Re: Adding Header an datetime when User prints excel Sheet. - XML / XSL Discussion Boards

Re: string manipulation in xsl - concatinating

led mike6-Mar-09 4:43

led mike

6-Mar-09 4:43

Try the XSLT Tutorials at www.w3schools.com, the are tailored to beginners.

Problem reading xml string and convert into a data table

Renukapadhamanaban4-Mar-09 12:38

Renukapadhamanaban

4-Mar-09 12:38

Hi,

I am wanted to convert/store a string, which has XML elements, into a dataset or data table. The string that I am reading from a service is the result of my query. I have written the following code,

{
DataSet ds = new DataSet();

StringReader stream = new StringReader(xmlString);

ds.ReadXml(stream);

DataTable dt = ds.Tables[0];

dg.DataSource = dt;

return dt;
}

I am receiving two different type of strings, they are

xmlString1:

<?xml version="1.0" encoding="UTF-8"?>
<nh:rows xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:nh="http://nhdb">
<nh:urnumber>0123456</nh:urnumber>
<nh:surname>xxxxx</nh:surname>
<nh:forename>yyy</nh:forename>
<nh:middlename></nh:middlename>
<nh:dob>1958-10-05</nh:dob>
<nh:address>Mxaxixlx</nh:address>
<nh:suburb>Prxxxx</nh:suburb>
<nh:state>Vxxx</nh:state>
<nh:postcode>3072</nh:postcode>
<nh:patientphone>phone</nh:patientphone>
<nh:gender>Female</nh:gender>
<nh:unit>Sxxxxx Oxxxx</nh:unit>
<nh:ward>Bxxx DPxx</nh:ward>
</nh:rows>

xmlString2:

<?xml version="1.0" encoding="UTF-8"?>
<nh:rows xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:nh="http://nhdb">
<nh:urnumber>0123450</nh:urnumber>
<nh:surname>Cxxxxxx</nh:surname>
<nh:forename>RxxxxxT</nh:forename>
<nh:middlename></nh:middlename>
<nh:dob>1954-07-19</nh:dob>
<nh:address>'ANNANDALE'</nh:address>
<nh:suburb></nh:suburb>
<nh:state></nh:state>
<nh:postcode></nh:postcode>
<nh:patientphone></nh:patientphone>
<nh:gender>Male</nh:gender>
<nh:unit></nh:unit>
<nh:ward></nh:ward>
<nh:urnumber>0123451</nh:urnumber>
<nh:surname>KROL</nh:surname>
<nh:forename>ANILLA</nh:forename>
<nh:middlename></nh:middlename>
<nh:dob>1958-12-10</nh:dob>
<nh:address>68A CLARENCE</nh:address>
<nh:suburb></nh:suburb>
<nh:state></nh:state>
<nh:postcode></nh:postcode>
<nh:patientphone></nh:patientphone>
<nh:gender>Female</nh:gender>
<nh:unit></nh:unit>
<nh:ward></nh:ward>
<nh:urnumber>0123452</nh:urnumber>
<nh:surname>GEYER</nh:surname>
<nh:forename>MAUREEN</nh:forename>
<nh:middlename></nh:middlename>
<nh:dob>1948-06-02</nh:dob>
<nh:address>19 JAYSON</nh:address>
<nh:suburb></nh:suburb>
<nh:state></nh:state>
<nh:postcode></nh:postcode>
<nh:patientphone></nh:patientphone>
<nh:gender>Female</nh:gender>
<nh:unit></nh:unit>
<nh:ward></nh:ward>
</nh:rows>

The above code works perfectly for the xmlString1 and it is converting a data table with single row of data for the cloumns urnumber, surname, forename, middlename, dob, address, suburb,state, postcode,patientPhone, gender, unit, ward.

When I am receiving the xmlString2, the data table is populated with only one column urnumber_Text(for all ur numbers in the xml string).

What am I missing here? Is there any other genuine way of doing it?

Thanks a Lot for your valuable time and invaluable advice.

Looking for advice,
Renu

Re: Problem reading xml string and convert into a data table

led mike5-Mar-09 4:22

led mike

5-Mar-09 4:22

Renukapadhamanaban wrote:
What am I missing here?

The XML is not formatted as required for generalized recognition of the records. Each set of records would need a parent element to form a collection. This appears to be absent in what you posted.

Renukapadhamanaban wrote:
Is there any other genuine way of doing it?

