as we know the iso 8583 message consist of 3 main component, MTI (message type indicator), Bitmaps and element. and for Base 24 they have special feature for header which adding 4 component at the beginning of the message.
MTI consist with 4 character at the begining of message (if not using base 24 Header). and represent the function of message. for example, = "0200” ,”0210'”,”0400”,”0800”
Bitmaps contains information which field on element will be used. in bitmaps can be only primary or primary with secondary bitmaps. To indicate the bitmaps have secondary bitmaps, just convert the first character(hexa) into binary. if the fisrt bit is “1” the meaning the bitmap have secondary bitmap.
like the message Below :
0210723A00010A80840018593600141001099999
0110000000100000001007021533000001191533
1006100706565656100609010224000000090136
0020100236C0102240000000
from message we can split message like below.
| component | Value |
| Message type indicator | 0210 ( 4 digit character from start of message) |
| Bitmaps( only have primary Bitmaps) | 723A00010A808400 because the binary of “7” (hexa) is 0111 the bitmap only have primary (16 hex character, if the bitmaps contain secondary the lenght of bitmpas is 32) |
| Data Element | 185936001410010999990110000000100000001007 |
after that we must unpack every bitmaps into binary. every character will generate 4 bit . the result of “723A00010A808400” in binary =
0111001000111010000000000000000100001010100000001000010000000000
from the result above we know the active bit is 2,3,4,7,11,12, 13, 15, 32, 37, 39, 41, 49, and 52 (the bitmap with “1” value, the first digit is for cheking the bitmap consist secondary or not)
with the manual iso 8583 the data element can be split like below :
Field value
2 -> 593600141001099999
3 -> 011000
4 -> 000010000000
7 -> 1007021533
11 -> 000001
12 -> 191533
13 -> 1006
15 -> 1007
32 -> 565656
37 -> 100609010224
39 -> 00
41 -> 00000901
49 -> 360
54 -> 100236C0102240000000
0 komentar:
Post a Comment