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
Related
Public Function SameStuff(s1 As String, s2 As String) As Boolean
Dim bad As Boolean
SameStuff = False
ary1 = Split(Replace(s1, " ", ""), ",")
ary2 = Split(Replace(s2, " ", ""), ",")
Length1 = UBound (ary1)
Length2 = UBound(ary2)
k=1
If Length1<= Length2 and Length1<>0 then
for i=0 to Length1-1
If ary1(i) = ary2(i) then
ary3(k,i) = ary1(i)
End If
Next i
k=k+1
else
Exit function
End If
End Function
Here I take value from Range("A1") - (has 3 words) and value of Range("A2") - (has 4 words). Split them both by finding space between words and store them in arrays. If length of one array is 3 and other is 4, 3 words from both the arrays will be compared. If 3 words are found to be same then Range("B1") and Range("B2") must both have the 3 word name i.e Range("A1").Value. I think this logic will work fine to find similar names like "ABC DEF HIJ " in A1 and "ABC DEF HIJ Limited" in A2.
I am not able to put it in code.
Word length will not remain the same i.e 3,4 .
Use a dictionary would be an easy alternative, you can use the .exists method to do this for you, you have to transfer the array (result of split() ) to a dictionary tho, but that's a loop, not too tricky. Or, you could leave one of the inputas as a string and split only 1, and use if strStringLeftAlone like "* " & strSection(x) & " *" or use instr, with the same idea as the search " " & strSection(x) & " " or find
This should work regardless how long the arrays are, i.e. no matter how many words (and spaces) there are in each of the strings to be compared. Notice I removed the k variable as it didn't seem to serve any purpose in the code. This solution does presuppose, however, that ONLY the LAST word in the two strings is different.
Public Function SameStuff(s1 As String, s2 As String) As Boolean
Dim sameBool As Boolean
Dim i As Long, Length1 As Long, Length2 As Long
Dim tempArr1 as String, tempArr2 as String
Dim ary1 as Variant, ary2 as Variant
ary1 = Split(Replace(s1, " ", ""), ",")
ary2 = Split(Replace(s2, " ", ""), ",")
Length1 = UBound (ary1)
Length2 = UBound(ary2)
If Length1 <= Length2 and Length1 > 0 then
For i=0 to Length1-1
tempArr1 = tempArr1 & ary1(i)
tempArr2 = tempArr2 & ary2(i)
Next i
If tempArr1 = tempArr2 then sameBool = True
End If
SameStuff = sameBool
End Function
Edit
Added some variable declarations to the code that I had forgotten, otherwise the code would not work with Option Explicit at the top of the module.
Let's say I have the following string within a cell:
E. Stark, T. Lannister, A. Martell, P Baelish, B. Dondarrion, and J. Mormont. Increased levels of nudity across Westeros contributes to its sporadic seasonal climate. Nat. Proc. Aca. Sci. (2011) 3: 142-149.
And I want to extract only the title from this. The approach I am considering is to write a script that says "Pull text from this string, but only if it is more than 50 characters long." This way it only returns the title, and not stuff like " Stark, T" and " Martell, P". The code I have so far is:
Sub TitleTest()
Dim txt As String
Dim Output As String
Dim i As Integer
Dim rng As Range
Dim j As Integer
Dim k As Integer
j = 5
Set rng = Range("A" & j) 'text is in cell A5
txt = rng.Value 'txt is string
i = 1
While j <= 10 'there are five references between A5 and A10
k = InStr(i, txt, ".") - InStr(i, txt, ". ") + 1 'k is supposed to be the length of the string returned, but I can't differenciate one "." from the other.
Output = Mid(txt, InStr(i, txt, "."), k)
If Len(Output) < 100 Then
i = i + 1
ElseIf Len(Output) > 10 Then
Output = Mid(txt, InStr(i, txt, "."), InStr(i, txt, ". "))
Range("B5") = Output
j = j + 1
End If
Wend
End Sub
Of course, this would work well if it wasn't two "." I was trying to full information from. Is there a way to write the InStr function in such a way that it won't find the same character twice? Am I going about this in the wrong way?
