VBA for Access lacks a simple Max(x,y) function to find the mathematical maximum of two or more values. I'm accustomed to having such a function already in the base API coming from other languages such as perl/php/ruby/python etc.
I know it can be done: IIf(x > y, x,y). Are there any other solutions available?
I'll interpret the question to read:
How does one implement a function in Access that returns the Max/Min of an array of numbers? Here's the code I use (named "iMax" by analogy with IIf, i.e., "Immediate If"/"Immediate Max"):
Public Function iMax(ParamArray p()) As Variant
' Idea from Trevor Best in Usenet MessageID rib5dv45ko62adf2v0d1cot4kiu5t8mbdp#4ax.com
Dim i As Long
Dim v As Variant
v = p(LBound(p))
For i = LBound(p) + 1 To UBound(p)
If v < p(i) Then
v = p(i)
End If
Next
iMax = v
End Function
Public Function iMin(ParamArray p()) As Variant
' Idea from Trevor Best in Usenet MessageID rib5dv45ko62adf2v0d1cot4kiu5t8mbdp#4ax.com
Dim i As Long
Dim v As Variant
v = p(LBound(p))
For i = LBound(p) + 1 To UBound(p)
If v > p(i) Then
v = p(i)
End If
Next
iMin = v
End Function
As to why Access wouldn't implement it, it's not a very common thing to need, seems to me. It's not very "databasy", either. You've already got all the functions you need for finding Max/Min across domain and in sets of rows. It's also not very hard to implement, or to just code as a one-time comparison when you need it.
Maybe the above will help somebody.
Calling Excel VBA Functions from MS Access VBA
If you add a reference to Excel (Tools → References → Microsoft Excel x.xx Object Library) then you can use WorksheetFunction to call most Excel worksheet functions, including MAX (which can also be used on arrays).
Examples:
MsgBox WorksheetFunction.Max(42, 1999, 888)
or,
Dim arr(1 To 3) As Long
arr(1) = 42
arr(2) = 1999
arr(3) = 888
MsgBox WorksheetFunction.Max(arr)
The first call takes a second to respond (actually 1.1sec for me), but subsequent calls are much more reasonable (<0.002sec each for me).
Referring to Excel as an object
If you're using a lot of Excel functions in your procedure, you may be able to improve performance further by using an Application object to refer directly to Excel.
For example, this procedure iterates a set of records, repeatedly using Excel's MAX on a Byte Array to determine the "highest" ASCII character of each record.
Option Compare Text
Option Explicit
'requires reference to "Microsoft Excel x.xx Object Library"
Public excel As New excel.Application
Sub demo_ListMaxChars()
'list the character with the highest ASCII code for each of the first 100 records
Dim rs As Recordset, mx
Set rs = CurrentDb.OpenRecordset("select myField from tblMyTable")
With rs
.MoveFirst
Do
mx = maxChar(!myField)
Debug.Print !myField, mx & "(" & ChrW(mx) & ")" '(Hit CTRL+G to view)
.MoveNext
Loop Until .EOF
.Close
End With
Set rs = Nothing 'always clean up your objects when finished with them!
Set excel = Nothing
End Sub
Function maxChar(st As String)
Dim b() As Byte 'declare Byte Array
ReDim b(1 To Len(st)) 'resize Byte Array
b = StrConv(st, vbFromUnicode) 'convert String to Bytes
maxChar = excel.WorksheetFunction.Max(b) 'find maximum Byte (with Excel function)
End Function
Because they probably thought that you would use DMAX and DMIN or the sql MAX and only working with the database in access?
Im also curious about why.. Its seems like a overkill to have to create a temp-table and add form values to the table and then run a DMAX or MAX-query on the table to get the result...
I've been known to create a small projMax() function just to deal with these. Not that VBA will probably ever be enhanced, but just in case they ever do add a proper Max (and Min) function, it won't conflict with my functions. BTW, the original poster suggests doing IIF... That works, but in my function, I usually throw a couple of Nz()'s to prevent a null from ruining the function.