Yes you could write code to transform the XML into the more standard form then it should load fine. This might be accomplished with XSLT but it might be simpler to use like a SAX parser or whatever the .NET equivalent is.

Re: Problem reading xml string and convert into a data table

Renukapadhamanaban9-Mar-09 15:37

Renukapadhamanaban

9-Mar-09 15:37

Hi Mike,

Thanks a lot for your valuable advice.

I tried with the parent element to each row and now it is working alright. Now I am working on XSLT to format the XML string.

THanks for your time.
Renu

Adding Header an datetime when User prints excel Sheet.

vikram_asv4-Mar-09 9:22

vikram_asv

4-Mar-09 9:22

I need to add header and datetime and orientation in excel when user print the excel document.

public static void excelprint(DataSet source, string fileName)
{
bool status = false;
try
{
System.IO.StreamWriter excelDoc;
excelDoc = new System.IO.StreamWriter(fileName);
const string startExcelXML = "<xml version>\r\n<Workbook " +
"xmlns=\"urn:schemas-microsoft-com:office:spreadsheet\"\r\n" +
" xmlns Blush | :O

=\"urn:schemas-microsoft-com:office:office\"\r\n " +
"xmlns:x=\"urn:schemas- microsoft-com:office:" +
"excel\"\r\n xmlns:ss=\"urn:schemas-microsoft-com:" +
"office:spreadsheet\">\r\n <Styles>\r\n " +
"<Style ss:ID=\"Default\" ss:Name=\"Normal\">\r\n " +
"<Alignment ss:Vertical=\"Bottom\"/>\r\n <Borders/>" +
"\r\n <Font/>\r\n <Interior/>\r\n <NumberFormat/>" +
"\r\n <Protection/>\r\n </Style>\r\n " +
"<Style ss:ID=\"BoldColumn\">\r\n <Font " +
"x:Family=\"Swiss\" ss:Bold=\"1\"/>\r\n </Style>\r\n " +
"<Style ss:ID=\"StringLiteral\">\r\n <NumberFormat" +
" ss:Format=\"@\"/>\r\n </Style>\r\n <Style " +
"ss:ID=\"Decimal\">\r\n <NumberFormat " +
"ss:Format=\"0.00\"/>\r\n </Style>\r\n " +
"<Style ss:ID=\"Integer\">\r\n <NumberFormat " +
"ss:Format=\"0\"/>\r\n </Style>\r\n <Style " +
"ss:ID=\"DateLiteral\">\r\n <NumberFormat " +
"ss:Format=\"mm/dd/yyyy;@\"/>\r\n </Style>\r\n " +
"</Styles>\r\n ";
const string endExcelXML = "</Workbook>";
int rowCount = 0;
int sheetCount = 1;

excelDoc.Write(startExcelXML);

for (int worksheet = 0; worksheet < source.Tables.Count; worksheet++)
{

excelDoc.Write("<Worksheet ss:Name=\"" + source.Tables[worksheet].TableName + "\">");
excelDoc.Write("<Table>");
excelDoc.Write("<Row>");

for (int x = 0; x < source.Tables[worksheet].Columns.Count; x++)
{
excelDoc.Write("<Cell ss:StyleID=\"BoldColumn\"><Data ss:Type=\"String\">");
excelDoc.Write(source.Tables[worksheet].Columns[x].ColumnName);
excelDoc.Write("</Data></Cell>");
}
excelDoc.Write("</Row>");

