I am trying to pick a substring from a string.
Example: I want to pick the table no from the below string starting from "T".
Changes in table T682 (SAP TABLE)
The position of the start of the table no is same ie 17th position, but the characters in a table name are not constant. So I basically want a substring that picks from the 17th position till it finds a space. I am using the below code but it doesn't seem to work.
Position = InStr(1, A, " ")
TableNo = Mid(A, 17, Position)
Here is what you need to do:
Dim A As String
Dim StartPosition As Integer
Dim EndPosition As Integer
A = "Changes in table T682 (SAP TABLE)"
StartPosition = InStr(1, A, "T")
EndPosition = InStr(StartPosition, A, " ")
tableno = Mid(A, StartPosition, EndPosition - StartPosition)
There are two mistakes in your code:
1. The first parameter of instr should be 18. You want to find the first space after the starting point
2. The last parameter of mid is the number of characters
So your code should look something like this:
Position = InStr(18, a, " ")
TableNo = Mid(a, 18, Position - 18)
The space before T682 is character 17, the table start at character 18 so we just need to find the space after that.
The GetTable function can also be used as a worksheet function, e.g. in cell B1 enter =GetTable(A1) to return T682.
Sub Test()
Debug.Print GetTable("Changes in table T682 (SAP TABLE)")
End Sub
Public Function GetTable(Target As Range) As String
GetTable = Mid(Target, 18, InStr(18, Target, " ") - 18)
End Function
Another way
s = "Changes in table T682 (SAP TABLE)"
Debug.Print Trim(Split(Split(s, "(")(0), "T")(1))
I am assuming the following. The format of the string will be constant i.e
There will be no other Capital T before (
There will be only 1 (
The number T* will be before the (
Related
I have a comment field with cells containing text like this:
Cancelled by user at 2018-01-03 03:11:57 without charge
I want to get the date and time information, but it may not always be in the 3rd/4th from last spaces, otherwise I might try to do some sort of complicated split of the cell. Is there an "in cell" way extract the date time information? Or will this need a VBA script? I prefer the former, but I'm trying to make a macro to simplify my life anyway, so VBA would work too.
I'd propose the following formula:
=MID(A1,FIND("at 20",A1)+3,19)
This would require that the date is always preceded by the word 'at' and the date string starts with 20.
You can try this function. It splits the string checking for items that have the first letter numeric, and builds a result string of just the date information.
Public Function ParseForDate(sCell As String) As String
Dim vSplit As Variant
Dim nIndex As Integer
Dim sResult As String
vSplit = Split(sCell, " ")
For nIndex = 0 To UBound(vSplit)
If IsNumeric(Left$(vSplit(nIndex), 1)) Then
sResult = sResult & vSplit(nIndex) & " "
End If
Next
ParseForDate = Trim$(sResult)
End Function
If you wanted to use it in a formula it would look something like this:
=ParseForDate(A1)
To use it in a VBA routine:
Dim s as String
s = ParseForDate(Range("A1"))
Non-VBA solution: (this is assuming the date format is always the same for all cells)
= MAX(IFERROR(DATEVALUE(MID(A1,ROW(INDEX($A:$A,1):INDEX($A:$A,LEN(A1)-19)),20)),0))
+MAX(IFERROR(TIMEVALUE(MID(A1,ROW(INDEX($A:$A,1):INDEX($A:$A,LEN(A1)-19)),20)),0))
Note this is an array formula, so you must press Ctrl+Shift+Enter instead of just Enter when typing this formula.
You will obviously then need to format the cell as a date and time, but this formula gets the numerical value that Excel uses for its internal date and time system.
Using a regex will enable you to fetch the date and time, irrespective of its placement in the string. The following solution will work if the date and time are of the same format as shown in the example string.