Both functions have problems with Null. I think this will be better.
Public Function iMin(ParamArray p()) As Variant
Dim vVal As Variant, vMinVal As Variant
vMinVal = Null
For Each vVal In p
If Not IsNull(vVal) And (IsNull(vMinVal) Or (vVal < vMinVal)) Then _
vMinVal = vVal
Next
iMin = vMinVal
End Function
I liked DGM's use of the IIF statement and David's use of the For/Next loop, so I am combining them together.
Because VBA in access does not have a strict type checking, I will be using varients to preserve all numerics, integer and decimal, and re-type the return value.
Kudos to HansUP for catching my parameter verification :)
Comments added to make code more friendlier.
Option Compare Database
Option Base 0
Option Explicit
Function f_var_Min(ParamArray NumericItems()) As Variant
If UBound(NumericItems) = -1 Then Exit Function ' No parameters
Dim vVal As Variant, vNumeric As Variant
vVal = NumericItems(0)
For Each vNumeric In NumericItems
vVal = IIf(vNumeric < vVal, vNumeric, vVal) ' Keep smaller of 2 values
Next
f_var_Min = vVal ' Return final value
End Function
Function f_var_Max(ParamArray NumericItems()) As Variant
If UBound(NumericItems) = -1 Then Exit Function ' No parameters
Dim vVal As Variant, vNumeric As Variant
vVal = NumericItems(0)
For Each vNumeric In NumericItems
vVal = IIf(vNumeric < vVal, vVal, vNumeric) ' Keep larger of 2 values
Next
f_var_Max = vVal ' Return final value
End Function
The only difference between the 2 functions is the order of vVal and vNumeric in the IIF statement.The for each clause uses internal VBA logic to handle the looping and array bounds checking, while "Base 0" starts the array index at 0.
You can call Excel functions in Access VBA:
Global gObjExcel As Excel.Application
Public Sub initXL()
Set gObjExcel = New Excel.Application
End Sub
Public Sub killXL()
gObjExcel.Quit
Set gObjExcel = Nothing
End Sub
Public Function xlMax(a As Double, b As Double) As Double
xlCeiling = gObjExcel.Application.Max(a, b)
End Function
You can do Worksheetfunction.max() or worksheetfunction.min() within Access VBA. Hope this helps.
Related
I'm working on a piece of code to extract the nominal size of a pipeline from it's tagname. For example: L-P-50-00XX-0000-000. The 50 would be it's nominal size (2") which I would like to extract. I know I could do it like this:
TagnameArray() = Split("L-P-50-00XX-0000-000", "-")
DNSize = TagnameArray(2)
But I would like it to be a function because it's a small part of my whole macro and I don't need it for all the plants I'm working on just this one. My current code is:
Sub WBDA_XXX()
Dim a As Range, b As Range
Dim TagnameArray() As String
Dim DNMaat As String
Dim DN As String
Set a = Selection
For Each b In a.Rows
IntRow = b.Row
TagnameArray() = Split(Cells(IntRow, 2).Value, "-")
DN = DNMaat(IntRow, TagnameArray())
Cells(IntRow, 3).Value = DN
Next b
End Sub
Function DNMaat(IntRow As Integer, TagnameArray() As String) As Integer
For i = LBound(TagnameArray()) To UBound(TagnameArray())
If IsNumeric(TagnameArray(i)) = True Then
DNMaat = TagnameArray(i)
Exit For
End If
Next i
End Function
However this code gives me a matrix expected error which I don't know how to resolve. I would also like to use the nominal size in further calculations so it will have to be converted to an integer after extracting it from the tagname. Does anyone see where I made a mistake in my code?
This is easy enough to do with a split, and a little help from the 'Like' evaluation.
A bit of background on 'Like' - It will return TRUE or FALSE based on whether an input variable matches a given pattern. In the pattern [A-Z] means it can be any uppercase letter between A and Z, and # means any number.
