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
Related
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 have records that include text between square brackets.
aaaaaa[aaaaa]
I need to erase that text, square brackets included.
The result would be:
aaaaaa
I'm trying this code:
Dim sqr as Integer
Dim origin as String
Dim result as String
InStr(origin,[)
I can find the first square bracket, but it does not do the job.
Since in your question you state that you wish to remove text within square brackets (including removing the brackets), I would suggest the following:
Function RemoveSqBracketText(strStr As String) As String
Dim lngId1 As Long
Dim lngId2 As Long
lngId1 = InStr(strStr, "[")
lngId2 = InStr(strStr, "]")
If lngId1 > 0 And lngId2 > 0 Then
RemoveSqBracketText = Left(strStr, lngId1 - 1) & RemoveSqBracketText(Mid(strStr, lngId2 + 1))
Else
RemoveSqBracketText = strStr
End If
End Function
This will recursively remove all instances of text enclosed in square brackets, and assumes that you only wish to remove text if it is enclosed within an opening and closing bracket.
Examples:
?RemoveSqBracketText("abc[123]")
abc
?RemoveSqBracketText("abc[123]def[ghi]")
abcdef
?RemoveSqBracketText("abc[123]defghi]")
abcdefghi]
You need to work out the index of the opening square bracket - InStr(origin, "[") (note the double quotes) is a good start.
Now you can loop from that index up to the end of the string, using the Mid$ function to inspect the character at the current index, until the closing bracket is located:
Dim currentPosition As Long
currentPosition = InStr(origin, "[")
If currentPosition = 0 Then
' no opening bracket. now what?
Else
Dim bracketedWord As String
For currentPosition = currentPosition + 1 To Len(origin)
If Mid$(origin, currentPosition, 1) <> "]" Then
bracketedWord = bracketedWord & Mid$(origin, currentPosition, 1)
Else
'found the closing bracket: we're done.
Exit For
End If
Next
End If
Or, you can use InStr to locate the [ opening brace and the closing brace ] positions, then compute the length of the substring between these two positions, and use the Mid$ function to pull the substring without looping.
Alternatively, with a reference to Microsoft VBScript Regular Expressions 5.5 you could use a simple regular expression:
Public Function FindBracketedWord(ByVal value As String) As String
Dim regex As RegExp
Set regex = new RegExp
regex.Pattern = "\[(\w+)\]" ' matches a square-bracketed "word", no spaces
Dim matches As MatchCollection
Set matches = regex.Execute(value)
If matches.Count <> 0 Then result = matches(0).SubMatches(0)
FindBracketedWord = result
End Function
I'm a programing student, so I've started with vb.net as my first language and I need some help.
I need to know how I delete excess white spaces between words in a sentence, only using these string functions: Trim, instr, char, mid, val and len.
I made a part of the code but it doesn't work, Thanks.
enter image description here
Knocked up a quick routine for you.
Public Function RemoveMyExcessSpaces(str As String) As String
Dim r As String = ""
If str IsNot Nothing AndAlso Len(str) > 0 Then
Dim spacefound As Boolean = False
For i As Integer = 1 To Len(str)
If Mid(str, i, 1) = " " Then
If Not spacefound Then
spacefound = True
End If
Else
If spacefound Then
spacefound = False
r += " "
End If
r += Mid(str, i, 1)
End If
Next
End If
Return r
End Function
I think it meets your criteria.
Hope that helps.
Unless using those VB6 methods is a requirement, here's a one-line solution:
TextBox2.Text = String.Join(" ", TextBox1.Text.Split(New Char() {" "c}, StringSplitOptions.RemoveEmptyEntries))
Online test: http://ideone.com/gBbi55
String.Split() splits a string on a specific character or substring (in this case a space) and creates an array of the string parts in-between. I.e: "Hello There" -> {"Hello", "There"}
StringSplitOptions.RemoveEmptyEntries removes any empty strings from the resulting split array. Double spaces will create empty strings when split, thus you'll get rid of them using this option.
String.Join() will create a string from an array and separate each array entry with the specified string (in this case a single space).
There is a very simple answer to this question, there is a string method that allows you to remove those "White Spaces" within a string.
Dim text_with_white_spaces as string = "Hey There!"
Dim text_without_white_spaces as string = text_with_white_spaces.Replace(" ", "")
'text_without_white_spaces should be equal to "HeyThere!"
Hope it helped!
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
How would you select the last part of a string starting at a specific character count.
For example I would like to get all text after the 3rd comma. but I get an error saying
"StartIndex cannot be less than zero."
Dim testString As String = "part, description, order, get this text, and this text"
Dim result As String = ""
result = testString.Substring(testString.IndexOf(",", 0, 3))
Heres my two cents:
string.Join(",", "aaa,bbb,ccc,ddd,eee".Split(',').Skip(2));
The code "testString.IndexOf(",", 0, 3)" does not find the 3rd comma. It find the first comma starting at position 0 looking at the first 3 positions (i.e. character positions 0,1,2).
If you want the part after the last comma use something like this:
Dim testString As String = "part, description, order, get this text"
Dim result As String = ""
result = testString.Substring(testString.LastIndexOf(",") + 1)
Note the +1 to move to the character after the comma. You should really also find the index first and add checks to confirm that the index is not -1 and index < testString.Length too.
Alternatives(I assume you want all the text after last comma):
Using LastIndexOf:
' You can add code to check if the LastIndexOf returns a positive number
Dim result As String = testString.SubString(testString.LastIndexOf(",")+1)
Regular Expressions:
Dim result As String = Regex.Replace(testString, "(.*,)(.*)$", "$2")
The third argument of indexOf is the number of charcters to search. You are searching for , starting at 0 for 3 characters - that is searching the string par for a comma which does not exist so the returned index is -1, hence your error. I think that you would need to use some recursion:
Dim testString As String = "part, description, order, get this text"
Dim index As Int32 = 0
For i As Int32 = 1 To 3
index = testString.IndexOf(","c, index + 1)
If index < 0 Then
' Not enough commas. Handle this.
End If
Next
Dim result As String = testString.Substring(index + 1)
The IndexOf function only finds the "First" of the specified character. The last parameter (in your case 3) specifies how many characters to examine and not the occurence.
Refer to Find Nth occurrence of a character in a string
The function specified here finds the Nth occurance of a character. Then use the substring function on the occurance returned.
Alternative , you can also use regular expression to find the nth occurance.
public static int NthIndexOf(this string target, string value, int n)
{
Match m = Regex.Match(target, "((" + value + ").*?){" + n + "}");
if (m.Success)
{
return m.Groups[2].Captures[n - 1].Index;
}
else
{
return -1;
}
}
I think this is what you are looking for
Dim testString As String = "part, description, order, get this text"
Dim resultArray As String() = testString.Split(New Char() {","c}, 3)
Dim resultString As String = resultArray(2)