Sunday, September 9, 2018

Communication 9020/9030




Was 06h received after you send 05h? If yes your wiring is correct. If no please check wiring and com setting.
Yes the wiring is same as 9450






If sending a job to library (same as 9Bh in 9450), you can use 9Fh instead of E3h. The string will be like this:
05
 9F
 00 55
00 01 39 30 32 30 53 41 4D 50 44 08 C9 60 30 30 30 30 30 30 30 30 30 39 39 39 39 39 39 39 39 39 30 31 00 00 00 01 30 39 30 39 30 39 30 39 30 39 30 39 30 39 30 39 30 39 30 39 30 39 30 39 30 39 30 39 30 39
0A
80 01
0D
 01
10
48 65 6C 6C 6F
10
01
 0D
80 01
0D
 6F

This is sending a message named “9020sample” with 1 line 7 dot message “Hello” to library position 001. Follow reply can be expected if download is successful:
06 C5 00 01 00 C4

For requesting print ACK, the PC sends following string to activate this function:
D8 00 00 D8
Then printer will send following ACK string after every printer:
CE 00 00 CE


9450

The problem may because you missed some process, some logic for 9450 is different from 9040.
When you use 9B command:
After PC send 9B to 9450, 9450 will return 06 and  C5 00 01 C4
Now the PC need to send 06 to 9450 again.
Please try to modify according to this point.

D8h
sending request of the print acknowledgement by the computer
mengirim permintaan pengakuan cetak oleh komputer
Ceh
sending  of the print acknowledgement by the printer
mengirim pengakuan cetak oleh printer



41h
print acknowledgement request from the computer
cetak permintaan pengakuan dari komputer


CODE 39 BACKGROUND INFORMATION




Code 39, the first alpha-numeric symbology to be developed, is still widely used-especially in non-retail environments. It is the standard bar code used by the United States Department of Defense, and is also used by the Health Industry Bar Code Council (HIBCC). Code 39 is also known as "3 of 9 Code" and "USD-3".


A typical Code 39 bar code is:




Code 39 is a discrete, variable-length symbology. It is self-checking in that a single print defect cannot transpose one character into another valid character.
Since Code 39 is self-checking, a check digit normally isn't necessary. However, in applications that require an extremely high level of accuracy a modulo 43 checksum digit may be added.
To calculate the optional checksum digit, follow the following steps.
  1. Take the value (0 through 42) of each character in the bar code. The start and stop characters are not included in the checksum calculation.
  2. Sum the value of each of the values of each of the characters described in step 1.
  3. Divide the result from step 2 by 43.
  4. The remainder from the division in step 3 is the checksum character that will be appended to the data message before the stop character.

Once the checksum digit has been calculated we know the entire message which must be encoded in the bars and spaces. Continuing with our example, we will encode, from zero, the Code 39 bar code we used in our example above: HI345678 with a checksum digit of 67.
In the following text, we will discuss the encoding of the bar code by considering that the number "1" represents a "dark" or "bar" section of the bar code whereas a "0" represents a "light" or "space" section of the bar code. Thus the numbers 1101 represents a double-wide bar (11), followed by a single-wide space (0), followed by a single-wide bar (1). This would be printed in the bar code as:



STRUCTURE OF A CODE 39 BARCODE
A Code 39 bar code has the following structure:
  1. A start character - the asterisk (*) character.
  2. Any number of characters encoded from the table below.
  3. An optional checksum digit calculated as described above and encoded from the table below.
  4. A stop character, which is a second asterisk character.

This table indicates how to encode each digit of a Code 39 bar code. Note that the "Width Encoding" column is expressed as "N" for narrow and "W" for wide while the "Barcode Encoding" column represents how the bar code will actually be encoded as described above in "Encoding the Symbol."
Keep in mind that each character begins and ends with a bar, thus the "bar code encoding" always starts and ends with a "1".