The code:
' Function declared to return variant strictly for returning a Null string or a Long
Public Function PipeSize(ByVal TagName As String) As Variant
' If TagName doesn't meet the tag formatting requirements, return a null string
If Not TagName Like "[A-Z]-[A-Z]-##-##[A-Z]-####-###" Then
PipeSize = vbNullString
Exit Function
End If
' This will hold our split pipecodes
Dim PipeCodes As Variant
PipeCodes = Split(TagName, "-")
' Return the code in position 2 (Split returns a 0 based array by default)
PipeSize = PipeCodes(2)
End Function
You will want to consider changing the return type of the function depending on your needs. It will return a null string if the input tag doesnt match the pattern, otherwise it returns a long (number). You can change it to return a string if needed, or you can write a second function to interpret the number to it's length.
Here's a refactored version of your code that finds just the first numeric tag. I cleaned up your code a bit, and I think I found the bug as well. You were declaring DNMAAT as a String but also calling it as a Function. This was likely causing your Array expected error.
Here's the code:
' Don't use underscores '_' in names. These hold special value in VBA.
Sub WBDAXXX()
Dim a As Range, b As Range
Dim IntRow As Long
Set a = Selection
For Each b In a.Rows
IntRow = b.Row
' No need to a middleman here. I directly pass the split values
' since the middleman was only used for the function. Same goes for cutting DN.
' Also, be sure to qualify these 'Cells' ranges. Relying on implicit
' Activesheet is dangerous and unpredictable.
Cells(IntRow, 3).value = DNMaat(Split(Cells(IntRow, 2).value, "-"))
Next b
End Sub
' By telling the function to expect a normal variant, we can input any
' value we like. This can be dangerous if you dont anticipate the errors
' caused by Variants. Thus, I check for Arrayness on the first line and
' exit the function if an input value will cause an issue.
Function DNMaat(TagnameArray As Variant) As Long
If Not IsArray(TagnameArray) Then Exit Function
Dim i As Long
For i = LBound(TagnameArray) To UBound(TagnameArray)
If IsNumeric(TagnameArray(i)) = True Then
DNMaat = TagnameArray(i)
Exit Function
End If
Next i
End Function
The error matrix expected is thrown by the compiler because you have defined DNMaat twice: Once as string variable and once as a function. Remove the definition as variable.
Another thing: Your function will return an integer, but you assigning it to a string (and this string is used just to write the result into a cell). Get rid of the variable DN and assign it directly:
Cells(IntRow, 3).Value = DNMaat(IntRow, TagnameArray())
Plus the global advice to use option explicit to enforce definition of all used variables and to define a variable holding a row/column number always as long and not as integer
I'm trying to use Application.Caller inside a Function (code below), but Excel returns a #VALUE and the background color is not set.
The personal function is called from an Excel cell. The idea is to map RGB values to color display in a "synchronous" fashion (i.e. without having to press a button).
When I run the following function through the debugger and step just before the instruction vCaller.Interior.Color = RGB(rlev, glev, blev), I can manually set the background color to green by pasting the exact same instruction in the execution console. So I'm puzzled as to why Excel is failing but VBA isn't.
Any clue ?
Public Function RGB_print(rlev As Integer, glev As Integer, blev As Integer)
As String
Dim vCaller As Variant
Set vCaller = Application.Caller
If TypeName(vCaller) = "Range" Then
vCaller.Interior.Color = RGB(rlev, glev, blev)
End If
RGB_print = ""
End Function
I completely agree with the comment from #Rory - I'd never use this code in my own projects, but I wanted to see anyway....
If in a normal module you create this function:
Public Function RGB_print(rlev As Integer, glev As Integer, blev As Integer)
Application.Volatile
End Function
Then in your sheet add this code:
Private Sub Worksheet_Calculate()
Dim rFormula As Range
Dim vForm As Variant
Dim sArguments As String
Dim sFormula As String
Dim rgblev As Variant
Set rFormula = Sheet1.Cells.SpecialCells(xlCellTypeFormulas)
For Each vForm In rFormula
If InStr(vForm.FormulaLocal, "RGB_print") <> 0 Then
sFormula = vForm.FormulaLocal
sArguments = Mid(sFormula, InStr(sFormula, "(") + 1, InStr(sFormula, ")") - InStr(sFormula, "(") - 1)
rgblev = Split(sArguments, ",")
vForm.Interior.Color = RGB(Evaluate(rgblev(0)), Evaluate(rgblev(1)), Evaluate(rgblev(2)))
End If
Next vForm
End Sub
This worked for formula such as:
=RGB_print(255,0,255) and =RGB_print(A5,B5,C5)
But again, find another way - this code has so many pitfalls I'll probably lose 100 reputation just for posting it.
