如何解决如何在 QB45 中检测鼠标滚轮
我使用以下代码在 QB45 中捕获左/右/中鼠标按钮和鼠标行/列:
(QB45 是 Microsoft Quick Basic v4.5)
我需要一种方法来检测 MouseWheel。我看了拉尔夫布朗的中断列表,运气不佳。
有什么想法吗?顺便说一句:我使用的是 Int 0x33。
我提交的代码适用于 Microsoft Quickbasic IDE,需要库文件 QB.QLB。
DECLARE SUB Mouse.Function (Var1,Var2)
DIM SHARED MouseX AS INTEGER,MouseY AS INTEGER
TYPE RegTypeX
AX AS INTEGER
BX AS INTEGER
CX AS INTEGER
DX AS INTEGER
BP AS INTEGER
SI AS INTEGER
DI AS INTEGER
Flags AS INTEGER
DS AS INTEGER
ES AS INTEGER
END TYPE
COMMON SHARED InregsX AS RegTypeX
COMMON SHARED OutregsX AS RegTypeX
DECLARE SUB InterruptX (N AS INTEGER,I AS RegTypeX,O AS RegTypeX)
CALL Mouse.Function(0,0) ' init mouse
CALL Mouse.Function(1,0) ' show mouse
DO
IF LEN(INKEY$) THEN
CALL Mouse.Function(2,0) ' hide mouse
EXIT DO
END IF
' read mouse button press
CALL Mouse.Function(3,0)
Var2 = INT((OutregsX.CX - 1) / 8 + 1)
Var3 = INT((OutregsX.DX - 1) / 8 + 1)
IF Var3 <> Mouse.Row OR Var2 <> Mouse.Column THEN
CALL Mouse.Function(2,0) ' hide mouse
Mouse.Row = Var3
Mouse.Column = Var2
PRINT Mouse.Row,Mouse.Column
CALL Mouse.Function(1,0) ' show mouse
END IF
Mouse.Button = False
CALL Mouse.Function(5,0)
IF (OutregsX.AX AND 1) = 1 THEN
IF OutregsX.BX > False THEN
Mouse.Button = -1
PRINT "Left-Click"
END IF
END IF
Mouse.Button2 = False
CALL Mouse.Function(5,1)
IF (OutregsX.AX AND 2) = 2 THEN
IF OutregsX.BX > False THEN
Mouse.Button2 = -1
PRINT "Right-Click"
END IF
END IF
Mouse.Button3 = False
CALL Mouse.Function(5,2)
IF (OutregsX.AX AND 4) = 4 THEN
IF OutregsX.BX > False THEN
Mouse.Button3 = -1
PRINT "Middle-Click"
END IF
END IF
LOOP
END
SUB Mouse.Function (Var1,Var2)
InregsX.AX = Var1
InregsX.BX = Var2
CALL InterruptX(&H33,InregsX,OutregsX)
END SUB
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