Code:
Sub getDateTime()
Dim objReg, matches, str
str = Sheet1.Cells(1, 1).Value 'Change this as per your requirements
Set objReg = CreateObject("vbscript.regexp")
objReg.Global = True
objReg.Pattern = "\d{4}(?:-\d{2}){2}\s*\d{2}(?::\d{2}){2}"
If objReg.test(str) Then
Set matches = objReg.Execute(str)
strResult = matches.Item(0)
MsgBox strResult
End If
End Sub
Click for Regex Demo
Regex Explanation:
\d{4} - matches 4 digits representing the year
(?:-\d{2}){2} - matches - followed by 2 digits. {2} in the end repeats this match 2 times. Once for getting MM and the next time for DD
\s* - matches 0+ whitespaces to match the space between the Date and Time
\d{2} - matches 2 digits representing the HH
(?::\d{2}){2} - matches : followed by 2 digits. The {2} in the end repeats this match 2 times. First time for matching the :MM and the next time for matching the :SS
Screenshots:
Output:
This will be good for about 90 years (using cell C3 for example):
Sub GetDate()
Dim s As String
s = Range("C3").Comment.Text
arr = Split(s, " ")
For i = LBound(arr) To UBound(arr)
If Left(arr(i), 2) = "20" Then
msg = arr(i) & " " & arr(i + 1)
MsgBox msg
Exit Sub
End If
Next i
End Sub
I am extracting text between parens in the active window title bar. That part is working great (thanks to some help I received here previously!). Now I want to create two separate macros - one that returns only the first name, and the other that returns only the last name.
My active window title bar looks something like this:
some text to the left (HENDERSON,TOM) some text to the right
(there is no space after the comma)
The last name macro works perfectly. It looks like this:
Sub a1LastName()
'Extract last name of patient from title bar (between parens)
Dim strPatientName As String
Dim OpenPosition As Integer '(open paren marker)
Dim closeposition As Integer '(close paren marker)
OpenPosition = InStr(ActiveDocument.ActiveWindow.Caption, "(")
closeposition = InStr(ActiveDocument.ActiveWindow.Caption, ")")
strPatientName = Mid(ActiveDocument.ActiveWindow.Caption, _
OpenPosition + 1, closeposition - OpenPosition - 1)
Dim c As Long
c = InStr(strPatientName, ",")
strPatientName = Left(strPatientName, c - 1)
Selection.TypeText strPatientName
End Sub
The second macro is identical to the first, except that the second-to-last line of code has a "Right" instead of a "Left" instruction:
Sub a1FirstName()
'Extract first name of patient from title bar (between parens)
Dim strPatientName As String
Dim OpenPosition As Integer '(open paren marker)
Dim closeposition As Integer '(close paren marker)
OpenPosition = InStr(ActiveDocument.ActiveWindow.Caption, "(")
closeposition = InStr(ActiveDocument.ActiveWindow.Caption, ")")
strPatientName = Mid(ActiveDocument.ActiveWindow.Caption, _
OpenPosition + 1, closeposition - OpenPosition - 1)
Dim c As Long
c = InStr(strPatientName, ",")
strPatientName = Right(strPatientName, c - 1)
Selection.TypeText strPatientName
End Sub
Here's my problem: The "first name" macro always returns the last name minus the first four characters, followed by the first name, instead of simply the first name.
The only examples I'm able to find anywhere in Google land are specifically for Excel. I have combined through my VBA manuals, and they all give similar examples as I have used for extracting the text to the right of a character.
What am I doing wrong?
You can use Split() to create an array out of the comma-separated parts of the text, then access either the first or second part:
Sub a1LastName()
Dim strPatientName As String
strPatientName = ParensContent(ActiveDocument.ActiveWindow.Caption)
If strPatientName Like "*,*" Then
Selection.TypeText Trim(Split(strPatientName, ",")(0))
End If
End Sub
Sub a1FirstName()
Dim strPatientName As String
strPatientName = ParensContent(ActiveDocument.ActiveWindow.Caption)
If strPatientName Like "*,*" Then
Selection.TypeText Trim(Split(strPatientName, ",")(1))
End If
End Sub
'Utility function: find and return text enclosed by ()
' Return empty string if no () found
Function ParensContent(txt) As String
Dim rv As String, pos As Long, pos2 As Long
If txt Like "*(*)*" Then
pos = InStr(1, txt, "(")
pos2 = InStr(pos, txt, ")")
rv = Mid(txt, pos + 1, (pos2 - pos) - 1)
End If
ParensContent = rv
End Function
Seems like everyone has keyed in on Split to get the first/second part of the name. You can also use Split to get rid of the parentheses. This works if you know that you will only (and always) have a single ( and ).
