I have a bit of an issue with the following piece of code.
Dim k As Integer
k = (TextBox1.Text - 750) \ 250
MsgBox("There should be " & k & " items")
Lets say that textbox1 has a value of 3050 the outcome would be 9.20 however my Mesagebox returns 9 which is not what I want. In fact, I want to see it without the rounding off
How do I do that?
\ is integer division in VB, so you need to use / instead.
See here (MSDN Documentation) for more information on VB operators.
Also, as is mentioned in the comments, you're storing k as an integer - use double (or whatever) instead.
So, how about:
Dim k As Double
Dim tbText as Double
If Double.TryParse(TextBox1.Text, tbText) Then
k = (tbText - 750) / 250
MsgBox("There should be " & k & " items")
End If
If you're certain that TextBox1.Text will be a number, you don't have to use TryParseit like I did - but I don't think you should ever trust a user to get it right...
Related
I'm wondering why the below code works as I hoped for, considering that I'm splitting a string into an array (that's also defined as a string), and afterwards comparing it in an arithmetic (numeric) way.
Option Explicit
Sub test()
Dim str As String, arr() As String
Dim num As Integer, i As Integer
str = "12 9 30"
num = 20
arr() = Split(str, " ")
For i = LBound(arr) To UBound(arr)
If arr(i) > num Then
MsgBox (arr(i) & " is larger than " & num)
End If
Next i
End Sub
As intended the msgBox within the if statement is fired, showing that:
12 isn't larger than 20
9 isn't larger than 20
30 is larger than 20
I didn't know/think that such comparison could work as hoped as i'm basically comparing a string to an integer. I assume there's something i'm not aware of, but in that case, what is it?
PS. I was a bit in doubt regarding which forum to post in, but based my choice on this meta question
For answer please refer to the following article: https://msdn.microsoft.com/en-us/library/aa263418(v=vs.60).aspx
In short if you compare string to numeric type variable, string variable is converted to double* type.
*double based on the information from VB .net comparison operators reference (https://learn.microsoft.com/en-us/dotnet/visual-basic/language-reference/operators/comparison-operators), VB 6.0, VBA and VBA .net are not the same things, however comparison logic should be the same.
VBA seems to be implicitly converting the data type during run-time.
Consider following code which also works.
Sub test2()
Dim str As String, arr() As String, num As String
Dim i As Integer
str = "12 9 30"
num = 12 '\\ Note the way number is being passed.
arr() = Split(str, " ")
For i = LBound(arr) To UBound(arr)
If arr(i) = num Then
MsgBox (arr(i) & " is equal to " & num)
End If
Next i
End Sub
And then below one where arithmetic operation is coercing it to be numeric at run-time.
Sub test3()
Dim str As String, arr() As String, num As String
Dim i As Integer
str = "12 9 30"
num = 12
arr() = Split(str, " ")
For i = LBound(arr) To UBound(arr)
If (arr(i) - num) > 0 Then
MsgBox (arr(i) & " is greater than " & num)
End If
Next i
End Sub
I know it will not answer your question fully but might explain why it is giving correct result. It is advisable to convert to correct data type rather than relying on defaults i.e.
If CInt(arr(i)) > num Then
I am getting a runtime error I don't understand in Excel 2011 for Mac under OS X 10.7.5. Here is a summary of the code:
Dim h, n, k as Integer
Dim report as Workbook
Dim r1 as Worksheet
Dim t, newline as String
Dim line() as String
newline = vbCr
'
' (code to get user input from a text box, to select a worksheet by number)
'
ReDim line(report.Sheets.Count + 10)
MsgBox "Array line has " & UBound(line) & " elements." '----> 21 elements
line = split(t, newline)
h = UBound(line)
MsgBox "Array line has " & h & " elements." '----> 16 elements
n = 0
MsgBox TypeName(n) '----> Integer
For k = h To 1 Step -1
If IsNumeric(line(k)) Then
n = line(k)
Exit For
End If
Next k
If n > 0 Then
MsgBox n '----> 7
MsgBox TypeName(n) '----> String
Set r1 = report.Sheets(n) '----> Runtime error "Subscript out of bounds"
So n is declared as an integer, but now VBA thinks it is a string and looks for a worksheet named "7". Is this a platform bug, or is there something I haven't learned yet?
It also surprises me that putting data into the dynamic array reduces its dimension, but perhaps that is normal, or perhaps for dynamic arrays Ubound returns the last used element instead of the dimension, although I have not seen that documented.
The first part of your question is answered by #ScottCraner in the comments - the correct syntax for declaring multiple strongly typed variables on one line is:
Dim h As Integer, n As Integer, k As Integer
'...
