I have some XML files that contain decimal numbers stored with '.' as decimal separator. I created Excel workbook that is used for editing data in that XML. MSXML is used to read the data from XML that are then displayed in worksheets. Once data is being stored I modify DOM structure with new values read from tables and I save XML.
The problem is that when I convert decimal numbers back to strings, to save them as values in XML, decimal separator that is specified by user in Windows regional settings is used.
I need a way to force storing decimal numbers with '.' as decimal separator in all cases. Is this possible?
How do you modify the DOM structure with the new values--programmatically? How?
Is it possible for you to use Format(Cell.value, "#0.0")? Also try Format(CDbl(Cell.Value), "#0.0"). If that doesn't work, can you just throw in a Replace(Value, ",", ".")? This should be okay if you're not using thousands separators.
Use the Str function. It recognizes only the period . as a valid decimal separator.
Dim d as Double
Dim s as String
d = 3.25
s = Str(d)
' s is now " 3.25"
Note that Str reserves a leading space for the sign of the number. If you don't want a leading space on your positive numbers, just Trim the result:
s = Trim(Str(d))
' s is now "3.25"
#Jean-Francois since we're talking about i18n you should use CStr ( for international applications ) function instead of Str ( where . dot is separator for floating number , not much use in French where comma , is a separator ex : 7,54)
Sub myFunction()
Dim num As Double
Dim str As String
num = 3.251
str = CStr(num) ' convert a number into a String
num = Val(str) ' convert a String into an integer
num = Trim(str) ' convert a String into a Variant
str = Trim(num) ' notice this works the other way around
End Sub
Related
I have the following strings from which I need to extract 6 digit numbers. Since these strings are generated by another software, they occur interchangeably and I cannot control it. Is there any one method that would extract both 6-digit numbers from each of these strings?
Branch '100235 to 100236 Ckt 1' specified in table 'East Contingency' for record with primary key = 21733 was not found in branch or transformer data.
Loadflow branch ID '256574_701027_1' defined in supplemental branch table was not found in branch or transformer input.
Transmission element from bus number 135415 to bus number 157062 circuit ID = 1 defined for corridor 'IESO-NYISO' was not found in input data
I don't know VBA, but I can learn it if it means I can get the 6 digit numbers using a single method.
thanks
I have been using LEFT(), RIGHT() & MID() previously, but it means manually applying the appropriate formula for individual string.
If you have Microsoft 365, you can use this formula:
=LET(arr,TEXTSPLIT(SUBSTITUTE(SUBSTITUTE(A1,"'"," "),"_"," ")," "),
FILTER(arr,ISNUMBER(-arr)*(LEN(arr)=6)))
Thanks to #TomSharpe for this shorter version, using an array constant within TEXTSPLIT to add on possible delimiters.
=LET(arr,TEXTSPLIT(A1,{"_"," ",","," ","'"}),FILTER(arr,(LEN(arr)=6)*ISNUMBER(-arr)))
Data
Output
An alternative is:
=LET(ζ,MID(A1,SEQUENCE(,LEN(A1)-5),6),ξ,MID(ζ,SEQUENCE(6),1),FILTER(ζ,MMULT(SEQUENCE(,6,,0),1-ISERR(0+ξ))=6))
A couple more suggestions (if you need them):
(1) Replacing all non-digit characters with a space then splitting the resulting string:
=LET(numbers,TEXTSPLIT(TRIM(REDUCE("",MID(A1,SEQUENCE(1,LEN(A1)),1),LAMBDA(a,c,IF(is.digit(c),a&c,a&" "))))," "),FILTER(numbers,LEN(numbers)=6))
Here I've defined a function is.digit as
=LAMBDA(c, IF(c = "", FALSE, AND(CODE(c) > 47, CODE(c) < 58)))
(tl;dr I quite like doing it this way because it hides the implementation details of is.digit and creates a rudimentary form of encapsulation)
(2) A UDF - based on the example here and called as
=RegexTest(A1)
Option Explicit
Function RegexTest(s As String) As Double()
Dim regexOne As Object
Dim theNumbers As Object
Dim Number As Object
Dim result() As Double
Dim i As Integer
Set regexOne = New RegExp
' Not sure how you would extract numbers of length 6 only, so extract all numbers...