for (int x = 0; x < source.Tables[worksheet].DefaultView.Count; x++)
{
rowCount++;
if (rowCount == 64000)
{
rowCount = 0;
sheetCount++;
excelDoc.Write("</Table>");
excelDoc.Write(" </Worksheet>");
excelDoc.Write("<Worksheet ss:Name=\"Sheet" + worksheet + "\">");
excelDoc.Write("<Table>");
}
excelDoc.Write("<Row>"); //ID=" + rowCount + "
for (int y = 0; y < source.Tables[worksheet].Columns.Count; y++)
{
System.Type rowType;
rowType = source.Tables[worksheet].DefaultView[x][y].GetType();
switch (rowType.ToString())
{
case "System.String":
//string XMLstring = x[y].ToString();
string XMLstring = source.Tables[worksheet].DefaultView[x][y].ToString();
XMLstring = XMLstring.Trim();
XMLstring = XMLstring.Replace("&", "&");
XMLstring = XMLstring.Replace(">", ">");
XMLstring = XMLstring.Replace("<", "<");
excelDoc.Write("<Cell ss:StyleID=\"StringLiteral\">" +
"<Data ss:Type=\"String\">");
excelDoc.Write(XMLstring);
excelDoc.Write("</Data></Cell>");
break;
case "System.DateTime":
// DateTime XMLDate = (DateTime)x[y];
DateTime XMLDate = (DateTime)source.Tables[worksheet].DefaultView[x][y];
string XMLDatetoString = "";
XMLDatetoString = XMLDate.Year.ToString() +
"-" +
(XMLDate.Month < 10 ? "0" +
XMLDate.Month.ToString() : XMLDate.Month.ToString()) +
"-" +
(XMLDate.Day < 10 ? "0" +
XMLDate.Day.ToString() : XMLDate.Day.ToString()) +
"T" +
(XMLDate.Hour < 10 ? "0" +
XMLDate.Hour.ToString() : XMLDate.Hour.ToString()) +
":" +
(XMLDate.Minute < 10 ? "0" +
XMLDate.Minute.ToString() : XMLDate.Minute.ToString()) +
":" +
(XMLDate.Second < 10 ? "0" +
XMLDate.Second.ToString() : XMLDate.Second.ToString()) +
".000";
excelDoc.Write("<Cell ss:StyleID=\"DateLiteral\">" +
"<Data ss:Type=\"DateTime\">");
excelDoc.Write(XMLDatetoString);
excelDoc.Write("</Data></Cell>");
break;
case "System.Boolean":
excelDoc.Write("<Cell ss:StyleID=\"StringLiteral\">" +
"<Data ss:Type=\"String\">");
//excelDoc.Write(x[y].ToString());
excelDoc.Write(source.Tables[worksheet].DefaultView[x][y].ToString());
excelDoc.Write("</Data></Cell>");
break;
case "System.Int16":
case "System.Int32":
case "System.Int64":
case "System.Byte":
excelDoc.Write("<Cell ss:StyleID=\"Integer\">" +
"<Data ss:Type=\"Number\">");
// excelDoc.Write(x[y].ToString());
excelDoc.Write(source.Tables[worksheet].DefaultView[x][y].ToString());
excelDoc.Write("</Data></Cell>");
break;
case "System.Decimal":
case "System.Double":
excelDoc.Write("<Cell ss:StyleID=\"Decimal\">" +
"<Data ss:Type=\"Number\">");
// excelDoc.Write(x[y].ToString());
excelDoc.Write(source.Tables[worksheet].DefaultView[x][y].ToString());
excelDoc.Write("</Data></Cell>");
break;
case "System.DBNull":
excelDoc.Write("<Cell ss:StyleID=\"StringLiteral\">" +
"<Data ss:Type=\"String\">");
excelDoc.Write("");
excelDoc.Write("</Data></Cell>");
break;
default:
throw (new Exception(rowType.ToString() + " not handled."));

}
}

excelDoc.Write("</Row>");
}
excelDoc.Write("</Table>");
excelDoc.Write(" </Worksheet>");

}
excelDoc.Write(endExcelXML);
excelDoc.Close();
status = true;
// MessageBox.Show("Data exported to excel file successfuly");
}

catch (Exception)
{
status = false;// MessageBox.Show("Error in exporting data to excel file");
}
}

Re: Adding Header an datetime when User prints excel Sheet.

Curtis Schlak.4-Mar-09 9:42

Curtis Schlak.

4-Mar-09 9:42

@vikram_asv:

Please repost and use the <pre> tag to allow us to easily read your code. Thanks!

"we must lose precision to make significant statements about complex systems."
-deKorvin on uncertainty

Re: Adding Header an datetime when User prints excel Sheet.

Ankur19784-Mar-09 13:26

Ankur1978

4-Mar-09 13:26

Hello Curtis,

I am sending u code which i used and modify with my criteria.I am able to export data to a different worksheets. But all i need is when user prints the excel sheet i need to put header, footer , date and orientation - landscape dynamically . I search but no luck may be anyone can help me about this . Here is the link
http://www.codeproject.com/KB/dotnet/ExportToExcel.aspx

Re: Adding Header an datetime when User prints excel Sheet.

Curtis Schlak.4-Mar-09 15:08

Curtis Schlak.

