I have a string called Policynumber
I need to know if it begins with a 1 or a 2 so I can create an if statement for if it starts with 1 do something and if it starts with 2 to do something else.
I keep finding ways to do this for string text but not for a variable.
I have tried doing the following:
If policynumber Like "1*" Then
display.text = policynumber
End If
I am simply looking for possible ways to know what the first character is and therefore determine if it's a 1 or a 2. When I try using the variable name or even a textbox.text I get no result in the display textbox so I know it's not working.
I would use VBA's ASC() function for this purpose. This function returns the ASCII code of the first character of a string. ASCII for the character 1 is 49 and 2 is 50. Therefore ...
Dim PolicyNumber As String
Dim n As Integer
PolicNumber = "1ABC-45678910"
n = ASC(PolicyNumber)
MsgBox "Policy number starts with a " & Chr(n)
For testing the result you can use either If n = 49 Then or If Chr(n) = "1" Then.
Related
I have a row with values something like this, How to extract first two digits that come after the text 'ABCD' to another cell, any formula or vba? There may be a few chars in between or sometimes none.
ABCD 10 sadkf sdfas
ABCD-20sdf asdf
ABCD 40
ABCD50 asdf
You can do this with a worksheet formula. No need for VBA.
Assuming you do not need to test for the presence of two digits:
=MID(A1,MIN(FIND({1,2,3,4,5,6,7,8,9,0},A1&"1234567890")),2)
If you need to test for the presence of two digits, you can try:
=IF(ISNUMBER(-RIGHT(MID(A1,MIN(FIND({1,2,3,4,5,6,7,8,9,0},A1&"1234567890")),2),1)),MID(A1,MIN(FIND({1,2,3,4,5,6,7,8,9,0},A1&"1234567890")),2),"Invalid")
In general, it is always a good idea to show some code in StackOverflow. Thus, you show that you have tried something and you give some directions for the answer.
Concerning the first two digits extract, there are many ways to do this. Starting from RegEx and finishing with a simple looping of the chars and checking each one of them.
This is the loop option:
Public Function ExtractTwoDigits(inputString As String) As Long
Application.Volatile
Dim cnt As Long
Dim curChar As String
For cnt = 1 To Len(inputString)
curChar = Mid(inputString, cnt, 1)
If IsNumeric(curChar) Then
If Len(ExtractTwoDigits) Then
ExtractTwoDigits = ExtractTwoDigits & curChar
Exit Function
Else
ExtractTwoDigits = curChar
End If
End If
Next cnt
ExtractTwoDigits = -1
End Function
Application.Volatile makes sure that the formula recalculates every time;
-1 is the answer if no two digits exist in the inputString;
IsNumeric checks whether the string inside is numeric;
As a further step, you may try to make the function a bit robust, extracting the first 1, 3, 4 or 5 digits, depending on a parameter that you put. Something like this =ExtractTwoDigits("tarato123ra2",4), returning 1232.
RegEx Version:
Public Function GetFirstTwoNumbers(ByVal strInput As String) As Integer
Dim reg As New RegExp, matches As MatchCollection
With reg
.Global = True
.Pattern = "(\d{2})"
End With
Set matches = reg.Execute(strInput)
If matches.Count > 0 Then
GetFirstTwoNumbers = matches(0)
Else
GetFirstTwoNumbers = -1
End If
End Function
You have to enable Microsoft Regular Expressions 5.5 under extras->references. The pattern (\d{2}) matches 2 digits, return value is the number, if not existing -1.
Note: it only extracts 2 successive numbers.
If you place this function into a module, you can use it like normal formula.
Here a great site to to get into regEx.
I need help with extracting 5-digit numbers only from one column to another in Excel 2010. These numbers can be in any position of the string (beginning of the string, anywhere in the middle, or at the end). They can be within brackets or quotes like:
(15478) or "15478" or '15478' or [15478]
I need to ignore any numbers that are less than 5 digits and include numbers that start with 1 or more leading zeros (like 00052, 00278, etc.) and ensure that leading zeros are copied over to the next column. Could someone help me with either creating a formula or UDF?
