FNC1 or Function 1 in EAN/UCC 128 barcodes

[waynesingh]

Hi all,

I don't really think this is where to post this question, but someone may have some ideas.

I have written a program that talks directly to the printers (datamax and intermecs). This all works fine, but was told recently that the barcodes I am putting out are Code 128 and not EAN128. I have been researching this today and have found that to make it a EAN128 barcode I need to give it a FNC1 at the start of the barcode (symbol control character). There are two purposes of FNC1 in a barcode, 1 being a field seperator for non-fixed length fields, and the 2nd being to interpret the barcode as EAN. To make it a field seperator is ASCII(29), but I am having a lot of trouble finding what the value of FNC1 is when used as a symbol control character within the start characters.

If anyone has done any work with this then please let me know how you passed this value to the printer.

I am using TCP/IP socket communication to talk to the printer through a TroyXCD(Serial server) device if that helps at all.

Thank you in advance for any help offered.

Wayne Singh

 

[2ffat]

As far as I know, FNC1 is ascii code 102 decimal or 86 hex. You may also need one of three start codes, 103(decimal)/87(hex) for start code A, 104(decimal)/88(hex) for start code B, or 105(decimal)/89(hex) for start code C. The stop code is 106(decimal).

The UCC/EAN-128 code is made up of a starting quiet zone (blank space), start character A, B, or C, a FNC 1 character, data, a symbol check character, a stop character, and an ending quite zone. That is if my memory serves me correctly.

I've moved all our bar code printing to dedicated bar code printers that format the code correctly without any extra input from me.


James P. Cottingham

http://www.ivcusa.com

 

[waynesingh]

Thanx mate,

I musn't of been thinking clearly, as we are using dedicated label printers, and they do actually do it for you from what I have read. I think I simply hve to change the Barcode type and the rest should be done for me.

Thank you anyway for taking the time to answer my question.
Wayne Singh