4-Mar-09 15:08

I don't use Microsoft Office, so I can't answer this directly. However, if you look at this article Dive into SpreadsheetML (Part 1 of 2)[^], they will show you how to set print options and what not.

Also, set the header and footer of an Excel spreadsheet to some distinct values, save it in the XML format, and look for those nodes.

Reverse engineering is the way to go, here.

"we must lose precision to make significant statements about complex systems."
-deKorvin on uncertainty

Re: Adding Header an datetime when User prints excel Sheet.

vikram_asv4-Mar-09 17:03

vikram_asv

4-Mar-09 17:03

Hello Curtis, I am again reposting the code with pre tag so that it can to easy to read what u say
I need to add header and datetime and orientation in excel when user print the excel document.I am not able to add datetime ,orientation,header programmatically .if u could help me greatly appreciated where i can add this items in my code so that i can see when i print excel document.

public static void excelprint(DataSet source, string fileName)
{
bool status = false;
try
{
System.IO.StreamWriter excelDoc;
excelDoc = new System.IO.StreamWriter(fileName);

const string startExcelXML =
<xml version="">
<workbook xmlns="urn:schemas-microsoft-com:office:spreadsheet">
xmlns Blush | :O

="urn:schemas-microsoft-com:office:office"
xmlns:x="urn:schemas-microsoft-com:office:excel"
xmlns:ss="urn:schemas-microsoft-com:office:spreadsheet">
<styles>

<alignment ss:vertical="Bottom">
<borders>
<font />
<interior />
<numberformat />
<protection />
</borders></alignment>

<font x:family="Swiss" ss:bold="1" xmlns:x="#unknown" />

<numberformat ss:format="@" />

<numberformat ss:format="0.0000" />

<numberformat ss:format="0" />

<numberformat ss:format="mm/dd/yyyy;@" />

const string endExcelXML = "</Workbook>";
int rowCount = 0;
int sheetCount = 1;

excelDoc.Write(startExcelXML);

for (int worksheet = 0; worksheet < source.Tables.Count; worksheet++)
{

excelDoc.Write("<Worksheet ss:Name=\"" + source.Tables[worksheet].TableName + "\">");
excelDoc.Write("<Table>");
excelDoc.Write("<Row>");

for (int x = 0; x < source.Tables[worksheet].Columns.Count; x++)
{
excelDoc.Write("<Cell ss:StyleID=\"BoldColumn\"><Data ss:Type=\"String\">");
excelDoc.Write(source.Tables[worksheet].Columns[x].ColumnName);
excelDoc.Write("</Data></Cell>");
}
excelDoc.Write("</Row>");

