PRB: Opening a ComboBox Prevents Timer Event Code from Executing when Timer.Interval is a Low Value
ID: Q238513
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The information in this article applies to:
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Microsoft Visual FoxPro for Windows, versions 5.0, 5.0a, 6.0
SYMPTOMS
Event code in a timer object is not executed at the interval specified in the timer object's Interval property.
CAUSE
This behavior occurs when the timer interval of the timer object is set to an extremely low value.
RESOLUTION
Set the timer object's Interval property to a higher value.
MORE INFORMATIONSteps to Reproduce Behavior
- Create a program file named Timedemo.prg using the following code:
PUBLIC xForm
xForm=CREATEOBJECT("TimerForm")
xForm.SHOW()
DEFINE CLASS TimerForm AS FORM
ncount=0
starttime=TIME()
ADD OBJECT Combo1 AS COMBOBOX WITH;
STYLE=2,;
FONTSIZE=8,;
HEIGHT=22,;
BOUNDTO=.T.,;
ROWSOURCETYPE=5
ADD OBJECT TextBox1 AS TEXTBOX WITH;
TOP=0,;
LEFT=150
ADD OBJECT Timer1 AS TIMER WITH;
INTERVAL=1,;
ENABLED=.T.
DIMENSION aArray(1)
ncount=0
PROCEDURE INIT
LOCAL tmpi,tmpnlen
tmpnlen=100
FOR tmpi=1 TO tmpnlen
DIMENSION THIS.aArray(tmpi)
THIS.aArray(tmpi)=tmpi
ENDFOR
THIS.Combo1.ROWSOURCE="thisform.aArray"
THIS.Combo1.REQUERY()
THIS.Combo1.LISTINDEX=1
THIS.Combo1.LEFT=5
THIS.starttime=TIME()
ENDPROC
PROCEDURE Timer1.TIMER
THISFORM.ncount=THISFORM.ncount+1
IF THISFORM.ncount=5*(1000/THISFORM.Timer1.INTERVAL)
THISFORM.CALCTIME()
THISFORM.ncount=0
THISFORM.starttime=TIME()
ENDIF
ENDPROC
PROCEDURE calctime
started=THISFORM.starttime
stoptime=TIME()
current_space=SELECT()
rec_count=RECCOUNT()
SELECT 0
IF LEN(ALLTRIM(started))=20
startda=VAL(LEFT(started,2))
STARTMO=VAL(SUBSTR(started,4,2))
startyr=VAL(SUBSTR(started,7,2))
IF UPPER(RIGHT(started,2))="PM"
starthr=VAL(SUBSTR(started,10,2))+12
ELSE
starthr=VAL(SUBSTR(started,10,2))
ENDIF
startmin=VAL(SUBSTR(started,13,2))
startsec=VAL(SUBSTR(started,16,2))
ELSE
starthr=VAL(LEFT(started,2))
startmin=VAL(SUBSTR(started,4,2))
startsec=VAL(RIGHT(started,2))
ENDIF
IF LEN(ALLTRIM(stoptime))>8
stopda=VAL(LEFT(stoptime,2))
stopmo=VAL(SUBSTR(stoptime,4,2))
stopyr=VAL(SUBSTR(stoptime,7,2))
IF UPPER(RIGHT(stoptime,2))="PM"
stophr=VAL(SUBSTR(stoptime,10,2))+12
ELSE
stophr=VAL(SUBSTR(stoptime,10,2))
ENDIF
stopmin=VAL(SUBSTR(stoptime,13,2))
stopsec=VAL(SUBSTR(stoptime,16,2))
ELSE
stophr=VAL(LEFT(stoptime,2))
stopmin=VAL(SUBSTR(stoptime,4,2))
stopsec=VAL(RIGHT(stoptime,2))
ENDIF
IF startsec>stopsec
stopsec=stopsec+60
stopmin=stopmin-1
ENDIF
IF startmin>stopmin
stopmin=stopmin+60
stophr=stophr-1
ENDIF
elapsedtime=REPLICATE("0",2-LEN(ALLTRIM(STR(stophr-starthr)))) + ;
ALLTRIM(STR(stophr-starthr)) + ":" + ;
REPLICATE("0",2-LEN(ALLTRIM(STR(stopmin-startmin)))) + ;
ALLTRIM(STR(stopmin-startmin)) + ":" + ;
REPLICATE("0",2-LEN(ALLTRIM(STR(stopsec-startsec)))) + ;
ALLTRIM(STR(stopsec-startsec))
IF !FILE('TIMETRAP.DBF')
CREATE TABLE timetrap (elapsed c(8), tinterval i)
ENDIF
INSERT INTO timetrap ;
VALUES (elapsedtime, THISFORM.Timer1.INTERVAL)
USE
SELECT (current_space)
THISFORM.TextBox1.VALUE=elapsedtime
THISFORM.TextBox1.REFRESH
ENDPROC
ENDDEFINE
- From the Command window, type:
DO Timedemo
- Observe that the value "00:00:05" appears in the text box after approximately five seconds.
- Click the combobox, move the mouse pointer down the center of the combobox and position the mouse pointer over the number 7.
- Hold the mouse pointer over the number 7 for approximately 45 seconds.
- Move the mouse away from the combobox and click on the form. Note the value displayed in the text box. The value is something other than 00:00:05. This value is greater than five seconds. The value may appear in the text box before you actually click on the form.
(c) Microsoft Corporation 1999, All Rights Reserved. Contributions by John Desch, Microsoft Corporation.
Additional query words:
Keywords : kbContainer kbCtrl kbVFp500 kbVFp500a kbVFp600 kbGrpFox kbDSupport
Version : WINDOWS:5.0,5.0a,6.0
Platform : WINDOWS
Issue type : kbprb
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