I have a Textbox field that takes in a string with a character limit of 10. I would like to implement a short hand version because there are a lot of zeros in the string that have to be entered. so an example of the string is T000028999. but id like to key in T28999 and have the zeros padded between the "T" and the "28999" and show up as the T000028999 string in the Textbox field.
Is this even possible?
I've tried searching examples on google and have only found ways to pad the beginning and end of the string.
You want to keep the first character, so you can use oldString.Chars(0) to get that.
You want the remainder of the string: oldString.Substring(1), and you can pad it to the width you require with a character of your choice with PadLeft, like this:
Dim newString = oldString.Chars(0) & oldString.Substring(1).PadLeft(9, "0"c)
It would be a good idea to check that oldString is at least 1 character long before doing that otherwise the .Chars(0) will give an error.
Alternatively you could insert a new string of the required quantity of "0"s:
Dim newString = oldString.Insert(1, New String("0"c, 10 - oldString.Length))
A good place to perform the formatting would be in the control's Validating event handler. (The TextChanged event handler would not be a good place because it would interfere with the user's typing.)
Refs:
String.Chars[Int32] Property
String.Substring Method
String.PadLeft Method
String.Insert(Int32, String) Method
String Constructors
Related
I have a TextBox in VB and a Command Button. I want to print the string upon clicking on command button.
I am using the following code, please tell what I am doing wrong:-
Dim name As String
name = Val(Text1.Text)
MsgBox ("Welcome" & Str(name))
When I input a string in Textbox and click on command button, result is:
Welcome 0
Leave out the val() around your Text1.Text, val() returns the numbers up to the first symbol it can't recognize as a number used in a String. See the documentation. I guess you used a 0 in your string in the TextField or no number at all, both would return 0.
Additionally, it is unnecessary to cast your String name to a String since it is already a String so you can also leave out the Str().
The val function returns the numeric representation of its argument, otherwise it returns "0". It's a bit hard these days to find the official VB6 documentation, but you may want to check: https://en.wikibooks.org/wiki/Visual_Basic/VB6_Command_Reference#Val
So, in your example, if you enter any number in the Text1 textbox control, you will see it in your message box. If you enter any text, you will get "Welcome 0", as you do now. Therefore, you have to remove the val function from your code, like:
Dim name As String
name = Text1.Text
MsgBox ("Welcome " & name)
maybe even simplifying it to:
MsgBox("Welcome " & Text1.Text)
So you declared a string varaible namewhich you want to fill with the text from the Text1box. So you need to spare the val(...)part.
Second, as namealready represents a string, leave out the strin the message box:
name = Text1.Text
MsgBox ("Welcome " & name)
I am wondering how to make something where if X=5 and Y=2, then have it output something like
Hello 2 World 5.
In Java I would do
String a = "Hello " + Y + " World " + X;
System.out.println(a);
So how would I do that in TI-BASIC?
You have two issues to work out, concatenating strings and converting integers to a string representation.
String concatenation is very straightforward and utilizes the + operator. In your example:
"Hello " + "World"
Will yield the string "Hello World'.
Converting numbers to strings is not as easy in TI-BASIC, but a method for doing so compatible with the TI-83+/84+ series is available here. The following code and explanation are quoted from the linked page:
:"?
:For(X,1,1+log(N
:sub("0123456789",ipart(10fpart(N10^(-X)))+1,1)+Ans
:End
:sub(Ans,1,length(Ans)-1?Str1
With our number stored in N, we loop through each digit of N and store
the numeric character to our string that is at the matching position
in our substring. You access the individual digit in the number by
using iPart(10fPart(A/10^(X, and then locate where it is in the string
"0123456789". The reason you need to add 1 is so that it works with
the 0 digit.
In order to construct a string with all of the digits of the number, we first create a dummy string. This is what the "? is used
for. Each time through the For( loop, we concatenate the string from
before (which is still stored in the Ans variable) to the next numeric
character that is found in N. Using Ans allows us to not have to use
another string variable, since Ans can act like a string and it gets
updated accordingly, and Ans is also faster than a string variable.
By the time we are done with the For( loop, all of our numeric characters are put together in Ans. However, because we stored a dummy
character to the string initially, we now need to remove it, which we
do by getting the substring from the first character to the second to
last character of the string. Finally, we store the string to a more
permanent variable (in this case, Str1) for future use.