regexOne.Pattern = "\d+"
regexOne.Global = True
regexOne.IgnoreCase = True
Set theNumbers = regexOne.Execute(s)
i = 1
For Each Number In theNumbers
'...Check the length of each number here
If Len(Number) = 6 Then
ReDim Preserve result(1 To i)
result(i) = CDbl(Number)
i = i + 1
End If
Next
RegexTest = result
End Function
Note - if you wanted to preserve leading zeroes you would need to omit the Cdbl() and return the numbers as strings. Returns an error if no 6-digit numbers are found.
I have a String in VBA with this text: < History Version="1.10" Client="TestClient001" >
I want to get this TestClient001 or anything that's inside Client="xxxx"
I made this code but it's not working
Client = MID(text,FIND("Client=""",text)+1,FIND("""",text)-FIND("Client=""",text)-1)
Is there a way to specifically get the text inside Client="xxxx"?
There's no such function as Find in VBA - that's a worksheet function. The VBA equivalent is InStr, but I don't think you need to use it here.
The best tool for extracting one string from another in VBA is often Split. It takes one string and splits it into an array based on a delimiting string. The best part is that the delimiter doesn't have to be a single character - you can make it an entire string. In this case, we'd probably do well with two nested Split functions.
Client = Split(Split(text,"Client=""")(1),Chr(34))(0)
The inner Split breaks your text string where it finds "Client="". The (1) returns array element 1. Then the outer Split breaks that returned text where it finds a " character, and returns array element 0 as the final result.
For better maintainability, you may want to use constants for your delimiters as well.
Sub EnclosedTextTest()
Const csFlag1 As String = "Client="""
Const csFlag2 As String = """"
Const csSource As String = "< History Version=""1.10"" Client=""TestClient001"" >"
Dim strClient As String
strClient = Split(Split(csSource, csFlag1)(1), csFlag2)(0)
Debug.Print strClient
End Sub
However, if the Split method doesn't work for you, we can use a method similar to the one you were using, with InStr. There are a couple of options here as well.
InStr will return the position in a string that it finds a matching value. Like Split, it can be given an entire string as its delimiter; however, if you use more than one character you need to account for the fact that it will return where it finds the start of that string.
InStr(1,text,"Client=""")
will return 26, the start of the string "Client="" in the text. This is one of the places where it's helpful to have your delimiter stored in a constant.
intStart = InStr(1,text,csFlag1)+len(csFlag1)
This will return the location it finds the start of the delimiter, plus the length of the delimiter, which positions you at the beginning of the text.
If you store this position in a variable, it makes the next part easier as well. You can use that position to run a second InStr and find the next occurrence of the " character.
intEnd = InStr(intStart,text,csFlag2)
With those values, you can perform your mid. You code overall will look something like this:
Sub InstrTextTest()
Const csFlag1 As String = "Client="""
Const csFlag2 As String = """"
Const csSource As String = "< History Version=""1.10"" Client=""TestClient001"" >"
Dim strClient As String
Dim intPos(0 To 1) As Integer
intPos(0) = InStr(1, csSource, csFlag1) + Len(csFlag1)
intPos(1) = InStr(intPos(0), csSource, csFlag2)
strClient = Mid(csSource, intPos(0), intPos(1) - intPos(0))
Debug.Print strClient
End Sub
This will work, but I prefer the Split method for ease of reading and reuse.
You can make use of Split function to split at character = then with last element of the resulting array remove character quotes and > with help of replace function and you will get the required output.