Thanks in advance,
EDIT: Another approach that might work (if possible), is if I could have one character be " any lower case letter." and ".". Would even this be possible? I can't find any example of how this could be achieved...
Here you go, it works exactly as you wish. Judging from your code I am sure that you can adapt it for your needs quite quickly:
Option Explicit
Sub ExtractTextSub()
Debug.Print ExtractText("E. Stark, T. Lannister, A. Martell, P Baelish, B. Dondarrion, and J. Mormont. Increased levels of nudity across Westeros contributes to its sporadic seasonal climate. Nat. Proc. Aca. Sci. (2011) 3: 142-149.")
End Sub
Public Function ExtractText(str_text As String) As String
Dim arr As Variant
Dim l_counter As Long
arr = Split(str_text, ".")
For l_counter = LBound(arr) To UBound(arr)
If Len(arr(l_counter)) > 50 Then
ExtractText = arr(l_counter)
End If
Next l_counter
End Function
Edit: 5 votes in no time made me improve my code a bit :) This would return the longest string, without thinking of the 50 chars. Furthermore, on Error handlaer and a constant for the point. Plus adding a point to the end of the extract.
Option Explicit
Public Const STR_POINT = "."
Sub ExtractTextSub()
Debug.Print ExtractText("E. Stark, T. Lannister, A. Martell, P Baelish, B. Dondarrion, and J. Mormont. Increased levels of nudity across Westeros contributes to its sporadic seasonal climate. Nat. Proc. Aca. Sci. (2011) 3: 142-149.")
End Sub
Public Function ExtractText(str_text As String) As String
On Error GoTo ExtractText_Error
Dim arr As Variant
Dim l_counter As Long
Dim str_longest As String
arr = Split(str_text, STR_POINT)
For l_counter = LBound(arr) To UBound(arr)
If Len(arr(l_counter)) > Len(ExtractText) Then
ExtractText = arr(l_counter)
End If
Next l_counter
ExtractText = ExtractText & STR_POINT
On Error GoTo 0
Exit Function
ExtractText_Error:
MsgBox "Error " & Err.Number & Err.Description
End Function
I want to count all Commas "," that occur only in selected text after that I will use Count as Integer to run the loop
My question is how do i Count , as following Image shows:
I Don't know how to use split and ubound. what is wrong with following code?
Sub CountComma()
Dim x As String, count As Integer, myRange As Range
Set myRange = ActiveDocument.Range(Selection.Range.Start, Selection.Range.End)
x = Split(myRange, ",")
count = UBound(x)
Debug.Print count
End Sub
A simple split will work.
x = Split("XXX,XXX,XXX,XXX,XX,XX", ",")
Count = UBound(x)
Debug.Print Count
B/c the array starts at zero you can take to Ubound number as is.
EDIT:
To use a range .
x = Split(Range("A1").Value, ",")
To break down the code.
Split("A string value","Delimiter to split the string by")
And if you want a single line of code than,
x = UBound(Split(myRange, ","))
your code is wrong in the initial declaration statement of x variable as of string type , since in the subsequent statement
with x = Split(myRange, ",")
you'd want x hold the return value of Split() function which is an array (see here), thus of Variant type
so you have to use
Dim x As Variant
But you can simplify your code as follows
Option Explicit
Sub CountComma()
Dim count As Integer
count = UBound(Split(Selection, ","))
Debug.Print count
End Sub
since:
you don't need any Range type variable to store Selection object into, being Selection the selected range already (see here)
you don't need the x Variant variable neither, feeding UBound()function (which expects an array as its first argument) directly with the Split() function which, as we saw above, returns just an array!
Finally I'd give out an alternative method of counting commas in a range
Sub CountComma()
Dim countAs Integer
count = Len(Selection) - Len(Replace(Selection, ",", ""))
Debug.Print count
End Sub
Thanks to KyloRen and Cindy Meister, Now I can use split and Ubound for Counting , in selection.text.
