Spinner 1

[cslawtom]

Hi All,

I would like to have a spinner starting from 0.5 with increment 0.5 for first and 1 for the rest.
That is

0.5 -> 1.0 -> 2.0 -> 3.0 and

3.0 -> 2.0 -> 1.0 -> 0.5

Is it possible? Many Thanks.

 

[webpager]

Not sure if it is possible with a spinner but it could be carried out using a drop down LISTBOX. Depends how many values you are to allow the user to select as to whether it is practical.
Keith

 

[cslawtom]

Thanks Keith

 

[Mike Gagnon]

CSLAWTOM

Here is an example of that (by using a textbox instead of using the spinner's entry box). Copy the following in a program to see how it works.:

PUBLIC oform1
oform1=NEWOBJECT("form1")
oform1.Show
RETURN
DEFINE CLASS form1 AS form
	DoCreate = .T.
	Caption = "Form1"
	Name = "Form1"
	ADD OBJECT spinner1 AS spinner WITH ;
		Height = 25, ;
		Left = 216, ;
		SpinnerHighValue = 100.00, ;
		SpinnerLowValue =   0.00, ;
		TabIndex = 2, ;
		Top = 72, ;
		Width = 20, ;
		Name = "Spinner1"
	ADD OBJECT text1 AS textbox WITH ;
		Alignment = 3, ;
		Value = 0, ;
		Height = 23, ;
		Left = 120, ;
		TabIndex = 1, ;
		Top = 73, ;
		Width = 97, ;
		Name = "Text1"
	PROCEDURE spinner1.DownClick
		DO CASE
		  CASE THISFORM.TEXT1.Value = 0
		      CASE THISFORM.TEXT1.Value = 0
		  CASE THISFORM.TEXT1.Value = .5
		    THISFORM.TEXT1.Value = 0
		  CASE THISFORM.TEXT1.Value = 1
		      THISFORM.TEXT1.Value = .5
		  OTHERWISE
		   THISFORM.TEXT1.Value = THISFORM.TEXT1.Value -1
        	ENDCASE
	ENDPROC
	PROCEDURE spinner1.UpClick
		DO CASE
		  CASE THISFORM.TEXT1.Value = 100
		     THISFORM.REXR1.VALUE = 100
		  CASE THISFORM.TEXT1.Value = 0
		    THISFORM.TEXT1.Value = .5
		  CASE THISFORM.TEXT1.Value = .5
		      THISFORM.TEXT1.Value = 1
		  OTHERWISE
		   THISFORM.TEXT1.Value = THISFORM.TEXT1.Value +1
		ENDCASE
	ENDPROC
ENDDEFINE

Mike Gagnon

If you want to get the best response to a question, please check out FAQ184-2483 first

 

[DSummZZZ]

You can use a spinner. After dropping the control on the form, use the following values:

Increment: 0.5
SpinnerLowValue: 0.5
SpinnerHighValue: whatever

Then put this code in the InteractiveChange event:

IF This.Value < 1
   This.Increment = .5
ELSE
   This.Increment = 1
ENDIF

Dave S.

 

[cslawtom]

Thanks Keith. It works well.

Thanks Dave. It works but 1.0 back to 0.5 has problem. Anyway I will try to solve the problem by following your advice.

 

[cslawtom]

Using Keith's method is better in my case. Thanks Keith again.