Code gives the main idea. You can use Split to get the part of the String which does not include the ( or ) and then do it again to get either side of the ,.
Sub t()
Dim str As String
str = "(Wall,Byron)"
Dim name_first As String
Dim name_last As String
'two splits total
name_last = Split(Split(str, "(")(1), ",")(0)
name_first = Split(Split(str, ")")(0), ",")(1)
'three split option, same code to remove parentheses
name_last = Split(Split(Split(str, "(")(1), ")")(0), ",")(0)
name_first = Split(Split(Split(str, "(")(1), ")")(0), ",")(1)
End Sub
The code above presents a two Split and a three Split option. The main difference is that the three Split variety uses the same code to strip off the parentheses, and the only change is which side of the , to grab. The two Split options take advantage of the fact that splitting on a comma removes one the parentheses for free. The indices there are a little more complicated.
EDIT it seems I didn,t understand well the first time. Attempt 2:
You want to remove everything right of the comma.
Mystr = Left(Mystr, Instr(Mystr, ","))
Mystr = Mid(Mystr, 2)
the second line uses mid to take everything after the first char (the parentesis)
You're looking for the Mid function.
http://www.excel-easy.com/vba/string-manipulation.html#mid
Use it like so:
Debug.Print Mid("abcde", InStr("abcde", "c"))
Will return "cde"
I used Instr to locate where it should start from instead of writing a number. To get it after the comma, use:
Debug.Print Mid(Mystring, InStr(Mystring, "," + 1))
Where Mystring would be your string. I wrote +1 so it starts after the comma.
I would like to take the count of character from one position until it reaches a Space Using VBA
Sub testing()
Dim YourText As String
YourText = "my name ismanu prasad"
Cells(1, 1).Value = Len(YourText)
End Sub
Above code will return 21 as output. But my scenario is bit different .I need the count of substring “manu” from the above string and output should be 4
Sub Display4thWord()
Dim Space As String
Dim YourText As String
Dim Begin4thWord As Integer
Dim End4thWord As Integer
YourText = "The first message box display a value"
Space = " "
'Find begin of 4th word.
Begin4thWord = InStr(InStr(InStr(1, YourText, Space, vbBinaryCompare) + 1, YourText, Space, vbBinaryCompare) + 1, YourText, Space, vbBinaryCompare)
'Find end of 4th word/begin of 5th word
End4thWord = InStr(Begin4thWord + 1, YourText, Space, vbBinaryCompare)
MsgBox (Begin4thWord)
'Display 4th word
MsgBox (Mid(YourText, Begin4thWord, End4thWord - Begin4thWord))
End Sub
You need to embed InStr function and use with Mid function.
Is it satisfaction you? This solution have hard code number number of word which it will return.
Declare two variables recordnocount & recordnocount
.position is the value we have
recordnocount = InStr(position + 20, text, " ")
recordnocount1 = recordnocount - (position + 20)
we can get the count .
Thankyou All
I have a string (for example: "Hello there. My name is John. I work very hard. Hello there!") and I am trying to find the number of occurrences of the string "hello there". So far, this is the code I have:
Dim input as String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase as String = "hello there"
Dim Occurrences As Integer = 0
If input.toLower.Contains(phrase) = True Then
Occurrences = input.Split(phrase).Length
'REM: Do stuff
End If
Unfortunately, what this line of code seems to do is split the string every time it sees the first letter of phrase, in this case, h. So instead of the result Occurrences = 2 that I would hope for, I actually get a much larger number. I know that counting the number of splits in a string is a horrible way to go about doing this, even if I did get the correct answer, so could someone please help me out and provide some assistance?
