There are several ways to achieve your goal IN TEXT MODE :
- Display char by char : this way you can choose one color for every character.
- Access screen memory : at segment 0B800:0.
- Display the whole string with the same color.
Next code does the job with the third option (the easiest). It was made with EMU8086:
.stack 100h
.data
Jan db " January ",13,10
db "Sun Mon Tue Wed Thu Fri Sat",13,10
db " 1 2 3 ",13,10
db " 4 5 6 7 8 9 10 ",13,10
db "11 12 13 14 15 16 17 ",13,10
db "18 19 20 21 22 23 24 ",13,10
db "25 26 27 28 29 30 31 "
color db 181
.code
;INITIALIZE DATA SEGMENT.
mov ax,@data
mov ds,ax
;DISPLAY STRING WITH COLOR.
mov es,ax ;ES SEGMENT MUST POINT TO DATA SEGMENT.
mov ah,13h ;SERVICE TO DISPLAY STRING WITH COLOR.
mov bp,offset Jan ;STRING TO DISPLAY.
mov bh,0 ;PAGE (ALWAYS ZERO).
mov bl,color
mov cx,201 ;STRING LENGTH.
mov dl,0 ;X (SCREEN COORDINATE).
mov dh,5 ;Y (SCREEN COORDINATE).
int 10h ;BIOS SCREEN SERVICES.
;FINISH THE PROGRAM PROPERLY.
mov ax,4c00h
int 21h
Notice I removed the $ signs (because 13h service requires string length, not $). For a different color just change the value (181) for the "color" variable in data segment.
To display diferent colors for diferent strings, copy-paste the display block for every string.
Let us know if it worked for you.
The formula to choose color goes like this:
text-background * 16 + text-color
Next are the colors :
Black = 0
Blue = 1
Green = 2
Cyan = 3
Red = 4
Magenta = 5
Brown = 6
LightGray = 7
DarkGray = 8
LightBlue = 9
LightGreen = 10
LightCyan = 11
LightRed = 12
LightMagenta = 13
Yellow = 14
White = 15
With the given formula, if you want red background with yellow text you need the color 78 :
4 * 16 + 14 = 78
Now, let's do it in GRAPHICS MODE :
.stack 100h
.data
Jan db " January ",13
db "Sun Mon Tue Wed Thu Fri Sat",13
db " 1 2 3 ",13
db " 4 5 6 7 8 9 10 ",13
db "11 12 13 14 15 16 17 ",13
db "18 19 20 21 22 23 24 ",13
db "25 26 27 28 29 30 31 ",0
color db 181
x db 0 ;SCREEN COORDINATE (COL).
y db 0 ;SCREEN COORDINATE (ROW).
.code
;INITIALIZE DATA SEGMENT.
mov ax,@data
mov ds,ax
;SWITCH SCREEN TO GRAPHICS MODE.
mov ah,0
mov al,13h ;320x240x256.
int 10H
mov di, offset jan
while:
call gotoxy ;SET CURSOR POSITION FOR CURRENT CHAR.
mov al, [ di ] ;CHAR TO DISPLAY.
cmp al, 13 ;IF CHAR == 13
je linebreak ;THEN JUMP TO LINEBREAK.
cmp al, 0 ;IF CHAR == 0
je finish ;THEN JUMP TO FINISH.
call char_display ;DISPLAY CHAR IN AL WITH "COLOR".
inc x ;NEXT CHARACTER GOES TO THE RIGHT.
jmp next_char
linebreak:
inc y ;MOVE TO NEXT LINE.
mov x, 0 ;X GOES TO THE LEFT.
next_char:
inc di ;NEXT CHAR IN "JAN".
jmp while
finish:
;WAIT FOR ANY KEY.
mov ah,7
int 21h
;FINISH THE PROGRAM PROPERLY.
mov ax,4c00h
int 21h
;-------------------------------------------------
;DISPLAY ONE CHARACTER IN "AL" WITH "COLOR".
proc char_display
mov ah, 9
mov bh, 0
mov bl, color ;ANY COLOR.
mov cx, 1 ;HOW MANY TIMES TO DISPLAY CHAR.
int 10h
ret
endp
;-------------------------------------------------
proc gotoxy
mov dl, x
mov dh, y
mov ah, 2 ;SERVICE TO SET CURSOR POSITION.
mov bh, 0 ;PAGE.
int 10h ;BIOS SCREEN SERVICES.
ret
endp
My graphics algorithm requires the use of char(13) for linebreaks (not 13,10) and the string to finish with 0.