Dim t As String, newline As String
So, I'll address the second part of your question specific to UBound - unless you've declared Option Base 1 at the top of the module, your arrays start at element 0 by default, not element 1. However, the Split function always returns a 0 based array (unless you split a vbNullString, in which case you get a LBound of -1):
Private Sub ArrayBounds()
Dim foo() As String
'Always returns 3, regardless of Option Base:
foo = Split("zero,one,two,three", ",")
MsgBox UBound(foo)
ReDim foo(4)
'Option Base 1 returns 1,4
'Option Base 0 (default) returns 0,3
MsgBox LBound(foo) & "," & UBound(foo)
End Sub
That means this line is extremely misleading...
h = UBound(line)
MsgBox "Array line has " & h & " elements."
...because the Array line actually has h + 1 elements, which means that your loop here...
For k = h To 1 Step -1
If IsNumeric(line(k)) Then
n = line(k)
Exit For
End If
Next k
...is actually skipping element 0. You don't really even need the h variable at all - you can just make your loop parameter this...
For k = UBound(line) To LBound(line) Step -1
If IsNumeric(line(k)) Then
n = line(k)
Exit For
End If
Next k
...and not have to worry what the base of the array is.
BTW, not asked, but storing vbCr as a variable here...
newline = vbCr
...isn't necessary at all, and opens the door for all kinds of other problems if you intend that a "newline" is always vbCr. Just use the pre-defined constant vbCr directly.
I have a macro that changes single quotes in front of a number to an apostrophe (or close single curly quote). Typically when you type something like "the '80s" in word, the apostrophe in front of the "8" faces the wrong way. The macro below works, but it is incredibly slow (like 10 seconds per page). In a regular language (even an interpreted one), this would be a fast procedure. Any insights why it takes so long in VBA on Word 2007? Or if someone has some find+replace skills that can do this without iterating, please let me know.
Sub FixNumericalReverseQuotes()
Dim char As Range
Debug.Print "starting " + CStr(Now)
With Selection
total = .Characters.Count
' Will be looking ahead one character, so we need at least 2 in the selection
If total < 2 Then
Return
End If
For x = 1 To total - 1
a_code = Asc(.Characters(x))
b_code = Asc(.Characters(x + 1))
' We want to convert a single quote in front of a number to an apostrophe
' Trying to use all numerical comparisons to speed this up
If (a_code = 145 Or a_code = 39) And b_code >= 48 And b_code <= 57 Then
.Characters(x) = Chr(146)
End If
Next x
End With
Debug.Print "ending " + CStr(Now)
End Sub
Beside two specified (Why...? and How to do without...?) there is an implied question – how to do proper iteration through Word object collection.
Answer is – to use obj.Next property rather than access by index.
That is, instead of:
For i = 1 to ActiveDocument.Characters.Count
'Do something with ActiveDocument.Characters(i), e.g.:
Debug.Pring ActiveDocument.Characters(i).Text
Next
one should use:
Dim ch as Range: Set ch = ActiveDocument.Characters(1)
Do
'Do something with ch, e.g.:
Debug.Print ch.Text
Set ch = ch.Next 'Note iterating
Loop Until ch is Nothing
Timing: 00:03:30 vs. 00:00:06, more than 3 minutes vs. 6 seconds.
Found on Google, link lost, sorry. Confirmed by personal exploration.
Modified version of #Comintern's "Array method":
Sub FixNumericalReverseQuotes()
Dim chars() As Byte
chars = StrConv(Selection.Text, vbFromUnicode)
Dim pos As Long
For pos = 0 To UBound(chars) - 1
If (chars(pos) = 145 Or chars(pos) = 39) _
And (chars(pos + 1) >= 48 And chars(pos + 1) <= 57) Then
' Make the change directly in the selection so track changes is sensible.
' I have to use 213 instead of 146 for reasons I don't understand--
' probably has to do with encoding on Mac, but anyway, this shows the change.
Selection.Characters(pos + 1) = Chr(213)
End If
Next pos
End Sub
Maybe this?
Sub FixNumQuotes()
Dim MyArr As Variant, MyString As String, X As Long, Z As Long
Debug.Print "starting " + CStr(Now)
For Z = 145 To 146
MyArr = Split(Selection.Text, Chr(Z))
For X = LBound(MyArr) To UBound(MyArr)
If IsNumeric(Left(MyArr(X), 1)) Then MyArr(X) = "'" & MyArr(X)
Next
MyString = Join(MyArr, Chr(Z))
Selection.Text = MyString
Next
Selection.Text = Replace(Replace(Selection.Text, Chr(146) & "'", "'"), Chr(145) & "'", "'")
Debug.Print "ending " + CStr(Now)
End Sub
I am not 100% sure on your criteria, I have made both an open and close single quote a ' but you can change that quite easily if you want.