Yet another idea:
Dim input As String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase As String = "Hello there"
Dim Occurrences As Integer = (input.Length - input.Replace(phrase, String.Empty).Length) / phrase.Length
You just need to make sure that phrase.Length > 0.
the best way to do it is this:
Public Function countString(ByVal inputString As String, ByVal stringToBeSearchedInsideTheInputString as String) As Integer
Return System.Text.RegularExpressions.Regex.Split(inputString, stringToBeSearchedInsideTheInputString).Length -1
End Function
str="Thisissumlivinginsumgjhvgsum in the sum bcoz sum ot ih sum"
b= LCase(str)
array1=Split(b,"sum")
l=Ubound(array1)
msgbox l
the output gives u the no. of occurences of a string within another one.
You can create a Do Until loop that stops once an integer variable equals the length of the string you're checking. If the phrase exists, increment your occurences and add the length of the phrase plus the position in which it is found to the cursor variable. If the phrase can not be found, you are done searching (no more results), so set it to the length of the target string. To not count the same occurance more than once, check only from the cursor to the length of the target string in the Loop (strCheckThisString).
Dim input As String = "hello there. this is a test. hello there hello there!"
Dim phrase As String = "hello there"
Dim Occurrences As Integer = 0
Dim intCursor As Integer = 0
Do Until intCursor >= input.Length
Dim strCheckThisString As String = Mid(LCase(input), intCursor + 1, (Len(input) - intCursor))
Dim intPlaceOfPhrase As Integer = InStr(strCheckThisString, phrase)
If intPlaceOfPhrase > 0 Then
Occurrences += 1
intCursor += (intPlaceOfPhrase + Len(phrase) - 1)
Else
intCursor = input.Length
End If
Loop
You just have to change the input of the split function into a string array and then delare the StringSplitOptions.
Try out this line of code:
Occurrences = input.Split({phrase}, StringSplitOptions.None).Length
I haven't checked this, but I'm thinking you'll also have to account for the fact that occurrences would be too high due to the fact that you're splitting using your string and not actually counting how many times it is in the string, so I think Occurrences = Occurrences - 1
Hope this helps
You could create a recursive function using IndexOf. Passing the string to be searched and the string to locate, each recursion increments a Counter and sets the StartIndex to +1 the last found index, until the search string is no longer found. Function will require optional parameters Starting Position and Counter passed by reference:
Function InStrCount(ByVal SourceString As String, _
ByVal SearchString As String, _
Optional ByRef StartPos As Integer = 0, _
Optional ByRef Count As Integer = 0) As Integer
If SourceString.IndexOf(SearchString, StartPos) > -1 Then
Count += 1
InStrCount(SourceString, _
SearchString, _
SourceString.IndexOf(SearchString, StartPos) + 1, _
Count)
End If
Return Count
End Function
Call function by passing string to search and string to locate and, optionally, start position:
Dim input As String = "Hello there. My name is John. I work very hard. Hello there!"
Dim phrase As String = "hello there"
Dim Occurrences As Integer
Occurrances = InStrCount(input.ToLower, phrase.ToLower)
Note the use of .ToLower, which is used to ignore case in your comparison. Do not include this directive if you do wish comparison to be case specific.
One more solution based on InStr(i, str, substr) function (searching substr in str starting from i position, more info about InStr()):
Function findOccurancesCount(baseString, subString)
occurancesCount = 0
i = 1
Do
foundPosition = InStr(i, baseString, subString) 'searching from i position
If foundPosition > 0 Then 'substring is found at foundPosition index
occurancesCount = occurancesCount + 1 'count this occurance
i = foundPosition + 1 'searching from i+1 on the next cycle
End If
Loop While foundPosition <> 0
findOccurancesCount = occurancesCount
End Function
As soon as there is no substring found (InStr returns 0, instead of found substring position in base string), searching is over and occurances count is returned.
Looking at your original attempt, I have found that this should do the trick as "Split" creates an array.