Here is a formula-based alternative that will extract the first 5 digit number found in cell A1. I tend to prefer reasonably simple formula solutions over VBA in most situations as formulas are more portable. This formula is an array formula and thus must be entered with Ctrl+Shift+Enter. The idea is to split the string up into every possible 5 character chunk and test each one and return the first match.
=MID(A1,MIN(IF(NOT(ISERROR(("1"&MID(A1,ROW(INDIRECT("R1C[1]:R"&(LEN(A1)-4)&"C[1]",FALSE)),5)&".1")*1))*ISERROR(MID(A1,ROW(INDIRECT("R1C[1]:R"&(LEN(A1)-4)&"C[1]",FALSE))+5,1)*1)*ISERROR(MID(A1,ROW(INDIRECT("R1C[1]:R"&(LEN(A1)-4)&"C[1]",FALSE))-1,1)*1),ROW(INDIRECT("R1C[1]:R"&(LEN(A1)-4)&"C[1]",FALSE)),9999999999)),5)
Let's break this down. First we have an expression I used twice to return an array of numbers from 1 up to 4 less than the length of your initial text. So if you have a string of length 10 the following will return {1,2,3,4,5,6}. Hereafter the below formula will be referred to as rowlist. I used R1C1 notation to avoid potential circular references.
ROW(INDIRECT("R1C[1]:R"&(LEN(A1)-4)&"C[1]",FALSE))
Next we will use that array to split the text into an array of 5 letter chunks and test each chunk. The test being performed is to prepend a "1" and append ".1" then verify the chunk is numeric. The prepend and append eliminate the possibility of white space or decimals. We can then check the character before and the character after to make sure they are not numbers. Hereafter the below formula will be referred to as isnumarray.
NOT(ISERROR(("1"&MID(A1,rowlist,5)&".1")*1))
*ISERROR(MID(A1,rowlist+5,1)*1)
*ISERROR(MID(A1,rowlist-1,1)*1)
Next we need to find the first valid 5 digit number in the string by returning the current index from a duplicate of the rowlist formula and returning a large number for non-matches. Then we can use the MIN function to grab that first match. Hereafter the below will be referred to as minindex.
MIN(IF(isnumarray,rowlist,9999999999))
Finally we need to grab the numeric string that started at the index returned by the MIN function.
MID(A1,minindex,5)
The following UDF will return the first five digit number in the string, including any leading zero's. If you need to detect if there is more than one five digit number, the modifications are trivial. It will return a #VALUE! error if there are no five-digit numbers.
Option Explicit
Function FiveDigit(S As String, Optional index As Long = 0) As String
Dim RE As Object
Set RE = CreateObject("vbscript.regexp")
With RE
.Pattern = "(?:\b|\D)(\d{5})(?:\b|\D)"
.Global = True
FiveDigit = .Execute(S)(index).submatches(0)
End With
End Function
As you may see from the discussion between Mark and myself, some of your specifications are unclear. But if you would want to exclude decimal numbers, when the decimal portion has five digits, then the regex pattern in my code above should be changed:
.Pattern = "(?:\d+\.\d+)|(?:\b|\D)(\d{5})(?:\b|\D)"
I just wrote this UDF for you , basic but will do it...
It will find the first 5 consecutive numbers in a string, very crude error checking so it just says Error if anything isn't right
Public Function GET5DIGITS(value As String) As String
Dim sResult As String
Dim iLen As Integer
sResult = ""
iLen = 0
For i = 1 To Len(value)
If IsNumeric(Mid(value, i, 1)) Then
sResult = sResult & Mid(value, i, 1)
iLen = iLen + 1
Else
sResult = ""
iLen = 0
End If
If iLen = 5 Then Exit For
Next
If iLen = 5 Then
GET5DIGITS = Format(sResult, "00000")
Else
GET5DIGITS = "Error"
End If
End Function
I have an AD username called "lastname-132" in Textbox1, this string is 12 long, so i want to add the username in to Textbox2, but shortened, in the textbox2 i only have a string length of only 10 available due to other tools this program is using, so i don't want to convert it all the time manually and want to just convert it automatically with a onleave event.