It splits the string to an array on chr(145), checks the first char of each element for a numeric and prefixes it with a single quote if found.
Then it joins the array back to a string on chr(145) then repeats the whole things for chr(146). Finally it looks through the string for an occurence of a single quote AND either of those curled quotes next to each other (because that has to be something we just created) and replaces them with just the single quote we want. This leaves any occurence not next to a number intact.
This final replacement part is the bit you would change if you want something other than ' as the character.
I have been struggling with this for days now. My attempted solution was to use a regular expression on document.text. Then, using the matches in a document.range(start,end), replace the text. This preserves formatting.
The problem is that the start and end in the range do not match the index into text. I think I have found the discrepancy - hidden in the range are field codes (in my case they were hyperlinks). In addition, document.text has a bunch of BEL codes that are easy to strip out. If you loop through a range using the character method, append the characters to a string and print it you will see the field codes that don't show up if you use the .text method.
Amazingly you can get the field codes in document.text if you turn on "show field codes" in one of a number of ways. Unfortunately, that version is not exactly the same as what the range/characters shows - the document.text has just the field code, the range/characters has the field code and the field value. Therefore you can never get the character indices to match.
I have a working version where instead of using range(start,end), I do something like:
Set matchRange = doc.Range.Characters(myMatches(j).FirstIndex + 1)
matchRange.Collapse (wdCollapseStart)
Call matchRange.MoveEnd(WdUnits.wdCharacter, myMatches(j).Length)
matchRange.text = Replacement
As I say, this works but the first statement is dreadfully slow - it appears that Word is iterating through all of the characters to get to the correct point. In doing so, it doesn't seem to count the field codes, so we get to the correct point.
Bottom line, I have not been able to come up with a good way to match the indexing of the document.text string to an equivalent range(start,end) that is not a performance disaster.
Ideas welcome, and thanks.
This is a problem begging for regular expressions. Resolving the .Characters calls that many times is probably what is killing you in performance.
I'd do something like this:
Public Sub FixNumericalReverseQuotesFast()
Dim expression As RegExp
Set expression = New RegExp
Dim buffer As String
buffer = Selection.Range.Text
expression.Global = True
expression.MultiLine = True
expression.Pattern = "[" & Chr$(145) & Chr$(39) & "]\d"
Dim matches As MatchCollection
Set matches = expression.Execute(buffer)
Dim found As Match
For Each found In matches
buffer = Replace(buffer, found, Chr$(146) & Right$(found, 1))
Next
Selection.Range.Text = buffer
End Sub
NOTE: Requires a reference to Microsoft VBScript Regular Expressions 5.5 (or late binding).
EDIT:
The solution without using the Regular Expressions library is still avoiding working with Ranges. This can easily be converted to working with a byte array instead:
Sub FixNumericalReverseQuotes()
Dim chars() As Byte
chars = StrConv(Selection.Text, vbFromUnicode)
Dim pos As Long
For pos = 0 To UBound(chars) - 1
If (chars(pos) = 145 Or chars(pos) = 39) _
And (chars(pos + 1) >= 48 And chars(pos + 1) <= 57) Then
chars(pos) = 146
End If
Next pos
Selection.Text = StrConv(chars, vbUnicode)
End Sub
Benchmarks (100 iterations, 3 pages of text with 100 "hits" per page):
Regex method: 1.4375 seconds
Array method: 2.765625 seconds
OP method: (Ended task after 23 minutes)
About half as fast as the Regex, but still roughly 10ms per page.
EDIT 2: Apparently the methods above are not format safe, so method 3:
Sub FixNumericalReverseQuotesVThree()
Dim full_text As Range
Dim cached As Long
Set full_text = ActiveDocument.Range
full_text.Find.ClearFormatting
full_text.Find.MatchWildcards = True
cached = full_text.End
Do While full_text.Find.Execute("[" & Chr$(145) & Chr$(39) & "][0-9]")
full_text.End = full_text.Start + 2
full_text.Characters(1) = Chr$(96)
full_text.Start = full_text.Start + 1
full_text.End = cached
Loop
End Sub
Again, slower than both the above methods, but still runs reasonably fast (on the order of ms).
