comma separated Value lists.Please let me know what you think or do with this information in the comments below.Kind regards,Stefan ThiemeReferences used:OpenOffice XML Filter - Create XSLT filters for import and export - http://user.services.openoffice.org/en/forum/viewtopic.php?f=45&t=3490SUN OpenOffice.org XML File Format 1.0 - http://xml.openoffice.org/xml_specification.pdf

    Read the article

  • Using SSIS to send a HTML E-Mail Message with built-in table of Counts.

    - by Kevin Shyr
    For the record, this can be just as easily done with a .NET class with a DLL call.  The two major reasons for this ending up as a SSIS package are: There are a lot of SQL resources for maintenance, but not as many .NET developers. There is an existing automated process that links up SQL Jobs (more on that in the next post), and this is part of that process.   To start, this is what the SSIS looks like: The first part of the control flow is just for the override scenario.   In the Execute SQL Task, it calls a stored procedure, which already formats the result into XML by using "FOR XML PATH('Row'), ROOT(N'FieldingCounts')".  The result XML string looks like this: <FieldingCounts>   <Row>     <CellId>M COD</CellId>     <Mailed>64</Mailed>     <ReMailed>210</ReMailed>     <TotalMail>274</TotalMail>     <EMailed>233</EMailed>     <TotalSent>297</TotalSent>   </Row>   <Row>     <CellId>M National</CellId>     <Mailed>11</Mailed>     <ReMailed>59</ReMailed>     <TotalMail>70</TotalMail>     <EMailed>90</EMailed>     <TotalSent>101</TotalSent>   </Row>   <Row>     <CellId>U COD</CellId>     <Mailed>91</Mailed>     <ReMailed>238</ReMailed>     <TotalMail>329</TotalMail>     <EMailed>291</EMailed>     <TotalSent>382</TotalSent>   </Row>   <Row>     <CellId>U National</CellId>     <Mailed>63</Mailed>     <ReMailed>286</ReMailed>     <TotalMail>349</TotalMail>     <EMailed>374</EMailed>     <TotalSent>437</TotalSent>   </Row> </FieldingCounts>  This result is saved into an internal SSIS variable with the following settings on the General tab and the Result Set tab:   Now comes the trickier part.  We need to use the XML Task to format the XML string result into an HTML table, and I used Direct input XSLT And here is the code of XSLT: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="html" indent="yes"/>   <xsl:template match="/ROOT">         <table border="1" cellpadding="6">           <tr>             <td></td>             <td>Mailed</td>             <td>Re-mailed</td>             <td>Total Mail (Mailed, Re-mailed)</td>             <td>E-mailed</td>             <td>Total Sent (Mailed, E-mailed)</td>           </tr>           <xsl:for-each select="FieldingCounts/Row">             <tr>               <xsl:for-each select="./*">                 <td>                   <xsl:value-of select="." />                 </td>               </xsl:for-each>             </tr>           </xsl:for-each>         </table>   </xsl:template> </xsl:stylesheet>    Then a script task is used to send out an HTML email (as we are all painfully aware that SSIS Send Mail Task only sends plain text) Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 using System; using System.Data; using Microsoft.SqlServer.Dts.Runtime; using System.Windows.Forms; using System.Net.Mail; using System.Net;   namespace ST_b829a2615e714bcfb55db0ce97be3901.csproj {     [System.AddIn.AddIn("ScriptMain", Version = "1.0", Publisher = "", Description = "")]     public partial class ScriptMain : Microsoft.SqlServer.Dts.Tasks.ScriptTask.VSTARTScriptObjectModelBase     {           #region VSTA generated code         enum ScriptResults         {             Success = Microsoft.SqlServer.Dts.Runtime.DTSExecResult.Success,             Failure = Microsoft.SqlServer.Dts.Runtime.DTSExecResult.Failure         };         #endregion           public void Main()         {             String EmailMsgBody = String.Format("<HTML><BODY><P>{0}</P><P>{1}</P></BODY></HTML>"                                                 , Dts.Variables["Config_SMTP_MessageSourceText"].Value.ToString()                                                 , Dts.Variables["InternalStr_CountResultAfterXSLT"].Value.ToString());             MailMessage EmailCountMsg = new MailMessage(Dts.Variables["Config_SMTP_From"].Value.ToString().Replace(";", ",")                                                         , Dts.Variables["Config_SMTP_Success_To"].Value.ToString().Replace(";", ",")                                                         , Dts.Variables["Config_SMTP_SubjectLinePrefix"].Value.ToString() + " " + Dts.Variables["InternalStr_FieldingDate"].Value.ToString()                                                         , EmailMsgBody);             //EmailCountMsg.From.             EmailCountMsg.CC.Add(Dts.Variables["Config_SMTP_Success_CC"].Value.ToString().Replace(";", ","));             EmailCountMsg.IsBodyHtml = true;               SmtpClient SMTPForCount = new SmtpClient(Dts.Variables["Config_SMTP_ServerAddress"].Value.ToString());             SMTPForCount.Credentials = CredentialCache.DefaultNetworkCredentials;               SMTPForCount.Send(EmailCountMsg);               Dts.TaskResult = (int)ScriptResults.Success;         }     } } Note on this code: notice the email list has Replace(";", ",").  This is only here because the list is configurable in the SQL Job Step at Set Values, which does not react well with colons as email separator, but system.Net.Mail only handles comma as email separator, hence the extra replace in the string. The result is a nicely formatted email message with count information:

