VB 6 Code Help Urgent

Suraj143 · Mar 7, 2013

I'm entering a number between 0-255 in the text box.I want to send this number to PIC via serial port when the command button clicks.

The problem is, it only send numbers between 1 to 9.From 10 onwards it won't send because the VB6 thinks it is more than 1 character.

I want to send the byte out.

Code:

Private Sub Command1_Click()
MSComm1.Output = Text1.Text
End Sub

Ian Rogers · Mar 7, 2013

So you need to send it like this

Code:

MSCom1.Output = chr(text1.text)

This way it will send the number as an unsigned character....ie if you type 48.... Then "0" (ascii 48 ) will be sent....

Suraj143 · Mar 7, 2013

Hi that worked,but the problem is the user always needs to write 48,49,50 etc...

I need to write 0,1,2,3 in the text box & needs to send data.

What VB code I have to write?

In simply user will write a number between 1-16 in the text box & the channel number must be saved that value.Thats it

Code:

Recieve_DATA	movf	RCREG,W			; yes, data from master
		andlw	b'01111111'
		movwf	RC_Buffer
		;
		movlw	30h
		subwf	RC_Buffer,W		; check channel number
		andlw	b'00011111'
		movwf	Channel_Byte		; *********

dougy83 · Mar 7, 2013

VB shouldn't be having any trouble sending multiple characters. e.g. mscomm1.Output = "123" will send three characters: '1', '2', '3'. The most likely cause to you not receiving multiple characters is due to your receiver code being invalid.

If you didn't want to have to worry about framing the data you send, you can do as Ian suggested and send a single character having a value between 0-255, i.e. mscomm1.Output = chr(val(text1.Text)). Then you only have to receive a single byte. The manner in which you extract data from that byte is up to you.

Ian Rogers · Mar 8, 2013

When you say you have multiple numbers ... Is there a delimiter? ie.. 1,2,3 separated by comma's.

If that is the case you need a simple parser.... lets assume we have the string "55,22,128,33" in text1.text

If you use

Code:

x = InStr( text1.text, ",")

X will contain the position of the first comma ie.. 2
Then strip off the last

Code:

1stno = left( text1.text,2)

should yield 1stno = "55"
re-adjust text1.text to get the second

Code:

y = InStr(x+1, text1.text, ",")

Then retrieve

Code:

2ndno =mid(x+1 text1.text,y)

should yield 2ndno = "22"
re-adjust text1.text to get the second

Code:

z = InStr(y+1, text1.text, ",")

Then retrieve

Code:

3rdno =mid(y+1 text1.text,z)

should yield 3rdno = "128"
And so on...

dougy83 · Mar 8, 2013

Ian Rogers said:
When you say you have multiple numbers ... Is there a delimiter? ie.. 1,2,3 separated by comma's.
If that is the case you need a simple parser.... lets assume we have the string "55,22,128,33" in text1.text

Just use the Split() function to convert the CSV into an array of strings (see http://www.vb6.us/tutorials/vb-string-array-functions-split-join-filter for example usage)

Ian Rogers · Mar 8, 2013

dougy83 said:
Just use the Split() function to convert the CSV into an array of strings (see http://www.vb6.us/tutorials/vb-string-array-functions-split-join-filter for example usage)

Thanks for that dougy..... It's a pain splitting strings in VB6....

Suraj143 · Mar 8, 2013

Hello friends I think you still did not get what I mean.

I'm entering a number in the text box.There is no any delimiter or commas.I'm entering a number between 1 to 16 thats it.

Ex:
I type 1 & I press command button then it must send 1
I type 2 & I press command button then it must send 2
I type 3 & I press command button then it must send 3
I type 10 & I press command button then it must send 10
I type 16 & I press command button then it must send 16

So this value must be saved in PIC thats it.The problem I'm facing is, from 1 to 9 the PIC will receive nicely & from 10 onwards it will send some other value.

Ian Rogers · Mar 8, 2013

So you are sending and receiving ascii (text)...... Then you'll have to program the pic to accept more than one byte... Unless you change your programming style to send binary...

Lourens · Mar 8, 2013

The 'number' in the textbox is actually a text string. Take that string and change it into a value. Will look something like:
NumberToSend = Val(TextBox.Text)

Suraj143 · Mar 8, 2013

Lourens said:
The 'number' in the textbox is actually a text string. Take that string and change it into a value. Will look something like:
NumberToSend = Val(TextBox.Text)

Ok I'll try with above

Suraj143 · Mar 8, 2013

Lourens said:
The 'number' in the textbox is actually a text string. Take that string and change it into a value. Will look something like:
NumberToSend = Val(TextBox.Text)

It still cannot send number 10 onwards

Code:

Private Sub Command1_Click()
Dim val2 As Byte
val2 = Val(Text2.Text)
MSComm1.Output =  val2 
End Sub

Pommie · Mar 8, 2013

Because of the way your pic code is written you need to send a single byte.

Try,

Code:

Private Sub Command1_Click()
    MSComm1.Output =  chr(Val(Text2.Text)+48)
End Sub

Mike.

Suraj143 · Mar 9, 2013

Pommie said:
Because of the way your pic code is written you need to send a single byte.

Try,

Code:

Private Sub Command1_Click() MSComm1.Output = chr(Val(Text2.Text)+48) End Sub

Mike.

Brilliant it worked well.Thats the code I wanted.

Thank you very much Mike.

Ian Rogers · Mar 9, 2013

OK!!!! I see so when you are sending ":" and ";" it come through as 10 and 11.... Novel idea... still if it works for you...

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