for (int x = 0; x < source.Tables[worksheet].DefaultView.Count; x++)
{
rowCount++;
if (rowCount == 64000)
{
rowCount = 0;
sheetCount++;
excelDoc.Write("</Table>");
excelDoc.Write(" </Worksheet>");
excelDoc.Write("<Worksheet ss:Name=\"Sheet" + worksheet + "\">");
excelDoc.Write("<Table>");
}
excelDoc.Write("<Row>"); //ID=" + rowCount + "
for (int y = 0; y < source.Tables[worksheet].Columns.Count; y++)
{
System.Type rowType;
rowType = source.Tables[worksheet].DefaultView[x][y].GetType();
switch (rowType.ToString())
{
case "System.String":
//string XMLstring = x[y].ToString();
string XMLstring = source.Tables[worksheet].DefaultView[x][y].ToString();
XMLstring = XMLstring.Trim();
XMLstring = XMLstring.Replace("&", "&");
XMLstring = XMLstring.Replace(">", ">");
XMLstring = XMLstring.Replace("<", "<");
excelDoc.Write("<Cell ss:StyleID=\"StringLiteral\">" +
"<Data ss:Type=\"String\">");
excelDoc.Write(XMLstring);
excelDoc.Write("</Data></Cell>");
break;
case "System.DateTime":
// DateTime XMLDate = (DateTime)x[y];
DateTime XMLDate = (DateTime)source.Tables[worksheet].DefaultView[x][y];
string XMLDatetoString = "";
XMLDatetoString = XMLDate.Year.ToString() +
"-" +
(XMLDate.Month < 10 ? "0" +
XMLDate.Month.ToString() : XMLDate.Month.ToString()) +
"-" +
(XMLDate.Day < 10 ? "0" +
XMLDate.Day.ToString() : XMLDate.Day.ToString()) +
"T" +
(XMLDate.Hour < 10 ? "0" +
XMLDate.Hour.ToString() : XMLDate.Hour.ToString()) +
":" +
(XMLDate.Minute < 10 ? "0" +
XMLDate.Minute.ToString() : XMLDate.Minute.ToString()) +
":" +
(XMLDate.Second < 10 ? "0" +
XMLDate.Second.ToString() : XMLDate.Second.ToString()) +
".000";
excelDoc.Write("<Cell ss:StyleID=\"DateLiteral\">" +
"<Data ss:Type=\"DateTime\">");
excelDoc.Write(XMLDatetoString);
excelDoc.Write("</Data></Cell>");
break;
case "System.Boolean":
excelDoc.Write("<Cell ss:StyleID=\"StringLiteral\">" +
"<Data ss:Type=\"String\">");
//excelDoc.Write(x[y].ToString());
excelDoc.Write(source.Tables[worksheet].DefaultView[x][y].ToString());
excelDoc.Write("</Data></Cell>");
break;
case "System.Int16":
case "System.Int32":
case "System.Int64":
case "System.Byte":
excelDoc.Write("<Cell ss:StyleID=\"Integer\">" +
"<Data ss:Type=\"Number\">");
// excelDoc.Write(x[y].ToString());
excelDoc.Write(source.Tables[worksheet].DefaultView[x][y].ToString());
excelDoc.Write("</Data></Cell>");
break;
case "System.Decimal":
case "System.Double":
excelDoc.Write("<Cell ss:StyleID=\"Decimal\">" +
"<Data ss:Type=\"Number\">");
// excelDoc.Write(x[y].ToString());
excelDoc.Write(source.Tables[worksheet].DefaultView[x][y].ToString());
excelDoc.Write("</Data></Cell>");
break;
case "System.DBNull":
excelDoc.Write("<Cell ss:StyleID=\"StringLiteral\">" +
"<Data ss:Type=\"String\">");
excelDoc.Write("");
excelDoc.Write("</Data></Cell>");
break;
default:
throw (new Exception(rowType.ToString() + " not handled."));

}
}

excelDoc.Write("</Row>");
}
excelDoc.Write("</Table>");
excelDoc.Write(" </Worksheet>");

}
excelDoc.Write(endExcelXML);
excelDoc.Close();
status = true;
// MessageBox.Show("Data exported to excel file successfuly");
}

catch (Exception)
{
status = false;// MessageBox.Show("Error in exporting data to excel file");
}
}

Re: Adding Header an datetime when User prints excel Sheet.

vikram_asv4-Mar-09 17:06

vikram_asv

4-Mar-09 17:06

xml version

/*
<xml version>
<Workbook xmlns="urn:schemas-microsoft-com:office:spreadsheet"
xmlns Blush | :O

="urn:schemas-microsoft-com:office:office"
xmlns:x="urn:schemas-microsoft-com:office:excel"
xmlns:ss="urn:schemas-microsoft-com:office:spreadsheet">
<Styles>
<Style ss:ID="Default" ss:Name="Normal">
<Alignment ss:Vertical="Bottom"/>
<Borders/>
<Font/>
<Interior/>
<NumberFormat/>
<Protection/>
</Style>
<Style ss:ID="BoldColumn">
<Font x:Family="Swiss" ss:Bold="1"/>
</Style>
<Style ss:ID="StringLiteral">
<NumberFormat ss:Format="@"/>
</Style>
<Style ss:ID="Decimal">
<NumberFormat ss:Format="0.0000"/>
</Style>
<Style ss:ID="Integer">
<NumberFormat ss:Format="0"/>
</Style>
<Style ss:ID="DateLiteral">
<NumberFormat ss:Format="mm/dd/yyyy;@"/>
</Style>
</Styles>
<Worksheet ss:Name="Sheet1">
</Worksheet>
</Workbook>
*/

Re: Adding Header an datetime when User prints excel Sheet.

AndyInUK4-Mar-09 23:28

AndyInUK

4-Mar-09 23:28

use pre tag mate as curtis asked you. Look around for pre tag before you post and post once again.

All the best

XML/XSL transformation question.

Maximilien4-Mar-09 9:08

Maximilien

4-Mar-09 9:08

Howdy'

I'm generating some XML files with MSXML; The text is all put on one single line;
so I'm trying to transform the document with some XSL "black" magic.

