xmldatareader?

  • Thread starter Thread starter bic
  • Start date Start date
B

bic

Hi,

Within .NET Framework there is System.Data.SqlClient namespace and within
that there is SqlDataReader object that has a read method. As shown in the
code this read method can read in desired dataset and store it in a control
and display it. Instead of SQL Server data I am dealing with XML data file
and with XSL formatter and would like to know a compable way of reading in
desired XML data or nodes. Thanks

-----SQLreader--------

Dim MyConnection As SqlConnection
Dim MyCommand As SqlCommand
Dim MyReader As SqlDataReader
Dim CT1 As String
Dim CaseTrackConnectionString As String
CaseTrackConnectionString =
ConfigurationManager.ConnectionStrings("CollectionConnectionString").ConnectionString
MyConnection = New SqlConnection(CaseTrackConnectionString)
MyCommand = New SqlCommand
CT1 = "SELECT * FROM Driver WHERE DriverID='" +
DriverDropDownList.SelectedValue.ToString + "'"
MyCommand.CommandText = CT1
MyCommand.CommandType = CommandType.Text
MyCommand.Connection = MyConnection
MyCommand.Connection.Open()
MyReader = MyCommand.ExecuteReader(CommandBehavior.CloseConnection)
While MyReader.Read()
If MyReader("DriverMobile") IsNot System.DBNull.Value Then
DriverMobileLabel.Text = MyReader("DriverMobile")
EndIf

EndWhile
MyCommand.Dispose()
MyConnection.Close()
MyConnection.Dispose()
 
Mark,
Thanks for the prompt reply. I saw in your referenced sample code only has
a XML file but has no XSL stylesheet file and wonder what I need to do
differently. Below is my xsl file which transform xml file into a subset, and
it also converts elements into attributes and sets a key.

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxsl="urn:schemas-microsoft-com:xslt"
exclude-result-prefixes="msxsl"<xsl:output method="xml" indent="yes"/>
hotelId
<xsl:key name="hotelId-key" match="Hotel" use="hotelId"/>
<xsl:template match="HotelAvailabilityListResults">
<HotelAvailabilityListResults>
<xsl:apply-templates select="Hotel"/>
</HotelAvailabilityListResults>
</xsl:template>

<xsl:template match="Hotel">
<xsl:copy>
<xsl:copy-of select="name | address1 | city | stateProvince | country |
postalCode | shortDescription | HotelProperty/RateInfo"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>


Thanks,
 
Mark,

Since I have a xsl stylesheet for transformation, I take it I can first
transform it to another xml file and then follow your suggested method.
Correct?
 
can you show me some code for returning FirstName and LastName based on
employeeID
i.e. SELECT * FROM employee WHERE employeeId=1 from the following xml file

<?xml version="1.0" encoding="utf-8" ?>
<Employees>
<Employee>
<EmployeeId>1</EmployeeId>
<FirstName>Tom</FirstName>
<LastName>Jones</LastName>
</Employee>
<Employee>
<EmployeeId>2</EmployeeId>
<FirstName>John</FirstName>
<LastName>Doe</LastName>
</Employee>
</Employees>

Thanks,
 
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