create a random aray for displying a mesage in a label

[Dennis W]

I want to display a differnt message evry time a button is clicked. so I set
up a aray of ms(12) so that I can asign 12 differnt strings to a labelbut I
get a error . First of all can I asign a string value to a label? Im new to
vb . I 'm still trying to shake the happits of basica or gwbasic trobo basic.
You get the Idea. So can someone get me on the right path here.
--
//Bart//

[Tom Shelton]

On Tue, 14 Sep 2004 19:33:02 -0700, Dennis W wrote:

I want to display a differnt message evry time a button is clicked. so I set
up a aray of ms(12) so that I can asign 12 differnt strings to a labelbut I
get a error . First of all can I asign a string value to a label? Im new to
vb . I 'm still trying to shake the happits of basica or gwbasic trobo basic.
You get the Idea. So can someone get me on the right path here.

Actually, in VB.NET (and VB6 too, unless you had option base 1 set) that
would have 13 items :) But anyway...

Dim ms() as string = {"first", "second", "third", "fourth", "fifth"}
dim rand As new Random()
dim i as integer = rand.next (0, ms.getupperboun d(0))

messagebox.show (ms(i))
--
Tom Shelton [MVP]

[One Handed Man \( OHM - Terry Burns \)]

Additionally to Tom's comments, you say you cannot assign to a label?, the
answer is yes

myLabel.Text = "YourString "

or

myLabel.Text = yourArray(index )

--

OHM ( Terry Burns )
. . . One-Handed-Man . . .
If U Need My Email ,Ask Me

Time flies when you don't know what you're doing

"Dennis W" <De*****@discus sions.microsoft .com> wrote in message
news:BD******** *************** ***********@mic rosoft.com...

I want to display a differnt message evry time a button is clicked. so I set up a aray of ms(12) so that I can asign 12 differnt strings to a labelbut I get a error . First of all can I asign a string value to a label? Im new to vb . I 'm still trying to shake the happits of basica or gwbasic trobo basic. You get the Idea. So can someone get me on the right path here.
--
//Bart//

[Chris Dunaway]

On Tue, 14 Sep 2004 21:04:29 -0600, Tom Shelton wrote:

On Tue, 14 Sep 2004 19:33:02 -0700, Dennis W wrote:

I want to display a differnt message evry time a button is clicked. so I set
up a aray of ms(12) so that I can asign 12 differnt strings to a labelbut I
get a error . First of all can I asign a string value to a label? Im new to
vb . I 'm still trying to shake the happits of basica or gwbasic trobo basic.
You get the Idea. So can someone get me on the right path here.

Actually, in VB.NET (and VB6 too, unless you had option base 1 set) that
would have 13 items :) But anyway...

Dim ms() as string = {"first", "second", "third", "fourth", "fifth"}
dim rand As new Random()
dim i as integer = rand.next (0, ms.getupperboun d(0))

messagebox.show (ms(i))

Just to point out one thing. The rand.next function that you show will get
a random number from 0 to 3. It does not include the upper bound. You
need to specify ms.getupperboun d(0,ms.getupper bound(0) + 1).

From the docs:

Return Value
A 32-bit signed integer greater than or equal to minValue and less than
maxValue; that is, the range of return values includes minValue but not
MaxValue. If minValue equals maxValue, minValue is returned.

--
Chris

dunawayc[AT]sbcglobal_lunch meat_[DOT]net

To send me an E-mail, remove the "[", "]", underscores ,lunchmeat, and
replace certain words in my E-Mail address.

[Tom Shelton]

In article <zb************ *************** **@40tude.net>, Chris Dunaway wrote:

On Tue, 14 Sep 2004 21:04:29 -0600, Tom Shelton wrote:

On Tue, 14 Sep 2004 19:33:02 -0700, Dennis W wrote:

I want to display a differnt message evry time a button is clicked. so I set
up a aray of ms(12) so that I can asign 12 differnt strings to a labelbut I
get a error . First of all can I asign a string value to a label? Im new to
vb . I 'm still trying to shake the happits of basica or gwbasic trobo basic.
You get the Idea. So can someone get me on the right path here.

Actually, in VB.NET (and VB6 too, unless you had option base 1 set) that
would have 13 items :) But anyway...

Dim ms() as string = {"first", "second", "third", "fourth", "fifth"}
dim rand As new Random()
dim i as integer = rand.next (0, ms.getupperboun d(0))

messagebox.show (ms(i))

Just to point out one thing. The rand.next function that you show will get
a random number from 0 to 3. It does not include the upper bound. You
need to specify ms.getupperboun d(0,ms.getupper bound(0) + 1).

From the docs:

Return Value
A 32-bit signed integer greater than or equal to minValue and less than
maxValue; that is, the range of return values includes minValue but not
MaxValue. If minValue equals maxValue, minValue is returned.

Yes, your correct... I always forget that. Actually, an easier
solution, would just to use ms.getupperboun d(0, ms.length).

--
Tom Shelton [MVP]