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Code 128 barcode specification

admin November 23, 2012 Barcode Specification, Knowledge Base No Comments

Example:

Code 128 is one of the more complex linear barcodes, with a greater capacity than Code 25 or Code 39, but requiring better quality images for reliable recognition.

Key features:

  • 4 different width of bar/space.
  • 3 symbol sets (A, B and C).
  • Symbol set C encodes pairs of digits and is one of the best ways to store large numbers in a linear barcode.
  • Ability to switch between symbol sets.
  • Barcodes include start and stop character plus a built in checksum.
  • Each character is made up of 3 bars and 3 spaces – except for the stop character which has 4 bars and 3 spaces.
  • The number of black bars should leave remainder 1 when divided by 3.

A Code 128 barcode is written as follows:

Start character (start A, start B or start C) (3 bars)
1 or more data characters (3 bars each)
Checksum character – calculated from the data characters (3 bars)
Stop character (4 bars)

Specification:

value A B C Pattern
0 SPACE SPACE 00 2 1 2 2 2 2
1 ! ! 01 2 2 2 1 2 2
2 “ “ 02 2 2 2 2 2 1
3 # # 03 1 2 1 2 2 3
4 $ $ 04 1 2 1 3 2 2
5 % % 05 1 3 1 2 2 2
6 & & 06 1 2 2 2 1 3
7 ‘ ‘ 07 1 2 2 3 1 2
8 ( ( 08 1 3 2 2 1 2
9 ) ) 09 2 2 1 2 1 3
10 * * 10 2 2 1 3 1 2
11 + + 11 2 3 1 2 1 2
12 , , 12 1 1 2 2 3 2
13 – – 13 1 2 2 1 3 2
14 . . 14 1 2 2 2 3 1
15 / / 15 1 1 3 2 2 2
16 0 0 16 1 2 3 1 2 2
17 1 1 17 1 2 3 2 2 1
18 2 2 18 2 2 3 2 1 1
19 3 3 19 2 2 1 1 3 2
20 4 4 20 2 2 1 2 3 1
21 5 5 21 2 1 3 2 1 2
22 6 6 22 2 2 3 1 1 2
23 7 7 23 3 1 2 1 3 1
24 8 8 24 3 1 1 2 2 2
25 9 9 25 3 2 1 1 2 2
26 : : 26 3 2 1 2 2 1
27 ; ; 27 3 1 2 2 1 2
28 < < 28 3 2 2 1 1 2
29 = = 29 3 2 2 2 1 1
30 > > 30 2 1 2 1 2 3
31 ? ? 31 2 1 2 3 2 1
32 @ @ 32 2 3 2 1 2 1
33 A A 33 1 1 1 3 2 3
34 B B 34 1 3 1 1 2 3
35 C C 35 1 3 1 3 2 1
36 D D 36 1 1 2 3 1 3
37 E E 37 1 3 2 1 1 3
38 F F 38 1 3 2 3 1 1
39 G G 39 2 1 1 3 1 3
40 H H 40 2 3 1 1 1 3
41 I I 41 2 3 1 3 1 1
42 J J 42 1 1 2 1 3 3
43 K K 43 1 1 2 3 3 1
44 L L 44 1 3 2 1 3 1
45 M M 45 1 1 3 1 2 3
46 N N 46 1 1 3 3 2 1
47 O O 47 1 3 3 1 2 1
48 P P 48 3 1 3 1 2 1
49 Q Q 49 2 1 1 3 3 1
50 R R 50 2 3 1 1 3 1
51 S S 51 2 1 3 1 1 3
52 T T 52 2 1 3 3 1 1
53 U U 53 2 1 3 1 3 1
54 V V 54 3 1 1 1 2 3
55 W W 55 3 1 1 3 2 1
56 X X 56 3 3 1 1 2 1
57 Y Y 57 3 1 2 1 1 3
58 Z Z 58 3 1 2 3 1 1
59 [ [ 59 3 3 2 1 1 1
60 \ \ 60 3 1 4 1 1 1
61 ] ] 61 2 2 1 4 1 1
62 ^ ^ 62 4 3 1 1 1 1
63 _ _ 63 1 1 1 2 2 4
64 NUL ` 64 1 1 1 4 2 2
65 SOH a 65 1 2 1 1 2 4
66 STX b 66 1 2 1 4 2 1
67 ETX c 67 1 4 1 1 2 2
68 EOT d 68 1 4 1 2 2 1
69 ENQ e 69 1 1 2 2 1 4
70 ACK f 70 1 1 2 4 1 2
71 BEL g 71 1 2 2 1 1 4
72 BS h 72 1 2 2 4 1 1
73 HT i 73 1 4 2 1 1 2
74 LF j 74 1 4 2 2 1 1
75 VT k 75 2 4 1 2 1 1
76 FF I 76 2 2 1 1 1 4
77 CR m 77 4 1 3 1 1 1
78 SO n 78 2 4 1 1 1 2
79 SI o 79 1 3 4 1 1 1
80 DLE p 80 1 1 1 2 4 2
81 DC1 q 81 1 2 1 1 4 2
82 DC2 r 82 1 2 1 2 4 1
83 DC3 s 83 1 1 4 2 1 2
84 DC4 t 84 1 2 4 1 1 2
85 NAK u 85 1 2 4 2 1 1
86 SYN v 86 4 1 1 2 1 2
87 ETB w 87 4 2 1 1 1 2
88 CAN x 88 4 2 1 2 1 1
89 EM y 89 2 1 2 1 4 1
90 SUB z 90 2 1 4 1 2 1
91 ESC { 91 4 1 2 1 2 1
92 FS | 92 1 1 1 1 4 3
93 GS } 93 1 1 1 3 4 1
94 RS ~ 94 1 3 1 1 4 1
95 US DEL 95 1 1 4 1 1 3
96 FNC 3 FNC 3 96 1 1 4 3 1 1
97 FNC 2 FNC 2 97 4 1 1 1 1 3
98 SHIFT B SHIFT A 98 4 1 1 3 1 1
99 CODE C CODE C 99 1 1 3 1 4 1
100 CODE B FNC 4 CODE B 1 1 4 1 3 1
101 FNC 4 CODE A CODE A 3 1 1 1 4 1
102 FNC 1 FNC 1 FNC 1 4 1 1 1 3 1
103 start A start A start A 2 1 1 4 1 2
104 start B start B start B 2 1 1 2 1 4
105 start C start C start C 2 1 1 2 3 2
106 stop stop stop 2 3 3 1 1 1 2

Note:

The CODE A, B and C symbols latch all following characters to the specified symbol set where as SHIFT A and SHIFT B just change the interpretation of the next character.

Checksum Calculation:

The checksum is calculated as follows:

  1. Multiply the value of each character by it’s position in the barcode (indexed from 1).
  2. Add up all the results from step 1.
  3. Add on the value of the start character.
  4. The checksum is remainder on division by 103.

For example, if you are encoding the string ABCD1234 in symbol set A then the checksum values will be:

Steps 1 and 2:

(33 * 1) + (34 * 2) + (35 * 3) + (36 * 4) + (17 * 5) + (18 * 6) + (19 * 7) + (20 * 8) = 836

Step 3:

836 + 103 = 939

Step 4:

939 mod 103 = 12

So the checksum character should have value 12.

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