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zebra-printerszplzpl-ii

ZPL How to Center Barcode (Code 128)

i'd like to know how to center barcode code 128. on the picture you should see that it is right now left justified. Label

my zpl:

^XA
^LH10,10
^FO0,0^XGE:SWESE001.GRF^FS
^FO440,0^XGE:SWESE000.GRF^FS
^FO0,70^FB550,50,0,C,0^AQN,25,30^FDSpraynozzle 50mm^FS
^FO0,130^BY2^BCN,30,Y,Y,N,N^FDS/N:941001-0114-0001^FS
^FO180,170^AQN,23,20^FDwww.swepro.com^FS
^XZ
Solution

  • I'm posting my solution in case someone is looking for it. As E_S mentions, in order to center a barcode in a label you have to calculate it by code following these steps:

    • Check your narrow bar width, in your case 2 (^BY2)
    • Find out your label total width in dots. For this you have to know what is your printer's resolution (eg: 8 dots / mm). so if you have a 80 mm wide label, 80 * 8 = 640 dots
    • Count each character in your barcode, including invocation codes and check digit as specified below. For information on invocation codes see: https://www.zebra.com/content/dam/zebra/manuals/en-us/software/zpl-zbi2-pm-en.pdf (Page 95)
    • Note that invocation codes (">:", ">5", etc.) count as one character, and that characters in mode C are stored in pairs. For more information on mode C, refer to http://en.wikipedia.org/wiki/Code_128
    • If your barcode is >:S/N:941001-0114-0001 you have to count [start code B] + [20 characters] + [check digit] = 22
    • If your barcode is >:S/N:>5941001>6->50114>6->50001 you have to count [start code B] + [4 characters for 'S/N:'] + [mode C invocation] + [3 characters for '941001'] + [mode B invocation] + [1 characters for '-'] + [mode C invocation] + [2 characters for '0114'] + [mode B invocation] + [1 characters for '-'] + [mode C invocation] + [2 characters for '0001'] + [check digit] = 20
    • Every character occupies 11 units mixing spaces and bars, with the exception of stop code that has 2 extra units (that is a total of 13)
    • Here comes the good stuff... The barcode width is: ((chars counted [22 or 20] * 11) + (stop char * 13)) * narrow bar width = 510 dots or 466 dots
    • Now all we have to do is round((label width - barcode width) / 2) and use that to position the barcode with ^FT

    That's it! Hope it helps someone!