Following is working Code:
Sub Count_Words()
Dim WrdArray() As String, myRange As String
myRange = ActiveDocument.Range(Selection.Range.Start, Selection.Range.End)
WrdArray() = Split(myRange, ", ")
MsgBox ("Total , in the string : " & UBound(WrdArray()))
End Sub
What is the VBA string interpolation syntax? Does it exist?
I would to to use Excel VBA to format a string.
I have a variable foo that I want to put in a string for a range.
Dim row as Long
row = 1
myString = "$row:$row"
I would like the $row in the string to be interpolated as "1"
You could also build a custom Format function.
Public Function Format(ParamArray arr() As Variant) As String
Dim i As Long
Dim temp As String
temp = CStr(arr(0))
For i = 1 To UBound(arr)
temp = Replace(temp, "{" & i - 1 & "}", CStr(arr(i)))
Next
Format = temp
End Function
The usage is similar to C# except that you can't directly reference variables in the string. E.g. Format("This will {not} work") but Format("This {0} work", "will").
Public Sub Test()
Dim s As String
s = "Hello"
Debug.Print Format("{0}, {1}!", s, "World")
End Sub
Prints out Hello, World! to the Immediate Window.
This works well enough, I believe.
Dim row as Long
Dim s as String
row = 1
s = "$" & row & ":$" & row
Unless you want something similar to Python's or C#'s {} notation, this is the standard way of doing it.
Using Key\Value Pairs
Another alternative to mimic String interpolation is to pass in key\value pairs as a ParamArray and replace the keys accordingly.
One note is that an error should be raised if there are not an even number of elements.
' Returns a string that replaced special keys with its associated pair value.
Public Function Inject(ByVal source As String, ParamArray keyValuePairs() As Variant) As String
If (UBound(keyValuePairs) - LBound(keyValuePairs) + 1) Mod 2 <> 0 Then
Err.Raise 5, "Inject", "Invalid parameters: expecting key/value pairs, but received an odd number of arguments."
End If
Inject = source
' Replace {key} with the pairing value.
Dim index As Long
For index = LBound(keyValuePairs) To UBound(keyValuePairs) Step 2
Inject = Replace(Inject, "{" & keyValuePairs(index) & "}", keyValuePairs(index + 1), , , vbTextCompare)
Next index
End Function
Simple Example
Here is a simple example that shows how to implement it.
Private Sub testingInject()
Const name As String = "Robert"
Const age As String = 31
Debug.Print Inject("Hello, {name}! You are {age} years old!", "name", name, "age", age)
'~> Hello, Robert! You are 31 years old!
End Sub
Although this may add a few extra strings, in my opinion, this makes it much easier to read long strings.
See the same simple example using concatenation:
Debug.Print "Hello, " & name & "! You are " & age & " years old!"
Using Scripting.Dicitionary
Really, a Scripting.Dictionary would be perfect for this since they are nothing but key/value pairs. It would be a simple adjustment to my code above, just take in a Dictionary as the parameter and make sure the keys match.
Public Function Inject(ByVal source As String, ByVal data As Scripting.Dictionary) As String
Inject = source
Dim key As Variant
For Each key In data.Keys
Inject = Replace(Inject, "{" & key & "}", data(key))
Next key
End Function
Dictionary example
And the example of using it for dictionaries:
Private Sub testingInject()
Dim person As New Scripting.Dictionary
person("name") = "Robert"
person("age") = 31
Debug.Print Inject("Hello, {name}! You are {age} years old!", person)
'~> Hello, Robert! You are 31 years old!
End Sub
Additional Considerations
Collections sound like they would be nice as well, but there is no way of accessing the keys. It would probably get messier that way.
If using the Dictionary method you might create a simple factory function for easily creating Dictionaries. You can find an example of that on my Github Library Page.
To mimic function overloading to give you all the different ways you could create a main Inject function and run a select statement within that.
