VikramSingh
04-04-2008, 12:22 AM
Hi Alain:
i have one XML input , and with the help of XSLT i have to bring the expected output.
here is my XML input:
<Attributes Protected="false" AddNewP="false" ReOrderP="false">
<ProtectedItem Protected="true">
<Value Name="FA1" DataType="Value" Filter="String">
<string>sdfg</string>
</Value>
</ProtectedItem>
<ProtectedItem Protected="true">
<Value Name="FA12" DataType="FileLink" Filter="*">
<string>D:\TimeSheet\2008\AVLIS_Timesheet_TFMS_VikramSingh_CW_13.xls</string>
</Value>
</ProtectedItem>
<ProtectedItem Protected="false">
<Value Name="FA13" DataType="WebLink" Filter="*">
<string>http://localhost:1799/CfgWS</string>
</Value>
</ProtectedItem>
<ProtectedItem Protected="true">
<Value Name="FA14" DataType="Attachment" Filter="*.*">
<ReferenceObjectPaco Name="corpora.txt" Date="2007-08-02T10:15:39.65625+05:30" Size="699" FilePath="C:\Documents and Settings\SINGHV\Desktop\Jan Kachra\corpora.txt" Modified="true" Zipped="true" ReferenceType="Attachment" DataType="Attachment">
<User />
<Reference Type="Attachment" Name="f7cf713e-dd90-4f5c-a9a1-637084e7385a" PacoRefName="Ref:FA[FA14]" />
<ReferenceValue xmlns:q8="http://www.w3.org/2001/XMLSchema" p7:type="q8:string" xmlns:p7="http://www.w3.org/2001/XMLSchema-instance">corpora.txt</ReferenceValue>
</ReferenceObjectPaco>
</Value>
</ProtectedItem>
<ProtectedItem Protected="false">
<Value Name="FA15" DataType="Standard" Filter="*.*">
<ReferenceObjectPaco Name="FA15" Date="2008-04-01T14:02:03.6875+05:30" UserName="System" Size="80384" FilePath="C:\Documents and Settings\leinera\Desktop\TFMS Training\TFMS-Glossary V1_1.xls" Modified="false" Zipped="false" ReferenceType="Standard" DataType="Standard" Filter="*.*">
<User />
<Reference Type="Standard" SubType="xls" Name="Glossary" PacoRefName="Ref:FA[FA15]" />
<ReferenceValue xmlns:q9="http://www.w3.org/2001/XMLSchema" p7:type="q9:string" xmlns:p7="http://www.w3.org/2001/XMLSchema-instance">Glossary</ReferenceValue>
</ReferenceObjectPaco>
</Value>
</ProtectedItem>
<ProtectedItem Protected="true">
<Value Name="FA16" DataType="Value" Filter="Integer">
<string>234</string>
</Value>
</ProtectedItem>
</Attributes>
using XSLT i have to bring the following output:
<FreeAttribute Name="FA1" DataType="Value" Filter="String" Value="sdfg" />
<FreeAttribute Name="FA12" DataType="FileLink" Filter="*" Value="D:\TimeSheet\2008\AVLIS_Timesheet_TFMS_VikramSingh_CW_13.xls" />
<FreeAttribute Name="FA13" DataType="WebLink" Filter="*" Value="http://localhost:1799/CfgWS" />
<FreeAttribute Name="FA14" DataType="Attachment" Filter="*.*" Value="" />
<FreeAttribute Name="FA15" DataType="Standard" Filter="*.*" Value="" />
<FreeAttribute Name="FA16" DataType="Value" Filter="Integer" Value="234" />
My question is what is to be written in XSLT so that i could bring the above output. i kindly request anyone to reply to this request as soon as possible
i have one XML input , and with the help of XSLT i have to bring the expected output.
here is my XML input:
<Attributes Protected="false" AddNewP="false" ReOrderP="false">
<ProtectedItem Protected="true">
<Value Name="FA1" DataType="Value" Filter="String">
<string>sdfg</string>
</Value>
</ProtectedItem>
<ProtectedItem Protected="true">
<Value Name="FA12" DataType="FileLink" Filter="*">
<string>D:\TimeSheet\2008\AVLIS_Timesheet_TFMS_VikramSingh_CW_13.xls</string>
</Value>
</ProtectedItem>
<ProtectedItem Protected="false">
<Value Name="FA13" DataType="WebLink" Filter="*">
<string>http://localhost:1799/CfgWS</string>
</Value>
</ProtectedItem>
<ProtectedItem Protected="true">
<Value Name="FA14" DataType="Attachment" Filter="*.*">
<ReferenceObjectPaco Name="corpora.txt" Date="2007-08-02T10:15:39.65625+05:30" Size="699" FilePath="C:\Documents and Settings\SINGHV\Desktop\Jan Kachra\corpora.txt" Modified="true" Zipped="true" ReferenceType="Attachment" DataType="Attachment">
<User />
<Reference Type="Attachment" Name="f7cf713e-dd90-4f5c-a9a1-637084e7385a" PacoRefName="Ref:FA[FA14]" />
<ReferenceValue xmlns:q8="http://www.w3.org/2001/XMLSchema" p7:type="q8:string" xmlns:p7="http://www.w3.org/2001/XMLSchema-instance">corpora.txt</ReferenceValue>
</ReferenceObjectPaco>
</Value>
</ProtectedItem>
<ProtectedItem Protected="false">
<Value Name="FA15" DataType="Standard" Filter="*.*">
<ReferenceObjectPaco Name="FA15" Date="2008-04-01T14:02:03.6875+05:30" UserName="System" Size="80384" FilePath="C:\Documents and Settings\leinera\Desktop\TFMS Training\TFMS-Glossary V1_1.xls" Modified="false" Zipped="false" ReferenceType="Standard" DataType="Standard" Filter="*.*">
<User />
<Reference Type="Standard" SubType="xls" Name="Glossary" PacoRefName="Ref:FA[FA15]" />
<ReferenceValue xmlns:q9="http://www.w3.org/2001/XMLSchema" p7:type="q9:string" xmlns:p7="http://www.w3.org/2001/XMLSchema-instance">Glossary</ReferenceValue>
</ReferenceObjectPaco>
</Value>
</ProtectedItem>
<ProtectedItem Protected="true">
<Value Name="FA16" DataType="Value" Filter="Integer">
<string>234</string>
</Value>
</ProtectedItem>
</Attributes>
using XSLT i have to bring the following output:
<FreeAttribute Name="FA1" DataType="Value" Filter="String" Value="sdfg" />
<FreeAttribute Name="FA12" DataType="FileLink" Filter="*" Value="D:\TimeSheet\2008\AVLIS_Timesheet_TFMS_VikramSingh_CW_13.xls" />
<FreeAttribute Name="FA13" DataType="WebLink" Filter="*" Value="http://localhost:1799/CfgWS" />
<FreeAttribute Name="FA14" DataType="Attachment" Filter="*.*" Value="" />
<FreeAttribute Name="FA15" DataType="Standard" Filter="*.*" Value="" />
<FreeAttribute Name="FA16" DataType="Value" Filter="Integer" Value="234" />
My question is what is to be written in XSLT so that i could bring the above output. i kindly request anyone to reply to this request as soon as possible