Anyone any idea how to write this?
So the End Result should look like this.
'String length can be 20 max.
Textbox1.Text = "lastname-123"
'some code to convert it to this:
'String length 10 max. Numbers and the "-" should stay the same, but remove letters if necessary.
Textbox2.Text = "lastna-123"
Here's the concept:
Split string based on '-' into 2 strings
In the example above: 'lastname' and '123'.
Check the length of the first string and cut if it is too long
the program checks 'lastname' and finds that it is too long, then
cuts it into 'lastna'
Combine 'lastna' and '123' back into a string
I hope this helps
Without more information, this will assume that there can be multiple hyphens, the number can be of variable length, and you can change the maximum length of the string by changing one variable.
Dim username As String = "lastname-123"
Dim max As Integer = 10
Dim lindex As Integer = username.LastIndexOf("-")
Dim numberLength As Integer = username.Length - lindex
Dim number As String = username.Substring(lindex)
Dim justName As String = username.Substring(0, lindex)
If justName.Length + numberLength >= max Then
username = justName.Substring(0, max - numberLength) & number
End If
If you are concentrating only on the restriction of length of characters to be accepted then you can use
Maxlength
property of the Textbox.
Ex: Maxlength="10"
restricts the Textbox to accept only 10 characters.
Try to make it fit with for example substring manipulation. See http://msdn.microsoft.com/en-us/library/dd789093.aspx for more info.
I am looking for assistance with my program. I have a user enter 6 digits; of these the input must be alpha-numeric. I have already done the TryParse method for the numbers, but I am looking for validation that the string contains an alpha.
I am aware you must use ASC but am unsure both on how to develop a range say Asc((Chr(65) <= Chr(90))) (between A-Z) and also to say (IF my input contains any of these values within the 6 characters, to return true. I keep getting an overload resolution and wish to know how to properly code so the variables are accurate.
This is a great place to use a regular expression
Dim input = ...
If Regex.IsMatch(input, "^\w+$") AndAlso input.Length = 6 Then
' It's a match
Else
' It's not a match
End If
This will match any string which consists only of letters that has length equal to 6
You can iterate through each char and check if it's a letter. If so, set a flag to true.
Dim containsAlpha Boolean = False
For i As Integer = 0 To input.Length - 1
If Char.IsLetter(input(i)) Then
containsAlpha = True
Exit For
End If
Next
Char.IsLetter will match Unicode alphabetic letters, so not just Latin A-Z (which may or may not be what you actually want).
This should be fairly simple but I'm having one of those days. Can anyone advise me as to how to replace the first and third occurrence of a character within a string? I have looked at replace but that cannot work as the string could be of different lengths. All I want to do is replace the first and third occurrence.
There is an overload of the IndexOf method which takes a start position as a parameter. Using a loop you'll be able to find the position of the first and third occurences. Then you could use a combination of the Remove and Insert methods to do the replacements.
You could also use a StringBuilder to do the replacements. The StringBuilder has a Replace method for which you can specify a start index and a number of characters affected.
aspiringCoder,
Perhaps something like this might be useful to you (in line with what Meta-Knight was talking about <+1>)
Dim str As String = "this is a test this is a test this is a test"
Dim first As Integer
Dim third As Integer
Dim base As Integer = 0
Dim i As Integer
While str.length > 0
If i = 0 Then
first = str.IndexOf("test")
else if i = 2 Then
third = base + str.IndexOf("test")
end if
base = base + str.IndexOf("test")
str = str.Remove(0, str.IndexOf("test") + "test".length -1 )
i++
End While
It might have a one-off error somewhere...but this should at least get you started.