Dim ssi(11) As String
For i = 0 To 10
If ssi(i) = "" Then ssi(i) = "0"
For j = 0 To Val(ssi(11)) + i
ssi(i) = xuh(Val(ssi(i)))
Next
Next
If ssi(11) = "2" Then
L_zz.Caption = Val(Left(ssi(0) & ssi(1) & ssi(2) & ssi(3) & ssi(4) & ssi(5) & ssi(6) & ssi(7), ssi(10)))
ElseIf ssi(11) = "3" Then
L_zz.Caption = Val(Left(ssi(0) & ssi(1) & ssi(2) & ssi(3) & ssi(4) & ssi(5) & ssi(6) & ssi(7), ssi(10))) * (-1)
End If
I am new here and new to VB as well.
I am trying to understand this double loop in vb code.
ssi(i) is defined as a String variable. and each element is assigned to a specific number in a String. Hope I told it clearly.
My problem with this loop is below.
Since i ranges from 0 to 10, what does this j mean? Does j mean the new ssi(1-10) or another whatever number?
I think the best way to answer your question about understanding a double loop is to try looking at something simpler.
The first program I always write in each new version of BASIC that comes along is a 12 times table.
I've modified it a bit below to be a 12 x 10 table for the purpose of illustrating for you how a double loop works ... hope it helps:
For x As Integer = 1 To 12
For y As Integer = 1 To 10
Console.Write(x * y)
Console.Write(vbTab)
Next
Console.WriteLine()
Next
I am writing a console application which requires user input of certain values. I want to disallow any letter input. The console automatically writes "Conversion from string "b" to type 'Integer' is not valid." but I want the console to display my personal message "Not a valid number, please try again." how can I do that?
I've tried many different keywords and phrases but none work. Maybe I'm doing it wrong (not unlikely) or maybe it's just meant for something else. Either way I need help.
To recap: User input in a console that only allows numbers not letters, and will display my message.
I didn't want to post my code since people always pick on it, but here it is. Please note that there MUST be an exception. It is homework and I NEED an exception and this is one way a user can screw up. Please don't tell me not to use an exception.
Module Module1
Sub Main()
Try
System.Console.WriteLine("Input up to 10 valid numbers to have them mathematically averaged.")
For Index = 0 To 9
Dim Input As IList
Input = Console.ReadLine()
Next Index
If ' here is where I want to add that numbers only Then
Throw New exception("Not a valid number, please try again.")
Else
System.Console.WriteLine("Now averaging numbers...")
Dim average As Double = (n + n + n + n + n + n + n + n + n + n) / 10
Console.WriteLine("The average of " & n & "," & n & "," & n & "," & n & "," & n & "," & n & "," & n & "," & n & "," & n & " and " & n & " is " & average & ".", "Calculation")
End If
Catch e As Exception
System.Console.WriteLine(e.Message)
End Try
End Sub
End Module
Dim inputString = Console.ReadLine
If Integer.TryParse(inputString, num) Then
DoSomething(num)
Else
Console.WriteLine("Not a valid number, please try again.")
End If
Here's one way to do it, honoring your requirements:
Module Module1
Sub Main()
System.Console.WriteLine("Input valid numbers seperated by spaces to have them mathematically averaged.")
Dim inputArray As String() = System.Text.RegularExpressions.Regex.Replace(Console.ReadLine().Trim(), "\s{2,}", " ").Split(New Char() {" "})
Dim values As New ArrayList
Dim sum As Integer
For i As Integer = 0 To inputArray.Length - 1
Try
sum = sum + Integer.Parse(inputArray(i), Globalization.NumberStyles.Integer)
values.Add(inputArray(i))
Catch ex As Exception
Console.WriteLine(String.Format("The value ""{0}"" is not a valid number and will be ignored. ExceptionMessage: {1}", inputArray(i), ex.Message))
End Try
Next
Dim average As Decimal = sum / values.Count
Console.WriteLine(vbCrLf)
Console.WriteLine(String.Format("The average of ""{0}"" is {1}", Join(values.ToArray, ", "), average))
Console.WriteLine(vbCrLf)
Main()
End Sub
End Module
Well for starters, the Try - Catch should be inside the input gathering For loop. As your program is written now, as soon as one wrong value is entered execution jumps to the catch and your program ends! If you try casting your input to an decimal you won't need to throw an exception manually, it will be thrown automatically if your input is not an decimal! Try casting the current console input as a decimal and then add that number to your list.
Dim inputNumber as Decimal = CDec(Console.ReadLine())
Input.Add(inputNumber)
Also n will be the same number every time so how you are doing it won't work (look up the basics of how a list works, how to add elements to a list and how to display list elements)