Occurrences = input.split(phrase).ubound
This is CaSe sensitive, so in your case the phrase should equal "Hello there", as there is no "hello there" in the input
Expanding on Sumit Kumar's simple solution, here it is as a one-line working function:
Public Function fnStrCnt(ByVal str As String, ByVal substr As String) As Integer
fnStrCnt = UBound(Split(LCase(str), substr))
End Function
Demo:
Sub testit()
Dim thePhrase
thePhrase = "Once upon a midnight dreary while a man was in a house in the usa."
If fnStrCnt(thePhrase, " a ") > 1 Then
MsgBox "Found " & fnStrCnt(thePhrase, " a ") & " occurrences."
End If
End Sub 'testit()
I don't know if this is more obvious?
Starting from the beginning of longString check the next characters up to the number characters in phrase, if phrase is not found start looking from the second character etc. If it is found start agin from the current position plus the number of characters in phrase and increment the value of occurences
Module Module1
Sub Main()
Dim longString As String = "Hello there. My name is John. I work very hard. Hello there! Hello therehello there"
Dim phrase As String = "hello There"
Dim occurences As Integer = 0
Dim n As Integer = 0
Do Until n >= longString.Length - (phrase.Length - 1)
If longString.ToLower.Substring(n, phrase.Length).Contains(phrase.ToLower) Then
occurences += 1
n = n + (phrase.Length - 1)
End If
n += 1
Loop
Console.WriteLine(occurences)
End Sub
End Module
I used this in Vbscript, You can convert the same to VB.net as well
Dim str, strToFind
str = "sdfsdf:sdsdgs::"
strToFind = ":"
MsgBox GetNoOfOccurranceOf( strToFind, str)
Function GetNoOfOccurranceOf(ByVal subStringToFind As String, ByVal strReference As String)
Dim iTotalLength, newString, iTotalOccCount
iTotalLength = Len(strReference)
newString = Replace(strReference, subStringToFind, "")
iTotalOccCount = iTotalLength - Len(newString)
GetNoOfOccurranceOf = iTotalOccCount
End Function
I know this thread is really old, but I got another solution too:
Function countOccurencesOf(needle As String, s As String)
Dim count As Integer = 0
For i As Integer = 0 to s.Length - 1
If s.Substring(i).Startswith(needle) Then
count = count + 1
End If
Next
Return count
End Function
I have a list of filenames in a spreadsheet in the form of "Smith, J. 010112.pdf". However, they're in the varying formats of "010112.pdf", "01.01.12.pdf", and "1.01.2012.pdf". How could I change these to one format of "010112.pdf"?
Personally I hate using VBA where worksheet functions will work, so I've worked out a way to do this with worksheet functions. Although you could cram this all into one cell, I've broken it out into a lot of independent steps in separate columns so you can see how it's working, step by step.
For simplicity I'm assuming your file name is in A1
B1 =LEN(A1)
determine the length of the filename
C1 =SUBSTITUTE(A1," ","")
replace spaces with nothing
D1 =LEN(C1)
see how long the string is if you replace spaces with nothing
E1 =B1-D1
determine how many spaces there are
F1 =SUBSTITUTE(A1," ",CHAR(8),E1)
replace the last space with a special character that can't occur in a file name
G1 =SEARCH(CHAR(8), F1)
find the special character. Now we know where the last space is
H1 =LEFT(A1,G1-1)
peel off everything before the last space
I1 =MID(A1,G1+1,255)
peel off everything after the last space
J1 =FIND(".",I1)
find the first dot
K1 =FIND(".",I1,J1+1)
find the second dot
L1 =FIND(".",I1,K1+1)
find the third dot
M1 =MID(I1,1,J1-1)
find the first number
N1 =MID(I1,J1+1,K1-J1-1)
find the second number
O1 =MID(I1,K1+1,L1-K1-1)
find the third number
P1 =TEXT(M1,"00")
pad the first number
Q1 =TEXT(N1,"00")
pad the second number
R1 =TEXT(O1,"00")
pad the third number
S1 =IF(ISERR(K1),M1,P1&Q1&R1)
put the numbers together
T1 =H1&" "&S1&".pdf"
put it all together
It's kind of a mess because Excel hasn't added a single new string manipulation function in over 20 years, so things that should be easy (like "find last space") require severe trickery.