In the end I got it thanks to the idea given by #alok and #Bigben
Dim cl() As String
Dim ClientCode As String
If (InStr(1, temp, "Client=", vbTextCompare) > 0) Then
cl = Split(temp, "=")
ClientCode = cl(UBound(cl))
ClientCode = Replace(ClientCode, """", "")
ClientCode = Replace(ClientCode, ">", "")
It's XML, so you could do this:
Dim sXML As String
sXML = "<History Version=""1.10"" Client=""TestClient001"">"
With CreateObject("MSXML.Domdocument")
.LoadXML Replace(sXML, ">", "/>") 'close the element
Debug.Print .FirstChild.Attributes.getnameditem("Client").Value
End With
Dim textstring as String,warray() as String
textstring=Range("A3").Value
warray=split(textstring,"")
If isDate(warray(0))=True Then
Range("A4").Value=warray(1)
End If
This code displays only one word, no other words are displayed.
Typically, the vba Split function is used to split a string into a multiple element array on a delimiter. If a space was used as the delimiter, a sentence or paragraph would be split into individual words.
You are providing a zero-length string (e.g. "" ) as the delimiter. From the msdn docs:
... If delimiter is a zero-length string, a single-element array containing the entire expression string is returned.
So if you want to actually split something into multiple pieces, you cannot use a zero-length string as the delimiter.
If you used a space as the delimiter and provided no other parameters, then your example would be split into four elements of the array. I suspect you only want to split off the time and leave the show title together. The limit parameter is used for this. A limit of 2 will split your original string into a time and a show title.
Option Explicit
Sub Macro3()
Dim textstring As String, warray() As String
textstring = Range("A3").Value
warray = Split(expression:=textstring, delimiter:=" ", limit:=2)
If IsDate(warray(0)) Then
Range("A4").Value = warray(0)
Range("A5").Value = warray(1)
End If
End Sub
I have a function which returns value as string.
Function Trimcell(cellvalue As varnant) As String
Trimcell = Replace(CStr(cellvalue), " ", "")
End Function
I want to change the data type string to long . Any help.
Change your function to this:
Function Trimcell(cellvalue As varnant) As Long
Trimcell = Val(Replace(CStr(cellvalue), " ", ""))
End Function
You have a spelling error - varnant instead of Variant.
A better option than using Replace is to use Val which removes blanks, tabs, and linefeed characters from a string and returns a Double. It also stops reading the string at the first non-numeric character apart from period (.) which it recognises as the decimal separator.
As you have declared cellvalue As Variant you shouldn't need CStr either.
Function Trimcell2(cellvalue As Variant) As Long
Trimcell = Val(cellvalue)
End Function
I have Excel 2007. I am trying to find the largest number in a cell that contains something like the following:
[[ E:\DATA\SQL\SY0\ , 19198 ],[ E:\ , 18872 ],[ E:\DATA\SQL\ST0\ , 26211 ],[ E:\DATA\SQL\ST1\ , 26211 ],[ E:\DATA\SQL\SD0\ , 9861 ],[ E:\DATA\SQL\SD1\ , 11220 ],[ E:\DATA\SQL\SL0\ , 3377 ],[ E:\DATA\SQL\SL1\ , 1707 ],[ E:\DATA\SQL_Support\SS0\ , 14375 ],[ E:\DATA\SQL_Support\SS1\ , 30711 ]]
I am not a coder but I can get by with some basic instructions. If there is a formula that can do this, great! If the best way to do this is some sort of backend code, just let me know. Thank you for your time.
I do have the following formula that almost gets me there:
=SUMPRODUCT(MID(0&A2,LARGE(INDEX(ISNUMBER(--MID(A2,ROW(INDIRECT("1:"&LEN($A$2))),1))*ROW(INDIRECT("1:"&LEN($A$2))),0),ROW(INDIRECT("1:"&LEN($A$2))))+1,1)*10^ROW(INDIRECT("1:"&LEN($A$2)))/10)
With a cell that contains a string like above, it will work. However, with a string that contains something like:
[[ E:\DATA\SQL\SY0\ , 19198.934678 ],[ E:\ , 18872.2567 ]]
I would end up with the value of 19198934678 as the largest value.
