Dim LineNo as Integer
LineNo = CStr(channel) 'This can have a value of 1 to 100
If LineNo = 1 then
Text1.Text = "Line one selected"
Elseif LineNo = 2 then
Text2.Text = "Line one selected"
'Etc etc
End if
I need to replace the number "1" in Text1.Text and every other TextBox with the value of LineNo? For example:
Text{LineNo}.Text
So I would not have to do a repeated "If" and have a smaller one line code like this:
Text{LineNo}.Text = "Line " & LineNo & " selected"
How would I do this?
Look into a Control array of text boxes. You could have txtLine(), for example, indexed by the channel number.
LineNo = CStr(channel)
txtLine(channel).Text = "Line " & LineNo & " selected"
To create the array, set the Index property of each of the text boxes to an increasing integer, starting at 0.
If you have a finite and relatively small number, you can use a property. I've used this approach with up to 30+ elements in a pinch. Super simple, easy pattern to recognize and replicate in other places. A bit if a pain if the number of elements changes in the future, but extensible nevertheless.
It uses the Choose statement, which takes an index N and returns the Nth element (1-based), hence, the check makes sure that N is > 0 and <= MAX (which you would configure).
Public Property Get TextBox txt(ByVal N As Long)
Const MAX As Long = 10
If N <= 0 || N > MAX Then Exit Property ' Will return a "Nothing". You could return the bound element if you prefer
set txt = Choose(Text1, Text2, Text3, Text4, Text5, Text6, Text7, Text8, Text9, Text10)
End Property
Then, you can simply reference them with the Property, much like an alias:
txt(1).Text = "Line 1 text"
txt(2).Text = "Line 2 text"
If you have an arbitrary number, then you are likely using a control array already, which is simpler because it can be referenced by Index already, so you can directly reference it.
If neither of these work for you and you have a very large number of controls, you can scan the Controls collection in a similar Property, attempting to match ctrl.Name with the pattern of your choice (e.g., matching the first 4 characters to the string "Text", thus matching Text1, Text2, etc, for an unlimited number). Theoretically, this should be future-proofed, but that's just theoretical, because anything can happen. What it does do for you is to encapsulate the lookup in a way that "pretends" to be a control array. Same syntax, just you control the value.
Public Property Get TextBox txt(ByVal N As Long)
Dim I As Long
For I = 0 To Controls.Count - 1 ' Controls is zero-based
' Perform whatever check you need to. Obviously, if you have a "Label" named
' "Text38", the assignment will throw an error (`TextBox = Label` doesn't work).
If Left(Controls(I).Name, 4) = "Text" Then
set txt = Controls(I)
End If
Next
' If you want the Property to never return null, you could uncomment the following line (preventing dereference errors if you insist on the `.Text` for setting/getting the value):
' If txt Is Nothing Then Set txt = Text1
End Property
Use the same way as above: txt(n).Text = "..."
Related
I am trying to add a random set of numbers to the end of a string. I'm still learning the basics of VBS but this has really tricked me and I can't seem to find anything online.
I've tried:
string2 = "hello" + (Rnd() * Len(VALID_TEXT)) + 1
And:
x = rnd*10
string2 = "hello" + x
What am I doing wrong?
All random number generators rely on an underlying algorithm, usually fed by what’s called a seed number. You can use the Randomize statement to create a new seed number to ensure your random numbers don’t follow a predictable pattern.
To get the random numbers, using rnd alone is not sufficient as you will keep on getting the same random number again and again. You have to use randomize to achieve the task as shown below:
Dim strTest:strTest = "Hello"
Dim intNoOfDigitsToAppend:intNoOfDigitsToAppend = 5
Randomize
Msgbox "String before appending: " & strTest
strTest = fn_appendRandomNumbers(strTest,intNoOfDigitsToAppend)
Msgbox "String before appending: " & strTest
function fn_appendRandomNumbers(strToAppend,intNoOfRandomDigits)
Dim i
for i=1 to intNoOfRandomDigits
strToAppend= strToAppend & int(rnd*10) 'rnd gives a random number between 0 and 1, something like 0.8765341. Then, we multiply it by 10 so that the number comes in the range of 0 to 9. In this case, it becomes 8.765341. After that, we use the int method to truncate the decimal part so that we are only left with the Integer part. In this case, 765341 is truncated and we are left with only the integer 8
next
fn_appendRandomNumbers = strToAppend
end function
Reference 1
Reference 2
& is the string concatenation character. + is an old compatability concat character and will error if you mix text and numbers. Use + for maths only.
I have to read between 2 values after asking the users if he want between the '' or not between.
Exemple if the user select 1 in the text
'Hi' My name is 'Kev'in and i'm happ'y' to be 'there'
he will have
'Hi' 'Kev' 'y' 'there'
in a text file. If he chose 2, he will have
My name is in and i'm happ to be
Right now I'm using
Do While objScriptFile.AtEndOfStream <> True
strCurrentLine = objScriptFile.ReadLine
intIsComment = InStr(1,strCurrentLine,"'")
If intIsComment > 0 Then
objCommentFile.Write strCurrentLine & vbCrLf
End If
Loop
Else
For now it's only reading both of the value (between '' and not between) but I have no idea how to change it.
That's pretty simple, provided the delimiter is unique. Split the line at ' and output either the even or the odd array elements, depending on whether 1 or 2 was chosen.
...
strCurrentLine = "'Hi' My name is 'Kev'in and i`m happ'y' to be 'there'"
arr = Split(strCurrentLine, "'")
For i = choice To UBound(arr) Step 2
objCommentFile.Write arr(i)
Next
...
