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Home > Archive > BizTalk Server Orchestration > September 2004 > Mapping Challenge ....Anyone??
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Mapping Challenge ....Anyone??
|
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| BTFun777 2004-09-28, 3:06 am |
| I have a source schema with looping records:
<SourceNode>
<A>
<B>
<c>
</SourceNode>
I have a destination schema
with
<DestinationNode>
<A1>
<B1>
<C1>
<A2>
<B2>
<C2>
<A3>
<B3>
<C3>
<A4>
<B4>
<C4>
<DestinationNode>
I need to take the first five repeating records in the source schema and
have them in the first record of destination
schema. The next five record in the source schema will then go into the
second record of the destination node.
So, if there are 18 records in the source node,
then there will be four destination records. The first three destination
record will have five records from source node, and
the fifth record in destination record will have the remaining two records
from source schema.
Any pointer will be very much appreciated.
K
| |
| Alan Smith 2004-09-28, 3:06 am |
| Hi,
Maybe it's easyer to use xslt, or C# to solve this one. If you have used the
XML classes in C#, it should be pretty easy to solve. There may be a good way
to solve this with looping in a map, Jeff Lynch has written some good posts
about this in his blog.
http://dotnetjunkies.com/WebLog/jlynch/
Regards,
Alan
"BTFun777" wrote:
> I have a source schema with looping records:
>
>
> <SourceNode>
> <A>
> <B>
> <c>
> </SourceNode>
>
>
>
> I have a destination schema
>
> with
>
> <DestinationNode>
> <A1>
> <B1>
> <C1>
> <A2>
> <B2>
> <C2>
> <A3>
> <B3>
> <C3>
> <A4>
> <B4>
> <C4>
> <DestinationNode>
>
> I need to take the first five repeating records in the source schema and
> have them in the first record of destination
> schema. The next five record in the source schema will then go into the
> second record of the destination node.
> So, if there are 18 records in the source node,
> then there will be four destination records. The first three destination
> record will have five records from source node, and
> the fifth record in destination record will have the remaining two records
> from source schema.
>
> Any pointer will be very much appreciated.
>
>
> K
>
>
>
| |
| Yossi Dahan 2004-09-28, 8:14 am |
| I suggest you'd a look at Itemfield ContentMaster, which is a strong data
transformation tool fully compatible with BizTalk
www.itemfield.com
Yossi Dahan
"BTFun777" <BTFun777@yahoo.com> wrote in message
news:%23LcYqGRpEHA.1300@TK2MSFTNGP12.phx.gbl...
> I have a source schema with looping records:
>
>
> <SourceNode>
> <A>
> <B>
> <c>
> </SourceNode>
>
>
>
> I have a destination schema
>
> with
>
> <DestinationNode>
> <A1>
> <B1>
> <C1>
> <A2>
> <B2>
> <C2>
> <A3>
> <B3>
> <C3>
> <A4>
> <B4>
> <C4>
> <DestinationNode>
>
> I need to take the first five repeating records in the source schema and
> have them in the first record of destination
> schema. The next five record in the source schema will then go into the
> second record of the destination node.
> So, if there are 18 records in the source node,
> then there will be four destination records. The first three destination
> record will have five records from source node, and
> the fifth record in destination record will have the remaining two records
> from source schema.
>
> Any pointer will be very much appreciated.
>
>
> K
>
>
| |
| Dave Howe 2004-09-28, 5:52 pm |
| Ah, yes I see what you mean now.
Let me take a look and get back to you.
Thanks,
Dave
This posting is provided "AS IS" with no warranties, and confers no
rights.
"BTFun777" <BTFun777@yahoo.com> wrote in message
news:eYX57sVpEHA.800@TK2MSFTNGP14.phx.gbl...
> Dave,
> Thanks for your help. Unfortunately, the problem is more complex than
> sample
> you sent. I've added my source and destination schema and a map which
> shows
> the complexity of
> the mapping I need to do.
>
> Below is a quick description; I also have the source schema, destination
> schema, and a sample instance message attached as a zipped file to this
> posting. I added the schema and sample to
> your original LoopingMap.zip file.
>
> MySourceLoop is my source schema
> MyDestination is my destination schema
> MyMap is the map for both
>
>
> LoanInfo in my source schema and it can occur 1 to many times
> LoanInfo node in my source is mapped to LoanRec node in my destination
> LoanRec in my Destination schema can occur 1 to many times
>
> For every 3 records in LoanInfo, there is one LoanRec created
> If there are 10 records in LoanInfo, then there are four LoanRec created.
>
> Of course, the first three LoanRec has 3 LoanInfo record each, then the
> last
> LoanRec has only one LoanInfo in it
>
>
>
> Thanks for your assistance.
>
> K
>
>
>
> "news.microsoft.com" <dhowe@online.microsoft.com> wrote in message
> news:uK71KmTpEHA.2764@TK2MSFTNGP11.phx.gbl...
> functoid.
> index
> rights.
> records
>
>
>
| |
| Dave Howe 2004-09-30, 10:40 am |
| Hi,
After spending some time on this it seems the best way to achieve what you
are loking for is to use the script functoid and create some XSLT as others
on here have suggested already.
Sorry couldn't be of more help, i thought there might be a way to do this
using the functoids.
Thanks,
Dave
This posting is provided "AS IS" with no warranties, and confers no rights.
"Dave Howe" <dhowe@online.microsoft.com> wrote in message
news:%2367sfYWpEHA.3572@TK2MSFTNGP10.phx.gbl...
> Ah, yes I see what you mean now.
>
> Let me take a look and get back to you.
> Thanks,
> Dave
>
> This posting is provided "AS IS" with no warranties, and confers no
> rights.
>
>
> "BTFun777" <BTFun777@yahoo.com> wrote in message
> news:eYX57sVpEHA.800@TK2MSFTNGP14.phx.gbl...
>
>
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