You can use this UDF:
Function MaxInString(rng As String) As Double
Dim splt() As String
Dim i&
splt = Split(rng)
For i = LBound(splt) To UBound(splt)
If IsNumeric(splt(i)) Then
If splt(i) > MaxInString Then
MaxInString = splt(i)
End If
End If
Next i
End Function
Put this in a module attached to the workbook. NOT in the worksheet or ThisWorkbook code.
Then you can call it like any other formula:
=MaxInString(A1)
If there is always a space before and after, you can use this formula. The formula is an array formula and must be confirmed by holding down ctrl + shift while hitting enter
With your string in A1:
=MAX(IFERROR(--TRIM(MID(SUBSTITUTE(A1," ",REPT(" ",99)),IF(seq=1,1,(seq-1)*99),99)),0))
seq is a defined name that refers to:
=ROW(INDEX(Sheet1!$1:$65536,1,1):INDEX(Sheet1!$1:$65536,255,1))
If a VBA UDF is preferable, I suggest the following. The Regex will match anything that might be a number. The number is expected to be in the format of iiii.dddd The integer part and the decimal point are both optional.
Option Explicit
Function LargestNumberFromString(S As String) As Double
Dim RE As Object, MC As Object, M As Object
Dim D As Double
Set RE = CreateObject("vbscript.regexp")
With RE
.Global = True
.Pattern = "\b[0-9]*\.?[0-9]+\b"
If RE.test(S) = True Then
For Each M In MC
D = IIf(D > CDbl(M), D, CDbl(M))
Next M
End If
End With
End Function
This is going to be a rather complex answer for someone with no programming background, so be prepared to spend a lot of time covering and researching this topic if you really wish to achieve a function that finds the largest number in a string in excel.
The solution, requires the use of VBA and Regular expressions
VBA is used in excel when there is a need for more complex functionality that just can't be achieved with the use of built in spreadsheet functions.
Regular expressions are a language used to tell programs how to extract useful information from texts, in this case we can extract all the numbers in your text. with the following regular expression.
(\d+.?\d*)/g
Which roughly means: Match one or more digits with an optional period and subsequent optional digits.
The program that will interpret this will do the following: Look for digits, if you see one, then that's a match, grab all contiguous digits and add them to the match. Once you find a character that is not a digit, start looking for new matches. if at any point you find a dot, add it to the match, but just once, and keep on looking for digits. Rinse and repeat until the end of the text.
You can test it here. In this case, the regex matches 19 numbers.
http://www.regextester.com/
Once you have a collection with the 19 matches (See link to regular expressions), all you would need to do is to loop over each of the matches to find out which of the numbers is the highest:
for each number in matches
if number > highestNumber then
highestNumber = number
end if
next
And highestNumber will be the the result! In order to have this code run in a simple custom function, you can follow this microsoft tutorial ( https://support.office.com/en-us/article/Create-Custom-Functions-in-Excel-2007-2f06c10b-3622-40d6-a1b2-b6748ae8231f?ui=en-US&rs=en-US&ad=US&fromAR=1 )
Where c is your string to find max from
Dim qwe() As String
qwe = Split(c, ", ")
maxed = 0
For x = LBound(qwe) To UBound(qwe)
qwe(x) = Left(qwe(x), InStr(1, qwe(x), " ", vbBinaryCompare))
On Error Resume Next
If CLng(qwe(x)) > maxed Then maxed = CLng(qwe(x))
Next x
MsgBox maxed
The error line is there to ignore when qwe(x) cannot be converted to a LONG number.
I must say this is very specific to your string format, for a more comprehensive doodad you'd want to have the split delimiter as a variable and possibly use the "IsNumeric" function to scan the entire string.