The value of choice is your users' selection (either 1 or 2).
Note that for this to work the strings must not contain apostrophes anywhere else. As #Ekkehard.Horner pointed out in his comment you can't use the delimiter character elsewhere in the text (i'm), because otherwise it would be impossible to distinguish where it was intended to be a delimiter and where not.
I ran into an issue with the Classic ASP VbScript InStr() function. As shown below, the second call to InStr() returns 1 when searching for an empty string in a non empty string. I'm curious why this is happening.
' InStr Test
Dim someText : someText = "So say we all"
Dim emptyString : emptyString = ""
'' I expect this to be true
If inStr(1,someText,"so",1) > 0 Then
Response.write ( "I found ""so""<br />" )
End If
'' I expect this to be false
If inStr(1, someText, emptyString, 1) > 0 Then
Response.Write( "I found an empty string<br />" )
End If
EDIT:
Some additional clarification: The reason for the question came up when debugging legacy code and running into a situation like this:
Function Go(value)
If InStr(1, "Option1|Option2|Option3", value, 1) > 0 Then
' Do some stuff
End If
End Function
In some cases function Go() can get called with an empty string. The original developer's intent was not to check whether value was empty, but rather, whether or not value was equal to one of the piped delimited values (Option1,Option2, etc.).
Thinking about this further it makes sense that every string is created from an empty string, and I can understand why a programming language would assume a string with all characters removed still contains the empty string.
What doesn't make sense to me is why programming languages are implementing this. Consider these 2 statements:
InStr("so say we all", "s") '' evaluates to 1
InStr("so say we all", "") '' evaluates to 1
The InStr() function will return the position of the first occurrence of one string within another. In both of the above cases, the result is 1. However, position 1 always contains the character "s", not an empty string. Furthermore, using another string function like Len() or LenB() on an empty string alone will result in 0, indicating a character length of 0.
It seems that there is some inconsistency here. The empty string contained in all strings is not actually a character, but the InStr() function is treating it as one when other string functions are not. I find this to be un-intuitive and un-necessary.
The Empty String is the Identity Element for Strings:
The identity element I (also denoted E, e, or 1) of a group or related
mathematical structure S is the unique element such that Ia=aI=a for
every element a in S. The symbol "E" derives from the German word for
unity, "Einheit." An identity element is also called a unit element.
If you add 0 to a number n the result is n; if you add/concatenate "" to a string s the result is s:
>> WScript.Echo CStr(1 = 1 + 0)
>> WScript.Echo CStr("a" = "a" & "")
>>
True
True
So every String and SubString contains at least one "":
>> s = "abc"
>> For p = 1 To Len(s)
>> WScript.Echo InStr(p, s, "")
>> Next
>>
1
2
3
and Instr() reports that faithfully. The docs even state:
InStr([start, ]string1, string2[, compare])
...
The InStr function returns the following values:
...
string2 is zero-length start
WRT your
However, position 1 always contains the character "s", not an empty
string.
==>
Position 1 always contains the character "s", and therefore an empty
string too.
I'm puzzled why you think this behavior is incorrect. To the extent that asking Does 'abc' contain ''? even makes sense, the answer has to be "yes": All strings contain the empty string as a trivial case. So the answer to your "why is this happening" question is because it's the only sane thing to do.
It is s correct imho. At least it is what I expect that empty string is part of any other string. But maybe this is a philosophical question. ASP does it so, so live with it. Practically speaking, if you need a different behavior write your own Method, InStrNotEmpty or something, which returns false on empty search string.
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
I have the following VBScript in a reusable action:
'Gather links
Browser("1").Navigate "http://InternalWebmail/something/inbox.nsf"
set oDesc = Description.Create()
oDesc("micclass").Value = "Link"
set links = Browser("1").Page("Webmail").ChildObjects(oDesc)
Dim links2
links2 = ""
'Filter out irrelevant links
For i = 0 To links.Count-1
If lcase(trim(links(i).GetROProperty("text"))) = lcase(trim(DataTable("ExpectedFrom", dtGlobalSheet))) Then
links2 = links2 + "," + links(i).GetROProperty("url")
End If
Next
Dim final
final = split(mid(links2,2),",") 'Remove leading comma and split into array
'For each link (i.e. for each E-mail received):
'Effectively giving a reusable action an input parameter, I hope
For i = 0 To final.Count - 1 'error: Object Required
DataTable("url","CheckHeader") = final(i)
RunAction "CheckHeader", oneIteration
Next
Everything runs just fine, until I get to the declaration of the loop at the bottom of the snippet. There, QTP gives me an error "Object Required" and refuses to elaborate.
i has a leading value of 58, though I've tried setting it to 0 prior to entering the loop.
final is an array of 6 strings, each a URL. All have a value.
If I msgbox(final(2)), I see the value of final(2) as being valid.
isobject(final(1)) = false
final(1) has the correct value
msgbox(final is nothing) and msgbox(final(1) is nothing) yield the same error.
It looks as if the array is null but somehow the array has members?
How is this possible?
What is QTP talking about?
In vbscript arrays don't have a Count property, you should use UBound
x = split("how now brown cow")
' MsgBox x.Count ' error
MsgBox UBound(x) ' 3
The reason .Count worked for the first loop is that ChildObjects does not return an array, it returns a COM collection object. That is also why you had to use the Set statement when assigning to links but not when assigning to final.