 

[Mike Gagnon]

Dave S

IF This.Value < 1
This.Increment = .5
ELSE
This.Increment = 1
ENDIF

The problem is, is that the spinner increments before it hit the interactive change of the spinner.
Mike Gagnon

If you want to get the best response to a question, please check out FAQ184-2483 first

 

[738262]

You can use a spinner. After dropping the control on the form, use the following values:

Increment: 0.5
SpinnerLowValue: 0.5
SpinnerHighValue: whatever
value=0.5

Then put this code in the InteractiveChange event:

IF this.Value>0.5
this.Increment=1
ENDIF
IF this.Value=0.5
this.Increment=0.5
endif

 

[Mike Gagnon]

738262

IF this.Value>0.5
this.Increment=1
ENDIF
IF this.Value=0.5
this.Increment=0.5
endif

Do you test this before posting it? In my experience this will not work.
[ol][li]You cannot set the value .5 in a spinner, and if you set it by code (which you can do), when you go from 3,2,1 it never reaches .5.[/li][/ol] Mike Gagnon

If you want to get the best response to a question, please check out FAQ184-2483 first

 

[DSummZZZ]

Mike,

I tested before I posted, and it worked. You can set the Increment property to .5, but you have to use 0.5 instead of just .5. Please note though, that I initially tested it using VFP 7. It didn't work properly using VFP 6 until I added this line in the Init event:

This.Value = 0.5
Dave S.

 

[Mike Gagnon]

DSummZZZ

I tested before I posted, and it worked. You can set the Increment property to .5, but you have to use 0.5 instead of just .5. Please note though, that I initially tested it using VFP 7. It didn't work properly using VFP 6 until I added this line in the Init event:

This.Value = 0.5

Have you tried to go from 0.5 (which you set as a default value) to 3 and back down to 0? Notice on the way back down you cannot get to 0.5.
Mike Gagnon

If you want to get the best response to a question, please check out FAQ184-2483 first

 

[DSummZZZ]

Here's my settings:

SpinnerHighValue = 10
SpinnerLowValue = 0.50
Value = 0.50

In the InteractiveChange event:

IF this.Value < 1 && ... in this case it would be 0
This.Increment = .5
ELSE
This.Increment = 1
ENDIF

So yes, I have gone from 0.5 to 10, then back from 10 to 0.5. Repeatedly.
Now, I also saved that control as a class, and opened it using the class browser. Here is the class code generated by VFP:

**************************************************
*-- Class:        sp1 (c:\program files\microsoft visual foxpro 7\sp1.vcx)
*-- ParentClass:  spinner
*-- BaseClass:    spinner
*-- Time Stamp:   11/21/02 08:00:00 AM
*
DEFINE CLASS sp1 AS spinner


	Height = 24
	Increment =   0.50
	Left = 492
	SpinnerHighValue =  10.00
	SpinnerLowValue =   0.50
	Top = 168
	Width = 69
	Value = 0.5
	Name = "Spinner1"


	PROCEDURE InteractiveChange
		IF this.Value < 1
		   This.Increment = .5
		ELSE
		   This.Increment = 1
		ENDIF
	ENDPROC


ENDDEFINE
*
*-- EndDefine: sp1
**************************************************

Granted, without a little more tweaking, you can't get to 0 but that would be doable too I'm sure.
Dave S.

 

[Mike Gagnon]

Granted, without a little more tweaking, you can't get to 0 but that would be doable too I'm sure.

I guess that would be a concern. Although the original poster did not specify it.

Mike Gagnon

If you want to get the best response to a question, please check out FAQ184-2483 first

 

[cslawtom]

Hi Mike,

738262's suggestion worked properly. I had tried but as I have more code required to add in, I used your method.

Thanks.

 

[jaa69]

I took this one step farther and it does work. This has a triple increment, plus up and down now works.

DO case
Case this.Value < 5 and this.Increment = 5
this.Increment = 1
this.value = 4.00
Case this.Value < 3 and this.Increment = 1
this.Increment = .5
this.value = 2.50
Case this.Value < 1 and this.Increment = .5
this.Increment = .25
this.value = .75
Case this.value < 1
thisform.spinner1.Increment = .25
Case this.value >= 1 and this.value < 3
thisform.spinner1.Increment = .5
Case this.value >= 3 .and. this.Value < 5
thisform.spinner1.Increment = 1
Case this.Value >= 5
this.Increment = 5

EndCase