    Read the article

  • DataContractSerializer and XSLT not Serializing Class Properties

    - by Russ Clark
    I've written a simple Employee class that I'm trying to serialize to an XDocument and then use XSLT to transform the document to a page that displays both the properties (Name and ID) from the Employee class, and an html form with 2 radio buttons (Approve and Reject) and a submit button. Here is the Employee class: [Serializable, DataContract(Namespace="XSLT_MVC.Controllers/")] public class Employee { [DataMember] public string Name { get; set; } [DataMember] public int ID { get; set; } public Employee() { } public Employee(string name, int id) { Name = name; ID = id; } public XDocument GetDoc() { XDocument doc = new XDocument(); var serializer = new DataContractSerializer(typeof(Employee)); using (var writer = doc.CreateWriter()) { serializer.WriteObject(writer, this); writer.Close(); } return doc; } } And here is the XSLT file: <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > <xsl:output method="html" indent="yes"/> <xsl:template match="/"> <html> <body> <xsl:value-of select="Employee/Name"/> <br /> <xsl:value-of select="Employee/ID"/> <br /> <form method="post" action="/Home/ProcessRequest?id={Employee/ID}"> <input id="Action" name="Action" type="radio" value="Approved"></input> Approved <br /> <input id="Action" name="Action" type="radio" value="Rejected"></input> Rejected <br /> <input type="submit" value="Submit"></input> </form> </body> </html> </xsl:template> </xsl:stylesheet> When I run this, all I get is the html form with the 2 radio buttons and the submit button, but not the properties from the Employee class. I saw a separate StackOverflow post that said I need to change the <xsl:template match="/"> to match on the namespace of my Employee class like this: <xsl:template match="/XSLT_MVC.Controllers"> but when I do that, now all I get are the Employee properties, and not the html form with the 2 radio buttons and the submit button. Does anyone know what needs to be done so that my transform will select and display both the Employee properties and the html form?

    Read the article

  • DataContractSerializer and XSLT

    - by Russ Clark
    I've got a simple Employee class that I'm trying to serialize to an XDocument and then use XSLT to transform the document to a page that displays both the properties (Name and ID) from the Employee class, and an html form with 2 radio buttons (Approve and Reject) and a submit button. Here is the Employee class: [Serializable, DataContract(Namespace="XSLT_MVC.Controllers/")] public class Employee { [DataMember] public string Name { get; set; } [DataMember] public int ID { get; set; } public Employee() { } public Employee(string name, int id) { Name = name; ID = id; } public XDocument GetDoc() { XDocument doc = new XDocument(); var serializer = new DataContractSerializer(typeof(Employee)); using (var writer = doc.CreateWriter()) { serializer.WriteObject(writer, this); writer.Close(); } return doc; } } And here is the XSLT file: <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > <xsl:output method="html" indent="yes"/> <xsl:template match="/"> <html> <body> <xsl:value-of select="Employee/Name"/> <br /> <xsl:value-of select="Employee/ID"/> <br /> <form method="post" action="/Home/ProcessRequest?id={Employee/ID}"> <input id="Action" name="Action" type="radio" value="Approved"></input> Approved <br /> <input id="Action" name="Action" type="radio" value="Rejected"></input> Rejected <br /> <input type="submit" value="Submit"></input> </form> </body> </html> </xsl:template> </xsl:stylesheet> When I run this, all I get is the html form with the 2 radio buttons and the submit button, but not the properties from the Employee class. I saw a separate StackOverflow post that said I need to change the <xsl:template match="/"> to match on the namespace of my Employee class like this: <xsl:template match="/XSLT_MVC.Controllers">, but when I do that, now all I get are the Employee properties, and not the html form with the 2 radio buttons and the submit button. Does anyone know what needs to be done so that my transform will select and display both the Employee properties and the html form?

    Read the article

< Previous Page | 14 15 16 17 18 19 20 21 22 23 24 25  | Next Page >