Here is all the code needed to do that if need be:
Public Function Inject(ByVal source As String, ParamArray data() As Variant) As String
Dim firstElement As Variant
assign firstElement, data(LBound(data))
Inject = InjectCharacters(source)
Select Case True
Case TypeName(firstElement) = "Dictionary"
Inject = InjectDictionary(Inject, firstElement)
Case InStr(source, "{0}") > 0
Inject = injectIndexes(Inject, CVar(data))
Case (UBound(data) - LBound(data) + 1) Mod 2 = 0
Inject = InjectKeyValuePairs(Inject, CVar(data))
Case Else
Err.Raise 5, "Inject", "Invalid parameters: expecting key/value pairs or Dictionary or an {0} element."
End Select
End Function
Private Function injectIndexes(ByVal source As String, ByVal data As Variant)
injectIndexes = source
Dim index As Long
For index = LBound(data) To UBound(data)
injectIndexes = Replace(injectIndexes, "{" & index & "}", data(index))
Next index
End Function
Private Function InjectKeyValuePairs(ByVal source As String, ByVal keyValuePairs As Variant)
InjectKeyValuePairs = source
Dim index As Long
For index = LBound(keyValuePairs) To UBound(keyValuePairs) Step 2
InjectKeyValuePairs = Replace(InjectKeyValuePairs, "{" & keyValuePairs(index) & "}", keyValuePairs(index + 1))
Next index
End Function
Private Function InjectDictionary(ByVal source As String, ByVal data As Scripting.Dictionary) As String
InjectDictionary = source
Dim key As Variant
For Each key In data.Keys
InjectDictionary = Replace(InjectDictionary, "{" & key & "}", data(key))
Next key
End Function
' QUICK TOOL TO EITHER SET OR LET DEPENDING ON IF ELEMENT IS AN OBJECT
Private Function assign(ByRef variable As Variant, ByVal value As Variant)
If IsObject(value) Then
Set variable = value
Else
Let variable = value
End If
End Function
End Function
Private Function InjectCharacters(ByVal source As String) As String
InjectCharacters = source
Dim keyValuePairs As Variant
keyValuePairs = Array("n", vbNewLine, "t", vbTab, "r", vbCr, "f", vbLf)
If (UBound(keyValuePairs) - LBound(keyValuePairs) + 1) Mod 2 <> 0 Then
Err.Raise 5, "Inject", "Invalid variable: expecting key/value pairs, but received an odd number of arguments."
End If
Dim RegEx As Object
Set RegEx = CreateObject("VBScript.RegExp")
RegEx.Global = True
' Replace is ran twice since it is possible for back to back patterns.
Dim index As Long
For index = LBound(keyValuePairs) To UBound(keyValuePairs) Step 2
RegEx.Pattern = "((?:^|[^\\])(?:\\{2})*)(?:\\" & keyValuePairs(index) & ")+"
InjectCharacters = RegEx.Replace(InjectCharacters, "$1" & keyValuePairs(index + 1))
InjectCharacters = RegEx.Replace(InjectCharacters, "$1" & keyValuePairs(index + 1))
Next index
End Function
I have a library function SPrintF() which should do what you need.
It replaces occurrences of %s in the supplied string with an extensible number of parameters, using VBA's ParamArray() feature.
Usage:
SPrintF("%s:%s", 1, 1) => "1:1"
SPrintF("Property %s added at %s on %s", "88 High St, Clapham", Time, Date) => ""Property 88 High St, Clapham added at 11:30:27 on 25/07/2019"
Function SprintF(strInput As String, ParamArray varSubstitutions() As Variant) As String
'Formatted string print: replaces all occurrences of %s in input with substitutions
Dim i As Long
Dim s As String
s = strInput
For i = 0 To UBound(varSubstitutions)
s = Replace(s, "%s", varSubstitutions(i), , 1)
Next
SprintF = s
End Function
Just to add as a footnote, the idea for this was inspired by the C language printf function.
I use a similar code to that of #natancodes except that I use regex to replace the occurances and allow the user to specifiy description for the placeholders. This is useful when you have a big table (like in Access) with many strings or translations so that you still know what each number means.