Ok, as an alternative to Darrent's very precise reply, I'm reposting Tim Williwam's comment : whether one may/should mix functions and macros is an important question and it has been discussed here. Bottom line is : you can but don't do it unless you know what you are doing and are prepared to face the consequences.
I can't reference a range with a sheet for my function like I can with =vlookup.
This works: =MVLOOKUP(a2,B:C,2,",",", ")
This isn't: =MVLOOKUP(a2,Sheet3!B:C,2,",",", ")
The code:
Public Function MVLookup(Lookup_Values, Table_Array As Range, Col_Index_Num As Long, Input_Separator As String, Output_Separator As String) As String
Dim in0, out0, i
in0 = Split(Lookup_Values, Input_Separator)
ReDim out0(UBound(in0, 1))
For i = LBound(in0, 1) To UBound(in0, 1)
out0(i) = Application.WorksheetFunction.VLookup(in0(i), Table_Array, Col_Index_Num, False)
Next i
MVLookup = Join(out0, Output_Separator)
End Function
I don't know basic and I'm not planning to learn it, I rarely even use excel, so sorry for the lame question. I guess basic is really "basic" it took me 30 minutes to get to this point from the reference(reading included), but other 60 minutes in frustration because the above problem.
Help me so I can go back to my vba free life!
EDIT: Although the code above worked after an excel restart, Jeeped gave me a safer solution and more universal functionality. Thanks for that.
I was not planning to use it on other than strings but thanks for the addition, I wrongly assumed there is a check for data type every time and type passed along in the background and vlookup acting accordingly. I have also learned how to set default values to function input variables.
See solution.
Thanks again, Jeeped!
You are confusing 1 with "1" and regardless of your personal distaste for VBA, I really don't know of any programming language that treats them as identical values (with the possible exception of a worksheet's COUNTIF function).
Public Function MVLookup(Lookup_Values, table_Array As Range, col_Index_Num As Long, _
Optional Input_Separator As String = ",", _
Optional output_Separator As String = ", ") As String
Dim in0 As Variant, out0 As Variant, i As Long
in0 = Split(Lookup_Values, Input_Separator)
ReDim out0(UBound(in0))
For i = LBound(in0) To UBound(in0)
If IsNumeric(in0(i)) Then
If Not IsError(Application.Match(Val(in0(i)), Application.Index(table_Array, 0, 1), 0)) Then _
out0(i) = Application.VLookup(Val(in0(i)), table_Array, col_Index_Num, False)
Else
If Not IsError(Application.Match(in0(i), Application.Index(table_Array, 0, 1), 0)) Then _
out0(i) = Application.VLookup(in0(i), table_Array, col_Index_Num, False)
End If
Next i
MVLookup = Join(out0, output_Separator)
End Function
When you Split a string into a variant array, you end up with an array of string elements. Granted, they look like numbers but they are not true numbers; merely textual representational facsimiles of true numbers. The VLOOKUP function does not treat them as numbers when the first column in your table_array parameter is filled with true numbers.
The IsNumeric function can reconize a string that looks like a number and then the Val function can convert that text-that-looks-like-a-number into a true number.
I've also added a quick check to ensure what you are looking for is actually there before you attempt to stuff the return value into an array.
Your split strings are one-dimensioned variant arrays. There is no need to supply the rank in the LBound / UBound functions.
Sample data on Sheet3 Results from MVLOOKUP
This is not a valid range reference ANYWHERE in Excel B:Sheet3!C.