The stylesheet (taken from some google-search):

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes" />
<xsl:template match="*">
   <xsl:copy>
      <xsl:copy-of select="@*" />
      <xsl:apply-templates />
   </xsl:copy>
</xsl:template>
</xsl:stylesheet>

This works well, except for one minor thing.

For empty "tags", the transformation transform :

<MyTag attribute="value"/>

<MyTag attribute="value"></MyTag >

I'm not quite familiar with XSL, and was wondering if there's a way to not create the empty closing tags ?

I hope it makes sens.
Thanks.
Max.

This signature was proudly tested on animals.

Re: XML/XSL transformation question.

Curtis Schlak.4-Mar-09 9:17

Curtis Schlak.

4-Mar-09 9:17

Is there a reason why you don't want the closing tag? Both of the nodes, the one with the /> and the one with the ></MyTag> have the same meaning in XML and should parse the same way....

"we must lose precision to make significant statements about complex systems."
-deKorvin on uncertainty

Re: XML/XSL transformation question.

Maximilien4-Mar-09 9:29

Maximilien

4-Mar-09 9:29

For no other reason than "aestheticism"

This signature was proudly tested on animals.

Re: XML/XSL transformation question.

Curtis Schlak.4-Mar-09 9:40

Curtis Schlak.

4-Mar-09 9:40

Yeah, I like it that way, too.

No, if you're using MSXML, there's no way to force it to output />. It's hard-coded to do that, as far as I know, having used it for quite a long time.

"we must lose precision to make significant statements about complex systems."
-deKorvin on uncertainty

AndyInUK4-Mar-09 3:59

AndyInUK

4-Mar-09 3:59

how can i setup a parser to log in to particular page and grab a cookie ?

Thanks
Andyyy

Re: cookies

Curtis Schlak.4-Mar-09 9:20

Curtis Schlak.

4-Mar-09 9:20

Most likely, Andy, you'll have to do the following in the programming language of your choice:

Create the content for the login POST request.
Stream the post request to the server over the applicable port.
Read the response.
Parse out the Set-Cookie lines from the HTTP response.

Here's the HTTP RFC[^] for reference.

"we must lose precision to make significant statements about complex systems."
-deKorvin on uncertainty

Re: cookies

AndyInUK4-Mar-09 23:35

AndyInUK

4-Mar-09 23:35

Thanks for your reply.
is there any psudo code which i can follow in xsl or xml ?
Cheers
Andyy Smile | :)

Re: cookies

Curtis Schlak.5-Mar-09 10:59

Curtis Schlak.

5-Mar-09 10:59

This really isn't an XML or XSL problem.

"we must lose precision to make significant statements about complex systems."
-deKorvin on uncertainty

Converting Word ML to XML

Tejabhiram3-Mar-09 21:26

Tejabhiram

3-Mar-09 21:26

Hey everyone

When i save a word document as an XML file it write a large amount of data into the xml document. I have used an XSL file to only capture the required data from the wordML to my own xml file. This customized xml file should have only the tags specified in the xslt. Look into the file below

<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>
- <pkg:package xmlns:pkg="http://schemas.microsoft.com/office/2006/xmlPackage">
- <pkg:part pkg:name="/_rels/.rels" pkg:contentType="application/vnd.openxmlformats-package.relationships+xml" pkg:padding="512">
- <pkg:xmlData>
- <Relationships xmlns="http://schemas.openxmlformats.org/package/2006/relationships">
<Relationship Id="rId3" Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/extended-properties" Target="docProps/app.xml" />
<Relationship Id="rId2" Type="http://schemas.openxmlformats.org/package/2006/relationships/metadata/core-properties" Target="docProps/core.xml" />
<Relationship Id="rId1" Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/officeDocument" Target="word/document.xml" />
</Relationships>
</pkg:xmlData>
</pkg:part>
- <pkg:part pkg:name="/word/_rels/document.xml.rels" pkg:contentType="application/vnd.openxmlformats-package.relationships+xml" pkg:padding="256">
- <pkg:xmlData>
- <Relationships xmlns="http://schemas.openxmlformats.org/package/2006/relationships">
<Relationship Id="rId3" Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/webSettings" Target="webSettings.xml" />
<Relationship Id="rId2" Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/settings" Target="settings.xml" />
<Relationship Id="rId1" Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/styles" Target="styles.xml" />
<Relationship Id="rId5" Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/theme" Target="theme/theme1.xml" />
<Relationship Id="rId4" Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/fontTable" Target="fontTable.xml" />
</Relationships>
</pkg:xmlData>
</pkg:part>
- <pkg:part pkg:name="/word/document.xml" pkg:contentType="application/vnd.openxmlformats-officedocument.wordprocessingml.document.main+xml">
- <pkg:xmlData>
- <w:document xmlns:ve="http://schemas.openxmlformats.org/markup-compatibility/2006" xmlns Blush | :O