Function Format(ByVal Source As String, ParamArray Replacements() As Variant) As String
Dim Replacement As Variant
Dim i As Long
For i = 0 To UBound(Replacements)
Dim rx As New RegExp
With rx
.Pattern = "{" & i & "(?::(.+?))?}"
.IgnoreCase = True
.Global = True
End With
Select Case VarType(Replacements(i))
Case vbObject
If Replacements(i) Is Nothing Then
Dim Matches As MatchCollection
Set Matches = rx.Execute(Source)
If Matches.Count = 1 Then
Dim Items As SubMatches: Set Items = Matches(0).SubMatches
Dim Default As String: Default = Items(0)
Source = rx.Replace(Source, Default)
End If
End If
Case vbString
Source = rx.Replace(Source, CStr(Replacements(i)))
End Select
Next
Format = Source
End Function
Sub TestFormat()
Debug.Print Format("{0:Hi}, {1:space}!", Nothing, "World")
End Sub
I have a macro I am trying to turn into a VBA Function or Query for adding leading zeros to a field.
For my circumstances, their needs to be 4 numeric digits plus any alphabetic characters that follow so a simple format query doesn't do the trick.
The macro I have uses Evaluate and =Match but I am unsure how this could be achieved in Access.
Sub Change_Number_Format_In_String()
Dim iFirstLetterPosition As Integer
Dim sTemp As String
For Each c In Range("A2:A100")
If Len(c) > 0 Then
iFirstLetterPosition = Evaluate("=MATCH(TRUE,NOT(ISNUMBER(1*MID(" & c.Address & ",ROW($1:$20),1))),0)")
sTemp = Left(c, iFirstLetterPosition - 1) 'get the leading numbers
sTemp = Format(sTemp, "0000") 'format the numbers
sTemp = sTemp & Mid(c, iFirstLetterPosition, Len(c)) 'concatenate the remainder of the string
c.NumberFormat = "#"
c.Value = sTemp
End If
Next
End Sub
In my database the field in need of formatting is called PIDNUMBER
EDIT:
To expand on why FORMAT doesnt work in my situation. Some PIDNUMBERS have an alpha character after the number that should not be counted when determining how many zeroes to add.
In example:
12 should become 0012
12A should become 0012A
When using format, it counts the letters as part of the string, so 12A would become 012A instead of 0012A as intended.
You could try:
Public Function customFormat(ByRef sString As String) As String
customFormat = Right("0000" & sString, 4 + Len(sString) - Len(CStr(Val(sString))))
End Function
Try utilize this function, if you only want this to be available in VBA, put Private in front of the Function:
Function ZeroPadFront(oIn As Variant) As String
Dim zeros As Long, sOut As String
sOut = CStr(oIn)
zeros = 4 - Len(sOut)
If zeros < 0 Then zeros = 0
ZeroPadFront = String(zeros, "0") & sOut
End Function
The Val() function converts a string to a number, and strips off any trailing non-numeric characters. We can use it to figure out how many digits the numeric portion has:
Function PadAlpha$(s$)
Dim NumDigs As Long
NumDigs = Len(CStr(Val(s)))
If NumDigs < 4 Then
PadAlpha = String$(4 - NumDigs, "0") & s
Else
PadAlpha = s
End If
End Function
? padalpha("12")
> 0012
? padalpha("12a")
> 0012a
Bill,
See if this will work. It seems like a function would better suit you.
Function NewPIDNumber(varPIDNumber As Variant) As String
Dim lngLoop As Long
Dim strChar As String
For lngLoop = 1 to Len(varPIDNumber)
strChar = Mid(varPIDNumber, lngLoop, 1)
If IsNumeric(strChar) Then
NewPIDNumber = NewPIDNumber & strChar
Else
Exit For
End If
Next lngLoop
If Len(NewPIDNumber) > 4 Then
MsgBox "Bad Data Maaaaan...." & Chr(13) & Chr(13) & "The record = " & varPIDNumber
Exit Function
End If
Do Until Len(NewPIDNumber) = 4
NewPIDNumber = "0" & NewPIDNumber
Loop
End Function
Data Result
012a 0012
12a 0012
12 0012
85 0085
85adfe 0085
1002a 1002
1002 1002