Either use B:C or Sheet3!B:C
Edit. Corrected as per Jeeped's comment.
I'm having some trouble with syntax options while writing a VBA Macro for Excel. In VBA you can call a method on an object in two different ways:
foo.bar(arg1, arg2)
or
foo.bar arg1, arg2
I absolutely detest the second sort of syntax because I find it lacks any sort of clarity, so I normally adhere to the first option. However, I've come across a situation where using the first option creates an error, while the second executes fine. (This may perhaps be an indicator of other problems in my code.) Here is the culprit code:
Function GetFundList() As Collection
Dim newFund As FundValues
Range("A5").Select
Set GetFundList = New Collection
While Len(Selection.Value)
Set newFund = New FundValues
' I set the fields of newFund and move Selection
The problem is in this next line:
GetFundList.Add newFund
Wend
End Function
FundValues is a class I created that is essentially just a struct; it has three properties which get set during the loop.
Basically, when I call GetFundList.Add(newFund) I get the following error:
Run-time error '438':
Object doesn't support this property or method
But calling GetFundList.Add newFund is perfectly fine.
Does anyone understand the intricacies of VBA well enough to explain why this is happening?
EDIT: Thanks much for the explanations!
Adding items to a collection is not defined as a function returning a value, but as a sub routine:
Public Sub Add( _
ByVal Item As Object, _
Optional ByVal Key As String, _
Optional ByVal { Before | After } As Object = Nothing _
)
When calling another sub routine by name and sending arguments (without adding the "Call" statement), you are not required to add parentheses.
You need to add parentheses when you call a function that returns a value to a variable.
Example:
Sub Test_1()
Dim iCnt As Integer
Dim iCnt_B As Integer
Dim iResult As Integer
iCnt = 2
iCnt_B = 3
fTest_1 iCnt, iResult, iCnt_B
End Sub
Public Function fTest_1(iCnt, iResult, iCnt_B)
iResult = iCnt * 2 + iCnt_B * 2
End Function
Sub Test_2()
Dim iCnt As Integer
Dim iCnt_B As Integer
Dim iResult As Integer
iCnt = 2
iCnt_B = 3
iResult = fTest_2(iCnt, iCnt_B)
End Sub
Public Function fTest_2(iCnt, iCnt_B)
fTest_2 = iCnt * 2 + iCnt_B * 2
End Function
Let me know if not clear.
This Daily Dose of Excel conversation will be helpful.
When you use the parentheses you are forcing VBA to evaluate what's inside them and adding the result to the collection. Since NewFund has no default property - I assume - the evaluation yields nothing, so can't be added. Without the parentheses it evaluates to the instance of the class, which is what you want.
Another example. This:
Dim coll As Collection
Set coll = New Collection
coll.Add Range("A1")
Debug.Print coll(1); TypeName(coll(1))
and this ...
coll.Add (Range("A1"))
Debug.Print coll(1); TypeName(coll(1))
... both yield whatever is in A1 in the debug.window, because Value is Range's default property. However, the first will yield a type of "Range", whereas the type in the 2nd example is the data type in A1. In other words, the first adds a range to the collection, the 2nd the contents of the range.
On the other hand, this works:
Dim coll As Collection
Set coll = New Collection
coll.Add ActiveSheet
Debug.Print coll(1).Name
... and this doesn't:
coll.Add (ActiveSheet)
Debug.Print coll(1).Name
because ActiveSheet has no default property. You'll get an runtime error 438, just like in your question.
Here's another way of looking at the same thing.
Let assume that cell A1 contains the string Hi!