="urn:schemas-microsoft-com:office:office" xmlns:r="http://schemas.openxmlformats.org/officeDocument/2006/relationships" xmlns:m="http://schemas.openxmlformats.org/officeDocument/2006/math" xmlns:v="urn:schemas-microsoft-com:vml" xmlns:wp="http://schemas.openxmlformats.org/drawingml/2006/wordprocessingDrawing" xmlns:w10="urn:schemas-microsoft-com:office:word" xmlns:w="http://schemas.openxmlformats.org/wordprocessingml/2006/main" xmlns:wne="http://schemas.microsoft.com/office/word/2006/wordml">
- <w:body>
.................................. so on and so forth

From this file i am only trying to extract the data that written in the document. It goes this way

Name Teja
Age 22

For this i have written a small xsl file which looks this way

<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:w="http://schemas.openxmlformats.org/wordprocessingml/2006/main"
exclude-result-prefixes="w"
version="1.0">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="w:p">
<Details> <xsl:value-of select="."/> </Details>
</xsl:template>
</xsl:stylesheet>

Now i am expecting my result to be an xml document with my own xml tags. This is how i want my document to look like

Name

Teja

Age

Now to look at the result i have been using xml notepad. But to convert WordML to XML i need a convertor. What do i need to use in order to do that?

Please help me out

Thanks in Advance
Teja

Re: Converting Word ML to XML

Lee Humphries3-Mar-09 21:55

Lee Humphries

3-Mar-09 21:55

Tejabhiram wrote:
But to convert WordML to XML i need a convertor

Do you mean something to read the WordML as XML to feed it to your XSLT, or do you mean something to actually run your XSLT?

For the former, use something like 7-zip to extract the source files from inside the docx - I'm guessing that you're already doing this though.

For the latter you can use the msxml commandline tool, or if you want to code it I'd use the .Net XSL compiled transform in your code. Or Saxon if you need something in the Java domain

I just love Koalas - they go great with Bacon.

Re: Converting Word ML to XML

Tejabhiram3-Mar-09 22:12

Tejabhiram

3-Mar-09 22:12

Hey Lee,

Ok let me explain it to u i detail. I need to write a .NET application which can help me open a normal xml file in word. Then when the user modifies the content, i need to store the modified content back to the original xml file. In order to do this, first i am writing a xslt to convert my xml file to word file. Then when the modification is done,I am saving the Word Document as WordML, writing another inverse xsl file to convert WordMl to the original xml format. In the XML to word process, the word itself is taking care of the conversion process. But in the reverse process, I am not able to figure out how to convert the file back to a normal xml file.

From you questions I analyuze that i am trying to do the latter part, i.e run my xsl. Now to do that, you have suggested me to use the .NET XSL compiled transform i my code. How do i do that? Where can I find information regarding this?

Please help me out

Thanks in Advance
Teja

"Why to be a MAN when you can be a SUCCESS?"

Re: Converting Word ML to XML

Lee Humphries3-Mar-09 23:10

Lee Humphries

3-Mar-09 23:10

OK,

Sorry for the shameless plug, but get the code for my article - Introduction to XPS[^] this includes code to use .Net Compiled Transform. Don't worry if you're a VBer I'm sure you can figure out the C#.

I just love Koalas - they go great with Bacon.

Re: Converting Word ML to XML

Tejabhiram3-Mar-09 23:34

Tejabhiram

3-Mar-09 23:34

Hey Lee
ya i googled it from ur earlier reply and was able to find the VB code and i just tested it for transformation. It worked. wow, i have these so called senior developers on my side and it dint strike them when i asked them for this transformation tool. K thanks a million.

Regards
Teja

XML to JSON using XSLT

Nishanthkumar26-Feb-09 19:29

Nishanthkumar

26-Feb-09 19:29

How do i convert an XML file to JSON(Java Script Object Notation) file using XSLT.
Help me......(new to XSLT).

Thanks in Advance

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