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BizTalk Training 3 – Customize Filename Dynamically

BizTalk Training 3 – Customize Filename Dynamically

November 24, 2007 Programming 10 Comments
This entry is part 3 of 5 in the series BizTalk Training

BizTalk Training

Customize Filename Dynamically

This tutorial shows how to customize output filename inside orchestration.

There are 5 tutorials in this training.

  1. BizTalk Training 1 – basic send/receive xml message
  2. BizTalk Training 2 – receive/send plain text
  3. BizTalk Training 3 – customize filename dynamically
  4. BizTalk Training 4 – using Functiods in Map
  5. BizTalk Training 5 – using FTP Adapter

Objectives

  • Learn how to change filename of the output file.

Prerequisites

  • BizTalkTraining 2 project.
  • Configuration on BizTalk Administration for BizTalkTraining application.

Topics

  1. Add variable in the orchestration.
  2. Add msgPerson2.
  3. Add Message Assignment to the orchestration.
  4. Test application.

Step-by-step

Suppose that I want to change name of outgoing file. As in training 2, the input filename is *.txt format but I want the outgoing message type as xml format and also add “output_” text before. E.g. incoming filename is “person1.txt”, outgoing filename will be “output_person1.xml”.

  1. Add variable in the orchestration.
    • Open Orchestration1 in BizTalkTraining. (Project from BizTalk Training 2).
      BizTalkTraining2 project
    • Switch to Orchestration View, right click on Variables and select New Variable.
      Add new variable
    • Click on the variable.
      1. On Variable Property, change Identifier to “fileName”.
      2. Change Type to “string”.
        Change variable's type
    • Now you’ve add variable name “fileName” as string to the Orchestration1.
  2. Add msgPerson2.
    • Create new Message on the orchestration name “msgPerson2” and set Message Type as the same as msgPerson (BizTalkTraining.FlatFilePersonSchema).
      Change Message Type
  3. Add Message Assignment to the orchestration.
    • Drag Message Assignment from Toolbox to bottom of the Orchestration1.
      Create Message Assignment
    • Click on ConstructMessage_1 object. Notice that you highlights entire Construct Message object.
    • On Construct Message Properties, check msgPerson2 in Message Constructed and change Name to “Construct msgPerson2”.
      Set Message Assignment's properties
    • Click on Message Assignment (the object inside Construct Message) and change Name to “Assign fileName” in Message Assignment Properties.
      Change Message Assignment's name
    • Double click on Message Assignment and fill in the code below and click OK. 
      msgPerson2 = msgPerson;
 
fileName = msgPerson2(FILE.ReceivedFileName);
fileName = System.IO.Path.GetFileName(fileName.Replace(".txt",".xml"));
 
msgPerson2(FILE.ReceivedFileName) = "output_" + fileName;

      The code above assigns msgPerson to msgPerson2 and assigns incoming filename to “filename” variable, modify string and assign back to “msgPerson2”.
      Write custom code on Message Assignment

    • Add new send component to the bottom of the orchestration and change Message to “msgPerson”.
      Add new Send component from Toolbox
    • Create new send logical port.
      1. Right click on Port Surface area, select New Configured Port.
        Create New Port
      2. Set name to “SendXMLPersonPort”.
      3. Set Port Type as following configurations.
        Port Type: Create a new Port Type
        Port Type Name: SendXMLPersonPortPortType
        Communication Pattern: One-Way
        Access Restrictions: Internal
        Configure Port Type
      4. Set Port Binding as below.
        Port direction: I’ll always be sending messages on this port.
        Port binding: Specify Now.
        Transport: FILE
        URI: C:\FILE2\OUT\%SourceFileName%
        Send pipeline: XMLTransmit
        Configure Port Binding
    • Connect green button from Send component to SendXMLPersonPort.
      Connect line between ports
    • Deploy the project.
      Deploy the project
  4. Test the application.
    • Open BizTalk Administration and restart BizTalk host.
      Restart the host server
    • Refresh BizTalkTraining application.
      Refresh the application
    • Test application by copy person1.txt from C:\FILE2\person2.txt to C:\FILE2\IN.
      Note: You must any delete files in C:\FILE2\OUT before.
    • You’ll see output in C:\FILE2\OUT.
      Result files
    • Person1.txt
      Person1.txt
    • Output_person1.xml
      Person1.xml

Resources

You can download the example source code (zip file) at here.
The zip file contains

  • BizTalkTraining folder which is the example project on Visual Studio 2005.
  • FILE2 folder which contains IN and OUT empty sub folders for test receive and send message and person1.txt which is an example text message.
  • binding.xml file which is a configuration file for import in BizTalk Administration.
Series Navigation<< BizTalk Training 2 – Receive/Send Plain TextBizTalk Training 4 – Using Functiods in Map >>

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10 Comments

  1. John September 25, 2008

    So where’s the output_person1.txt file? I just see the person1.txt and person1.xml files.

  2. linglom September 26, 2008

    It was my mistake. The output should be output_person1.xml not person1.xml. I’ll correct it soon.

  3. Tahir December 17, 2008

    Very nice! It helped me and also solved my problem too.
    Thanks

  4. Priyan October 4, 2009

    Nice Article. Thanks a lot linglom!

  5. Nigel Bachmann January 8, 2010

    Excellent article, exactly what I wanted.

  6. Foon Lam January 20, 2010

    I’ve tried out the example and the reason why the output filename is person1.xml and not output_person1.xml is because in the message assignment, “msgPerson2(FILE.ReceivedFileName);” will return you the full path and filename “C:\FILE2\OUT\person1.txt” and not just “person1.txt”, so when you append “output_” to the filename, you get “output_C:\FILE2\OUT\person1.xml”. The send port only uses what’s after the last “\”, so it still ends up as “person1.xml”. To correct this you should use

    filename = System.IO.Path.GetFileName(filename.Replace(“.txt”,”.xml”));
    filename = “output_” + filename;

    instead of just

    filename = filename.Replace(“.txt”,”.xml”);
    filename = “output_” + filename;

    This resolution came courtesy of Thiago Almeida,MVP. If you are interested in reading some of his articles, you can get to his site at “http://connectedthoughts.wordpress.com/”

  7. linglom January 21, 2010

    Hi, Foon Lam
    Thanks for correcting, it was my fault.
    I have fixed it already.

  8. Poojith July 25, 2010

    Hi,

    I guess this method works only when we use FILE adapter as the receive port. My question is how to name the file dynamically for WCF adapter receive ports?

    Hope you can help me with this.

    Thank you in advance.

  9. Raman May 22, 2012

    I had run the solution but I am getting only .txt file and not the .xml file. I am not getting any error message. I guess the control is not moving after Send_1. please help.

  10. Sohail July 19, 2012

    Hi,

    I am getting below error while assigning datatype of variable .

    Attempted to read or write protected memory. This is often an indication that other memory is corrupt.

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