I have a function which finds a string between to given strings. Here is an example:
midReturn("?sender=", "&to=", mailItem.HTMLBody)
The above should get the string between ?sender and &to but the & is converted to &.
Is there a way i can look for both occurrences? I did a statement to check this but it doesn't work:
Dim from = midReturn("?sender=", "&to=", mailItem.HTMLBody)
If from < 1 Then
from = midReturn("?sender=", "&to=", mailItem.HTMLBody)
End If
Related
I have this String:
"[" & vbCrLf & " ""APPLE""" & vbCrLf & "]"
The only thing I need is APPLE.
I tried a few options with Split, Trim, Left and more, but they didn't work very well.
Thank you very much!
As the comments above have said, there's not enough information to give an answer without making assumptions, which could be wrong. I've assumed you want to extract the value between two quotation marks, regardless of what else is before or after.
If that's what you want, try this:
Dim Result As String = Nothing
Dim source As String = $"[{vbCrLf}""Apple""{vbCrLf}]"
Dim FirstQuote As Integer = source.IndexOf("""")
If FirstQuote > -1 And source.Length > FirstQuote Then
Dim SecondQuote As Integer = source.IndexOf("""", FirstQuote + 1)
If SecondQuote > FirstQuote Then
Result = source.Substring(FirstQuote + 1, SecondQuote - FirstQuote - 1)
End If
End If
If Result Is Nothing Then
'Handle Invalid Format
Else
'Process Result
End If
You would need to modify that so that you passed your source string, rather than defining it in the code. If you wanted to extract multiple words from a single string in the same format, just set FirstQuote = SecondQuote + 1, check that doesn't exceed the length of the source string and loop through again.
I am going to assume that you probably just need to get the first occurance of a string (in this case "apple") within square-brackets using split and so:
Dim AppleString As String = "This is an [Apple] or etc [...]"
console.WriteLine(AppleString.split("[")(1).split("]")(0).trim())
⚠️ This is not a solution for all purposes !!!
I have a variable with a string...and I want to know if it contains any value other than single quote, comma and a space ("', ") I'm using vba in excel.
for example, i have a varible strA = "'test', 'player'"
I want to check to see if strA has any characters other than "', " (single quote, comma and space).
Thanks
Here is a strategy based on Count occurrences of a character in a string
I don't have vba handy, but this should work. The idea is to remove all these characters and see if anything is left. text represents your string that is being tested.
Dim TempS As String
TempS = Replace(text, " " , "")
TempS = Replace(TempS, "," , "")
TempS = Replace(TempS, "'" , "")
and your result is Len(TempS>0)
Another approach is to use recursion by having a base case of false if the string is empty, if the first character is one of the three call ourselves on the rest of the string, or if not the value is true. Here is the code
function hasOtherChars(s As String) As Boolean
hasOtherChars=false
if (len(s)=0) then
exit function
end if
Dim asciiSpace As Integer
asciiSpace = Asc(" ")
Dim asciiComma As Integer
asciiComma= Asc(",")
Dim asciiApostrophe As Integer
asciiApostrophe = Asc("'")
Dim c as Integer
c = Asc(Mid$(s, 1, 1))
if ((c=asciiSpace) or (c=asciiComma) or (c=asciiApostrophe)) then
hasOtherChars = hasOtherChars(Mid$(s,2))
else
hasOtherChars=true
end if
End function
Again I am borrowing from the other thread.
I'm rooky for VBA. I have some problem about reversing my data on VBA-Excel. My data is "3>8 , 6>15 , 26>41 (each data on difference cells)" that i could reverse "3>8" to "8>3" follow my requirement by using function reverse. But i couldn't reverse "6>15" and "26>41" to "15>6" and "41>26". It will be "51>6" and "14>62" that failure, I want to be "15>6" and "41>26".
Reverse = StrReverse(Trim(str))
Help me for solve my issue please and thank for comment.
You first need to find the position of the ">" in the cell. you do this by taking the contents of the cell and treating it as a String and finding the ">"
This is done in the line beginning arrowPosition. This is the integer value of the position of the ">" in you original string
Next use Left to extract the text up to the ">" and Right to extract the text after the ">"
Then build a new String of rightstr & ">" & leftStr.
Note I input my data from Sheet1 B5 but you can just use any source as long as it is a String in the correct format.
Sub Test()
Dim myString As String
myString = Sheets("Sheet1").Range("B5")
Debug.Print myString
Debug.Print reverseString(myString)
End Sub
Function reverseString(inputString As String) As String
Dim leftStr As String
Dim rightStr As String
Dim arrowPosition As Integer
arrowPosition = InStr(1, inputString, ">")
leftStr = Left(inputString, arrowPosition - 1)
rightStr = Right(inputString, Len(inputString) - arrowPosition)
reverseString = rightStr & ">" & leftStr
End Function
just because you look for a VBA, you can add this function into your code:
Function rev(t As String) As String
s = Split(t, ">", 2)
rev = s(1) & ">" & s(0)
End Function
of course only if you have to reverse 2 number, otherwise you'll loop the "s", but the function would lose its usefulness
I'm looking for a way in VB to find a string between two characters,
"(" and ")".
For example, for the string...
"THIS IS (ONE) AND THIS IS (TWO)"
....I would like for a variable to store the characters between the
second set of parenthesis, e.g.
strMyString = "TWO".
But if the string to search only contains one set of parenthesis, to
store this instead. e.g.
strFirstString = "THIS IS (ONE)"
strMyString = "ONE"
As a preliminary answer, you can use this function to find the string within the last pair or brackets in your test string. If the brackets are in the wrong order or if brackets are missing it will throw an exception.
Private Function StringInLastBracketPair(testString As String) As String
Dim startBracket, endBracket As Integer
startBracket = testString.LastIndexOf("(") + 1
endBracket = testString.LastIndexOf(")")
If startBracket >= endBracket Or startBracket = 0 Or endBracket = -1 Then
Throw New System.Exception("String is not formatted properly : " & testString)
End If
StringInLastBracketPair = stringToTest.Substring(startBracket, endBracket - startBracket)
End Function
I have a strange problem here. In my code, variable b string, has the value "Test Test Test". This value we can see while debugging the variable as well as in the text visualizer.
Now the problem is, if I show the same string using Messagebox, the value is just "Test". What can I do here to get the complete value.
I am converting from an ebcdic encoded bytes to corresponding utf8 string and doing the above operation. Any thoughts. below is my sample code.
Dim hex As String = "e385a2a300000000e385a2a3000000e385a2a3"
Dim raw As Byte() = New Byte((hex.Length / 2) - 1) {}
Dim i As Integer
For i = 0 To raw.Length - 1
raw(i) = Convert.ToByte(hex.Substring((i * 2), 2), &H10)
Next i
Dim w As String = System.Text.Encoding.GetEncoding(37).GetString(raw)
Dim raw1 As Byte() = Encoding.UTF8.GetBytes(w)
Dim b As String = Encoding.UTF8.GetString(raw1)
MessageBox.Show(b)
Look at the byte array. You have 4 ASCII 0's after each "Test". ASCII character code 0 corresponds to nul, which is a string termination sequence. If you want spaces instead of nulls there...
Dim b As String = Encoding.UTF8.GetString(raw1).Replace(Chr(0), " ")
It is possible that the string "b" might contains some control character.
To test a control char in string.
For Each p As Char In b
MsgBox(p & " " & Char.IsControl(p) & " " & AscW(p))
Next
Use String#Replace to replace control chars.
b = b.Replace(ChrW(0), " ")
MsgBox(b)