CHECK
VALUE
ASCII
CHAR
WIDTH
ENCODING
BARCODE
ENCODING
CHECK
VALUE
ASCII
CHAR
WIDTH
ENCODING
BARCODE
ENCODING
0
0
NNNWWNWNN
101001101101
22
M
WNWNNNNWN
110110101001
1
1
WNNWNNNNW
110100101011
23
N
NNNNWNNWW
101011010011
2
2
NNWWNNNNW
101100101011
24
O
WNNNWNNWN
110101101001
3
3
WNWWNNNNN
110110010101
25
P
NNWNWNNWN
101101101001
4
4
NNNWWNNNW
101001101011
26
Q
NNNNNNWWW
101010110011
5
5
WNNWWNNNN
110100110101
27
R
WNNNNNWWN
110101011001
6
6
NNWWWNNNN
101100110101
28
S
NNWNNNWWN
101101011001
7
7
NNNWNNWNW
101001011011
29
T
NNNNWNWWN
101011011001
8
8
WNNWNNWNN
110100101101
30
U
WWNNNNNNW
110010101011
9
9
NNWWNNWNN
101100101101
31
V
NWWNNNNNW
100110101011
10
A
NNWWNNWNN
110101001011
32
W
WWWNNNNNN
110011010101
11
B
NNWNNWNNW
101101001011
33
X
NWNNWNNNW
100101101011
12
C
WNWNNWNNN
110110100101
34
Y
WWNNWNNNN
110010110101
13
D
NNNNWWNNW
101011001011
35
Z
NWWNWNNNN
100110110101
14
E
WNNNWWNNN
110101100101
36
-
NWNNNNWNW
100101011011
15
F
NNWNWWNNN
101101100101
37
.
WWNNNNWNN
110010101101
16
G
NNNNNWWNW
101010011011
38
SPACE
NWWNNNWNN
100110101101
17
H
WNNNNWWNN
110101001101
39
$
NWNWNWNNN
100100100101
18
I
NNWNNWWNN
101101001101
40
/
NWNWNNNWN
100100101001
19
J
NNNNWWWNN
101011001101
41
+
NWNNNWNWN
100101001001
20
K
WNNNNNNWW
110101010011
42
%
NNNWNWNWN
101001001001
21
L
NNWNNNNWW
101101010011
n/a
*
NWNNWNWNN
100101101101
If a Code 39 bar code starts with a space, that bar code will be appended to any previous code 39 bar code in the buffer and the system will wait for additional bar code(s). If a Code 39 bar code doesn't start with a space, the bar code will be appended to any previous code 39 bar codes and the entire message will be delivered to the application.
In other words, if a code 39 bar code has additional bar codes to follow, it must start with a space-if the bar code is the last bar code in the message it must not start with a space.

We will now code the example we used above, TEST8052. In this case we will not use a check digit.
  1. The START character (*): 100101101101.
  2. The digit "T": enocded as 101011011001.
  3. The digit "E": enocded as 110101100101.
  4. The digit "S": enocded as 101101011001.
  5. The digit "T": enocded as 101011011001.
  6. The digit "8": enocded as 110100101101.
  7. The digit "0": enocded as 101001101101.
  8. The digit "5": enocded as 110100110101.
  9. The digit "2": enocded as 101100101011.
  10. The STOP character (*): 100101101101.
This is shown in the following graphical representation where the bar code has been sectioned-off into areas that reflect each of the 10 components just mentioned.




NOTE: In the above encoding example note that there is an inter-character space between each character. This is not listed in the list of 10 components, but there is an inter-character space between each character. This inter-character space is represented in the graphic by the white space separating the grey areas.

EXTENDED CODE 39 ENCODING TABLE
It is possible, using Code 39's "Full ASCII Mode" to encode all 128 ASCII characters. This is accomplished by using the $, /, %, and + symbols as "shift" characters. Those characters combined with the single character that follows indicate which Full ASCII character is to be used.
ASCII
ENCODING
ASCII
ENCODING
ASCII
ENCODING
ASCII
ENCODING
NUL
%U
SP
Space
@
%V
`
%W
SOH
$A
!
/A
A
A
a
+A
STX
$B
"
/B
B
B
b
+B
ETX
$C
#
/C
C
C
c
+C
EOT
$D
$
/D
D
D
d
+D
ENQ
$E
%
/E
E
E
e
+E
ACK
$F
&
/F
F
F
f
+F
BEL
$G
'
/G
G
G
g
+G
BS
$H
(
/H
H
H
H
H
HT
$I
)
/I
I
I
i
+I
LF
$J
*
/J
J
J
j
+J
VT
$K
+
/K
K
K
k
+K
FF
$L
,
/L
L
L
l
+L
CR
$M
-
-
M
M
m
+M
SO
$N
.
.
N
N
n
+N
SI
$O
/
/O
O
O
o
+O
DLE
$P
0
0
P
P
p
+P
DC1
$Q
1
1
Q
Q
q
+Q
DC2
$R
2
2
R
R
r
+R
DC3
$S
3
3
S
S
s
+S
DC4
$T
4
4
T
T
t
+T
NAK
$U
5
5
U
U
u
+U
SYN
$V
6
6
V
V
v
+V
ETB
$W
7
7
W
W
w
+W
CAN
$X
8
8
X
X
x
+X
EM
$Y
9
9
Y
Y
y
+Y
SUB
$Z
:
/Z
Z
Z
z
+Z
ESC
%A
;
%F
[
%K
{
%P
FS
%B
<< 
%G
\
%L
|
%Q
GS
%C
=
%H
]
%M
}
%R
RS
%D
> 
%I
^
%N
~
%S
YS
%E
?
%J
_
%O
DEL
%T, %X, %Y, %Z