Here's a screenshot of a simple four-step method based on Excel commands and formulas, as suggested in a comment to the answered post (with a few changes)...
This function below works. I've assumed that the date is in ddmmyy format, but adjust as appropriate if it's mmddyy -- I can't tell from your example.
Function FormatThis(str As String) As String
Dim strDate As String
Dim iDateStart As Long
Dim iDateEnd As Long
Dim temp As Variant
' Pick out the date part
iDateStart = GetFirstNumPosition(str, False)
iDateEnd = GetFirstNumPosition(str, True)
strDate = Mid(str, iDateStart, iDateEnd - iDateStart + 1)
If InStr(strDate, ".") <> 0 Then
' Deal with the dot delimiters in the date
temp = Split(strDate, ".")
strDate = Format(DateSerial( _
CInt(temp(2)), CInt(temp(1)), CInt(temp(0))), "ddmmyy")
Else
' No dot delimiters... assume date is already formatted as ddmmyy
' Do nothing
End If
' Piece it together
FormatThis = Left(str, iDateStart - 1) _
& strDate & Right(str, Len(str) - iDateEnd)
End Function
This uses the following helper function:
Function GetFirstNumPosition(str As String, startFromRight As Boolean) As Long
Dim i As Long
Dim startIndex As Long
Dim endIndex As Long
Dim indexStep As Integer
If startFromRight Then
startIndex = Len(str)
endIndex = 1
indexStep = -1
Else
startIndex = 1
endIndex = Len(str)
indexStep = 1
End If
For i = startIndex To endIndex Step indexStep
If Mid(str, i, 1) Like "[0-9]" Then
GetFirstNumPosition = i
Exit For
End If
Next i
End Function
To test:
Sub tester()
MsgBox FormatThis("Smith, J. 01.03.12.pdf")
MsgBox FormatThis("Smith, J. 010312.pdf")
MsgBox FormatThis("Smith, J. 1.03.12.pdf")
MsgBox FormatThis("Smith, J. 1.3.12.pdf")
End Sub
They all return "Smith, J. 010312.pdf".
You don't need VBA. Start by replacing the "."s with nothing:
=SUBSTITUTE(A1,".","")
This will change the ".PDF" to "PDF", so let's put that back:
=SUBSTITUTE(SUBSTITUTE(A1,".",""),"pdf",".pdf")
Got awk? Get the data into a text file, and
awk -F'.' '{ if(/[0-9]+\.[0-9]+\.[0-9]+/) printf("%s., %02d%02d%02d.pdf\n", $1, $2, $3, length($4) > 2 ? substr($4,3,2) : $4); else print $0; }' your_text_file
Assuming the data are exactly as what you described, e.g.,
Smith, J. 010112.pdf
Mit, H. 01.02.12.pdf
Excel, M. 8.1.1989.pdf
Lec, X. 06.28.2012.pdf
DISCLAIMER:
As #Jean-FrançoisCorbett has mentioned, this does not work for "Smith, J. 1.01.12.pdf". Instead of reworking this completely, I'd recommend his solution!
Option Explicit
Function ExtractNumerals(Original As String) As String
'Pass everything up to and including ".pdf", then concatenate the result of this function with ".pdf".
'This will not return the ".pdf" if passed, which is generally not my ideal solution, but it's a simpler form that still should get the job done.
'If you have varying extensions, then look at the code of the test sub as a guide for how to compensate for the truncation this function creates.
Dim i As Integer
Dim bFoundFirstNum As Boolean
For i = 1 To Len(Original)
If IsNumeric(Mid(Original, i, 1)) Then
bFoundFirstNum = True
ExtractNumerals = ExtractNumerals & Mid(Original, i, 1)
ElseIf Not bFoundFirstNum Then
ExtractNumerals = ExtractNumerals & Mid(Original, i, 1)
End If
Next i
End Function
I used this as a testcase, which does not correctly cover all your examples:
Sub test()
MsgBox ExtractNumerals("Smith, J. 010112.pdf") & ".pdf"
End Sub