Function SomeFunc(item1, item2)
SomeFunc = 4
End Function
Sub Mac()
' here in both of the following two lines of code,
' item1 will be Variant/Object/Range, while item2 will be Variant/String:
SomeFunc Range("A1"), (Range("A1"))
Let i = SomeFunc(Range("A1"), (Range("A1")))
'this following is a compile syntax error
SomeFunc(Range("A1"), (Range("A1")))
' while here in both the following two lines of code,
' item1 will be Variant/String while item2 will be Variant/Object/Range:
SomeFunc ((Range("A1")), Range("A1")
Let j = SomeFunc((Range("A1")), Range("A1"))
'this following is a compile syntax error
SomeFunc((Range("A1")), Range("A1"))
Set r = Range("A1") ' sets r to Variant/Object/Range
Set r = (Range("A1")) ' runtime error 13, type mismatch; cannot SET r (as reference) to string "Hi!" -- Strings are not objects in VBA
Set r = Range("A1").Value ' runtime error (same)
Let r = Range("A1") ' set r to "Hi!" e.g. contents of A1 aka Range("A1").Value; conversion to value during let = assignment
Let r = (Range("A1")) ' set r to "Hi!" e.g. contents of A1 aka Range("A1").Value; conversion to value by extra ()'s
Let r = Range("A1").Value ' set r to "Hi!" by explicit use of .Value
End Sub
I only add this to help illustrate that there are two things going on here, which could be conflated.
The first is that the () in an expression that converts the item to its Value property as stated above in other answers.
The second is that functions invoked with intent to capture or use the return value require extra () surrounding the whole argument list, whereas functions (or sub's) invoked without intent to capture or use the return value (e.g. as statements) must be called without those same () surrounding the argument list. These surrounding () do not convert the argument list using .Value. When the argument list has only one parameter, this distinction can be particularly confusing.
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
Which features of the VBA language are either poorly documented, or simply not often used?
This trick only works in Access VBA, Excel and others won't allow it. But you can make a Standard Module hidden from the object browser by prefixing the Module name with an underscore. The module will then only be visible if you change the object browser to show hidden objects.
This trick works with Enums in all vb6 based version of VBA. You can create a hidden member of an Enum by encasing it's name in brackets, then prefixing it with an underscore. Example:
Public Enum MyEnum
meDefault = 0
meThing1 = 1
meThing2 = 2
meThing3 = 3
[_Min] = meDefault
[_Max] = meThing3
End Enum
Public Function IsValidOption(ByVal myOption As MyEnum) As Boolean
If myOption >= MyEnum.[_Min] Then IsValidOption myOption <= MyEnum.[_Max]
End Function
In Excel-VBA you can reference cells by enclosing them in brackets, the brackets also function as an evaluate command allowing you to evaluate formula syntax:
Public Sub Example()
[A1] = "Foo"
MsgBox [VLOOKUP(A1,A1,1,0)]
End Sub
Also you can pass around raw data without using MemCopy (RtlMoveMemory) by combining LSet with User Defined Types of the same size:
Public Sub Example()
Dim b() As Byte
b = LongToByteArray(8675309)
MsgBox b(1)
End Sub
Private Function LongToByteArray(ByVal value As Long) As Byte()
Dim tl As TypedLong
Dim bl As ByteLong
tl.value = value
LSet bl = tl
LongToByteArray = bl.value
End Function
Octal & Hex Literals are actually unsigned types, these will both output -32768:
Public Sub Example()
Debug.Print &H8000
Debug.Print &O100000
End Sub
As mentioned, passing a variable inside parenthesis causes it to be passed ByVal:
Sub PredictTheOutput()
Dim i&, j&, k&
i = 10: j = i: k = i
MySub (i)
MySub j
MySub k + 20
MsgBox Join(Array(i, j, k), vbNewLine), vbQuestion, "Did You Get It Right?"
End Sub
Public Sub MySub(ByRef foo As Long)
foo = 5
End Sub
You can assign a string directly into a byte array and vice-versa:
Public Sub Example()
Dim myString As String
Dim myBytArr() As Byte
myBytArr = "I am a string."
myString = myBytArr
MsgBox myString
End Sub
"Mid" is also an operator. Using it you overwrite specific portions of strings without VBA's notoriously slow string concatenation:
Public Sub Example1()
''// This takes about 47% of time Example2 does:
Dim myString As String
myString = "I liek pie."
Mid(myString, 5, 2) = "ke"
Mid(myString, 11, 1) = "!"