Once converted to a string, you can simply use the + operator to concatenate your string literals with the converted number strings.
You should also take a look at a similar Stack Overflow question which addresses a similar issue.
For this issue you can use the toString( function which was introduced in version 5.2.0. This function translates a number to a string which you can use to display numbers and strings together easily. It would end up like this:
Disp "Hello "+toString(Y)+" World "+toString(X)
If you know the length of "Hello" and "World," then you can simply use Output() because Disp creates a new line after every statement.
I have a string called indicators, that the original developer of this application used to store single characters to indicate certain components of a model. I need to change the 7th character in the string, which I tried to do with the following code:
indicators[6] = "R"
The problem, I discovered quickly, was that the string is not always 7 characters long. For example, I have one set of values with U 2, that I need to convert to U 2 R (adding an additional space after the 2). Is there an easy way to force character count with Ruby?
use String.ljust(integer, padstr=' ')
If integer is greater than the length of [the receiver], returns a new String of
length integer with [the return value] left justified and padded with padstr;
otherwise, returns [an unmodified version of the receiver].
indicators = indicators.ljust(7)
indicators[6] = "R"
I've recently been trying to validate a user input so that only letters from the alphabet are accepted, how would I do this? I know how to validate user input for most things, but this one line of code for letters is really troubling me.
You can check the contents of a field with this function:
function validate theString
return matchText(theString,"^[a-zA-Z]+$")
end validate
^[a-zA-Z]+$ is a regular expression. ^indicates the beginning of a string, the brackets equal one char and the expression inside the bracket determine a set of characters. The + means that all following characters have to be equal to the preceding (set of) character(s). The $ indicates the end of the string. In other words, according to this expression, all characters must be of the set a up to and including z or A up to and including Z.
matchText() is a LiveCode function, which checks if the string in the first parameter matches the regular expression in the second parameter. Put the validate() function somewhere at card or stack level and call it from a field in a rawKeyUp handler:
on rawKeyUp
if not validate(the text of me) then
beep
answer "Sorry, that's wrong"
end if
end rawKeyUp
You could also check beforehand:
on keyDown theKey
if validate(theKey) then
pass keyDown
end if
end keyDown
This method is slightly verbose. You could also put the matchText function in a keyDown handler of your field.
I am trying to convert letters to numbers.
I have a sub which ensures only numbers are put into the textbox.
My questions is will the following code work. I have a textbox(for numbers) and combobbox(for letters)
Dim sha As String
Dim stringposition As Long
Dim EngNumber As Long
sha = "abcdefghifjklmnopqrstuvwxyz"
stringposition = InStr(1, sha, Mid(1, cmbEngletter.Text, 1))
MsgBox "stringposition"
EngNumber = (txtManuNo.Text * 10) + stringposition
My only question above would be will the multiplication work with a .text. I believe it won't because it is a string. Please advise then on how to deal with a situation.
You can use CLng() to convert a string to a Long variable
CLng() will throw an error though if it doesn't like the contents of the string (for example if it contains a non-numeric character), so only use it when you are certain your string will only contain numbers
More forgiving is it to use Val() to convert a string into a numeric variable (a Double by default)
I also suggest you look into the following functions:
Asc() : returns the ASCII value of a character
Chr$() : coverts an ASCII value into a character
Left$() : returns the first characters of a string
CStr() : convert a number into a string
I think in your code you mean to show the contents of your variable stringposition instead of the word "stringposition", so you should remove the ""
I do wonder though what you are trying to accomplish with your code, but applying the above to your code gives:
Dim sha As String
Dim stringposition As Long
Dim EngNumber As Long
sha = "abcdefghifjklmnopqrstuvwxyz"
stringposition = InStr(1, sha, Left$(cmbEngletter.Text, 1))
MsgBox CStr(stringposition)
EngNumber = (Val(txtManuNo.Text) * 10) + stringposition
I used Val() because I am not certain your txtManuNo will contain only numbers
To ensure an user can only enter numbers you can use the following code:
Private Sub txtManuNo_KeyPress(KeyAscii As Integer)
Select Case KeyAscii
Case vbKeyBack
'allowe backspace
Case vbKey0 To vbKey9
'allow numbers
Case Else
'refuse any other input
KeyAscii = 0
End Select
End Sub
An user can still input non-numeric charcters with other methods though, like copy-paste via mouse actions, but it is a quick and easy first filter