Valhalla Legends Archive

I want to use quotations in a string. for Example:

frmMain.txtName.Text = ""Whatever Name Is""
Name = frmMain.txtName.Text
Print #1, "doc = " & Name & ";"

This errors - because i can't use double quotes like this - can anyone give me a way around this?

Edit: link didn't work

Quote
Embed Quotation Marks
You use quotation marks in VB to define strings, but how do you include them in your output? Use whichever of these methods works the best for you:

   Dim strUseChr As String
   Dim strUseVar As String
   Dim strUseDbl As String
   
   Const Quote As String = """
   
   strUseChr = "Hello " & Chr$(34) & "VB" & _
      Chr$(34) & " World!"
   strUseVar = "Hello " & Quote & "VB" & _
      Quote & " World!"
   strUseDbl = "Hello "VB"" World!"
   
   Debug.Print strUseChr
   Debug.Print strUseVar
   Debug.Print strUseDbl

Each one prints:

   Hello "VB" World!

Source: http://www.devx.com/tips/Tip/16241

eww..

msgbox "Hello ""VB"" world!" is by far the best.

Quote from: Dyndrilliac on November 13, 2003, 07:15 PM
I want to use quotations in a string. for Example:

Code Select Expand

frmMain.txtName.Text = """Whatever Name Is"""
Name = frmMain.txtName.Text
Print #1, "doc = " & Name & ";"

This errors - because i can't use double quotes like this - can anyone give me a way around this?

Fixed.

Now I have a whole new set of problems - whenever I put in soemthing like this:wHeight = txtHeight.Text it gives me "RunTime Error "13" - Type Mismatch", and I don't know why, but I need that so the user can define the number of the wHeight Integer.

It's Declared as : Dim wHeight As Integer

Quote from: Dyndrilliac on November 13, 2003, 08:00 PM
Now I have a whole new set of problems - whenever I put in soemthing like this:

Code Select Expand

wHeight = txtHeight.Text it gives me "RunTime Error "13" - Type Mismatch", and I don't know why, but I need that so the user can define the number of the wHeight Integer.

It's Declared as :

Code Select Expand

Dim wHeight As Integer

wHeight = CInt(txtHeight)

CInt() returns Integer of Expression.

Edit -- You may want to check if contents of txtHeight is actually a number first (you could use IsNumeric for this).

Quote from: Spht on November 13, 2003, 08:06 PM

Quote from: Dyndrilliac on November 13, 2003, 08:00 PM
Now I have a whole new set of problems - whenever I put in soemthing like this:

Code Select Expand

wHeight = txtHeight.Text it gives me "RunTime Error "13" - Type Mismatch", and I don't know why, but I need that so the user can define the number of the wHeight Integer.

It's Declared as :

Code Select Expand

Dim wHeight As Integer

Code Select Expand

wHeight = CInt(txtHeight)

CInt() returns Integer of Expression.

Edit -- You may want to check if contents of txtHeight is actually a number first (you could use IsNumeric for this).

How's CInt different from val?

Quote from: iago on November 13, 2003, 10:05 PM
How's CInt different from val?

Val() returns Double, CInt returns Integer (which is what he wants). Val has other functionality, such as stripping out non-numeric characters, and formatting string identifiers (i.e., Val("&HA") would return 10). If he wants that functionality, I suppose he could do something like wHeight = CInt(Val(txtHeight)).