MsgBox myString
End Sub
Public Sub Example2()
Dim myString As String
myString = "I liek pie."
myString = "I li" & "ke" & " pie" & "!"
MsgBox myString
End Sub
There is an important but almost always missed feature of the Mid() statement. That is where Mid() appears on the left hand side of an assignment as opposed to the Mid() function that appears in the right hand side or in an expression.
The rule is that if the if the target string is not a string literal, and this is the only reference to the target string, and the length of segment being inserted matches the length of the segment being replaced, then the string will be treated as mutable for the operation.
What does that mean? It means that if your building up a large report or a huge list of strings into a single string value, then exploiting this will make your string processing much faster.
Here is a simple class that benefits from this. It gives your VBA the same StringBuilder capability that .Net has.
' Class: StringBuilder
Option Explicit
Private Const initialLength As Long = 32
Private totalLength As Long ' Length of the buffer
Private curLength As Long ' Length of the string value within the buffer
Private buffer As String ' The buffer
Private Sub Class_Initialize()
' We set the buffer up to it's initial size and the string value ""
totalLength = initialLength
buffer = Space(totalLength)
curLength = 0
End Sub
Public Sub Append(Text As String)
Dim incLen As Long ' The length that the value will be increased by
Dim newLen As Long ' The length of the value after being appended
incLen = Len(Text)
newLen = curLength + incLen
' Will the new value fit in the remaining free space within the current buffer
If newLen <= totalLength Then
' Buffer has room so just insert the new value
Mid(buffer, curLength + 1, incLen) = Text
Else
' Buffer does not have enough room so
' first calculate the new buffer size by doubling until its big enough
' then build the new buffer
While totalLength < newLen
totalLength = totalLength + totalLength
Wend
buffer = Left(buffer, curLength) & Text & Space(totalLength - newLen)
End If
curLength = newLen
End Sub
Public Property Get Length() As Integer
Length = curLength
End Property
Public Property Get Text() As String
Text = Left(buffer, curLength)
End Property
Public Sub Clear()
totalLength = initialLength
buffer = Space(totalLength)
curLength = 0
End Sub
And here is an example on how to use it:
Dim i As Long
Dim sb As StringBuilder
Dim result As String
Set sb = New StringBuilder
For i = 1 to 100000
sb.Append CStr( i)
Next i
result = sb.Text
VBA itself seems to be a hidden feature. Folks I know who've used Office products for years have no idea it's even a part of the suite.
I've posted this on multiple questions here, but the Object Browser is my secret weapon. If I need to ninja code something real quick, but am not familiar with the dll's, Object Browser saves my life. It makes it much easier to learn the class structures than MSDN.
The Locals Window is great for debugging as well. Put a pause in your code and it will show you all the variables, their names, and their current values and types within the current namespace.
And who could forget our good friend Immediate Window? Not only is it great for Debug.Print standard output, but you can enter in commands into it as well. Need to know what VariableX is?
?VariableX
Need to know what color that cell is?
?Application.ActiveCell.Interior.Color
In fact all those windows are great tools to be productive with VBA.
It's not a feature, but a thing I have seen wrong so many times in VBA (and VB6): Parenthesis added on method calls where it will change semantics:
Sub Foo()
Dim str As String
str = "Hello"
Bar (str)
Debug.Print str 'prints "Hello" because str is evaluated and a copy is passed
Bar str 'or Call Bar(str)
Debug.Print str 'prints "Hello World"
End Sub
Sub Bar(ByRef param As String)
param = param + " World"
End Sub
Hidden Features
Although it is "Basic", you can use OOP - classes and objects
You can make API calls
Possibly the least documented features in VBA are those you can only expose by selecting "Show Hidden Members" on the VBA Object Browser. Hidden members are those functions that are in VBA, but are unsupported. You can use them, but microsoft might eliminate them at any time. None of them has any documentation provided, but you can find some on the web. Possibly the most talked about of these hidden features provides access to pointers in VBA. For a decent writeup, check out; Not So Lightweight - Shlwapi.dll
Documented, but perhaps more obscure (in excel anyways) is using ExecuteExcel4Macro to access a hidden global namespace that belongs to the entire Excel application instance as opposed to a specific workbook.
You can implement interfaces with the Implements keyword.
Dictionaries. VBA is practically worthless without them!
Reference the Microsoft Scripting Runtime, use Scripting.Dictionary for any sufficiently complicated task, and live happily ever after.
The Scripting Runtime also gives you the FileSystemObject, which also comes highly recommended.
Start here, then dig around a bit...
http://msdn.microsoft.com/en-us/library/aa164509%28office.10%29.aspx
Typing VBA. will bring up an intellisense listing of all the built-in functions and constants.
With a little work, you can iterate over custom collections like this:
' Write some text in Word first.'
Sub test()
Dim c As New clsMyCollection
c.AddItems ActiveDocument.Characters(1), _
ActiveDocument.Characters(2), _
ActiveDocument.Characters(3), _
ActiveDocument.Characters(4)
Dim el As Range
For Each el In c
Debug.Print el.Text
Next
Set c = Nothing
End Sub
Your custom collection code (in a class called clsMyCollection):
Option Explicit
Dim m_myCollection As Collection
Public Property Get NewEnum() As IUnknown
' This property allows you to enumerate
' this collection with the For...Each syntax
' Put the following line in the exported module
' file (.cls)!'
'Attribute NewEnum.VB_UserMemId = -4
Set NewEnum = m_myCollection.[_NewEnum]
End Property
Public Sub AddItems(ParamArray items() As Variant)
Dim i As Variant
On Error Resume Next
For Each i In items
m_myCollection.Add i
Next
On Error GoTo 0
End Sub
Private Sub Class_Initialize()
Set m_myCollection = New Collection
End Sub
Save 4 whole keystrokes by typing debug.? xxx instead of debug.print xxx.
Crash it by adding: enum foo: me=0: end enum to the top of a module containing any other code.
Support for localized versions, which (at least in the previous century) supported expressions using localized values. Like Pravda for True and Fałszywy (not too sure, but at least it did have the funny L) for False in Polish... Actually the English version would be able to read macros in any language, and convert on the fly. Other localized versions would not handle that though.
FAIL.
The VBE (Visual Basic Extensibility) object model is a lesser known and/or under-utilized feature. It lets you write VBA code to manipulate VBA code, modules and projects. I once wrote an Excel project that would assemble other Excel projects from a group of module files.
The object model also works from VBScript and HTAs. I wrote an HTA at one time to help me keep track of a large number of Word, Excel and Access projects. Many of the projects would use common code modules, and it was easy for modules to "grow" in one system and then need to be migrated to other systems. My HTA would allow me to export all modules in a project, compare them to versions in a common folder and merge updated routines (using BeyondCompare), then reimport the updated modules.
The VBE object model works slightly differently between Word, Excel and Access, and unfortunately doesn't work with Outlook at all, but still provides a great capability for managing code.
IsDate("13.50") returns True but IsDate("12.25.2010") returns False
This is because IsDate could be more precisely named IsDateTime. And because the period (.) is treated as a time separator and not a date separator. See here for a full explanation.
VBA supports bitwise operators for comparing the binary digits (bits) of two values. For example, the expression 4 And 7 evaluates the bit values of 4 (0100) and 7 (0111) and returns 4 (the bit that is on in both numbers.) Similarly the expression 4 Or 8 evaluates the bit values in 4 (0100) and 8 (1000) and returns 12 (1100), i.e. the bits where either one is true.
Unfortunately, the bitwise operators have the same names at the logical comparison operators: And, Eqv, Imp, Not, Or, and Xor. This can lead to ambiguities, and even contradictory results.
As an example, open the Immediate Window (Ctrl+G) and enter:
? (2 And 4)
This returns zero, since there are no bits in common between 2 (0010) and 4 (0100).
Deftype Statements
This feature exists presumably for backwards-compatibility. Or to write hopelessly obfuscated spaghetti code. Your pick.