String Methods


 

Strings

  • a series of characters treated as a single unit.
  • may include letters, digits, and various special characters such as +, -, *, /, $, and others.
  • VB allows character strings up to 2 billion characters in length.
  • String literals are enclosed in double quotation marks such as: "CIS 220"
  • MSDN Reference
  • Strings are declared simply by using the String type name when you declare the variable. For instance:
Dim myVarString As String

String Properties

String.Length -- determines the number of characters in a String. Returns an Integer value.

  • stringLength = cityName.Length
  • Length returns 0 if there are no characters in the string. .

String Methods

Comparing Strings

All characters are represented as numeric codes, and lexicographic comparison of two strings compares the numeric codes of the individual characters in the strings. 

The Unicode character set assigns a numeric value to every character.  For example, an "A" has the value 65, while "a" has the value 97. Subsequent characters increase in value, such as "b" which has a value of 98. See ASCII table.

String comparisons are performed on a character-by-character basis. 

Example: the String "Hello" is not equal to the String "hello".

Example: "zebra" < "zeus"

Strings can be compared using the standard relational (<, <=, >, >=) and equality (=, <>) operators.

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Strings can also be compared using the String.Compare method.

String.Compare (String, String)  or

String.Compare (String, String, Boolean)

                                  Return Values for Compare

Return Value Comparison
< 0
0
> 0		 string1 < string2
string1 = string2
string1 > string2
  • String.Compare (string1, string2) -  performs a case-sensitive comparison. 
  • String.Compare (string1, string2, False) -  performs a case-sensitive comparison.  
  • String.Compare (string1, string2, True) -  performs a comparison that is not case-sensitive.

Sample Code:

Dim myStr1, myStr2 As String
Dim myComp As Integer
myStr1 = "ABCD"
myStr2 = "abcd"

' The two strings sort equally. Returns 0
myComp = String.Compare(myStr1, myStr2, True)

' myStr1 comes before myStr2. Returns -1.
myComp = String.Compare(myStr1, myStr2, False)

Locating Characters and Substrings in Strings

String.IndexOf -- returns the first position (Integer) of a string inside another string

  • position = String.IndexOf ( SubString [, StartPosition])
  • returns -1 if substring is not found
  • String is the string being searched.
  • SubString is the string that is searched for in String.
  • StartPosition is a numeric expression that sets the start position for the search of SubString in String.  (Index numbering starts from zero.)  If StartPosition is omitted, IndexOf begins searching at the first position.
  • The search for SubString is case-sensitive.

' Checks to be sure name has been entered in Lastname, Firstname format
tempName = txtName.Text
If tempName.IndexOf(",") = 0 Then
        lblMessage.Text = "Please separate names with a comma."
        txtName.Focus
Else
        name = tempName
End If

testString = "papoose"
testString.IndexOf("p")      ' returns 0
testString.IndexOf("poo")  ' returns 2

Extracting Substrings from Strings

String.Substring -- returns a specified number of characters from the any part of a String.

  • strResult = String.Substring (Start [, Length])
  • strResult receives the return string
  • String is the string from which characters are being returned.
  • Start is the character position in the string at which the string to be extracted begins.  Start is zero-based.
  • Length is the number of characters to be returned.  If this is omitted then the entire string from Start onward is returned.
  • testString = "Hello World")
  • strResult = testString.Substring (4, 3) returns "o W"
  • strResult = testString.Substring (4) returns "o World"

exercise: Convert a name in the form Moose, Bullwinkle J. to Bullwinkle J. Moose.

Step 1: Locate the end of the last name

  • locate the comma using IndexOf function
  • commaLocation = name.IndexOf(",")

Step 2: Extract first name and middle initial, if any

  • Starting in the location after the comma location, extract the remainder of the string using Mid.
  • firstName = name.Substring (commaLocation +1)

Step 3: Extract the Lastname

  • Starting at the beginning of the name, extract all characters up to the comma using Left.
  • lastName = name.Substring (0, commaLocation -1) 

Concatenating Strings

Strings can be combined through the process of concatenation.   Concatenation is performed by using the ampersand (&).  The plus sign (+) was used in previous releases of VB and should no longer be used.

    S1 = "Blue"
    S2 = "berry"
    S3 = S1 & S2

The above statement concatenates (or appends) S2 to the right of S1 to create an entirely new string, S3, containing "Blueberry."  S1 and S2 are unchanged.  

exercise: fullName = firstName & " " & lastName

You can also use the String.Concat method: fullname = String.Concat(firstName, " ", lastName)

Case Conversion

String.ToUpper -- converts a String to all uppercase characters and returns a new String value. 

  • strResult = testString.ToUpper()

String.ToLower -- converts a String to all lowercase characters and returns a new String value. 

  • strResult = testString.ToLower()

Whitespace Removal

String.Trim -- returns a String that contains a copy of a given String with any leading and trailing spaces removed.

  • strResult = testString.Trim()

String.TrimStart -- returns a String that contains a copy of a given String with any leading spaces removed.

  • strResult = testString.TrimStart()

String.TrimEnd -- returns a String that contains a copy of a given String with any trailing spaces removed.

  • strResult = testString.TrimEnd()

exercise: fullName = fullName.Trim()

The Trim methods can also be used to remove all occurrences of a set of specified characters from the beginning and end of a string.

Dim str1 As [String] = "*;|@123***456@|;*"
Dim delim As [String] = "|*;@"
Dim str2 As [String] = str1.Trim(delim.ToCharArray()  ' is set to "123***456"

Miscellaneous String Methods

String.Replace - finds and then replaces or removes occurrences of the "find" string within a target string.  

  • strResult = targetString.Replace ( oldValue, newValue )
  • strResult receives the string returned from the function
  • targetString is the string to be altered
  • oldValue is the string to be replaced in targetString
  • newValue is the string to replace all occurrences of oldValue.

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String.Insert -- inserts a specified instance of a string at a specified index position in this instance. 

  • strResult = origString.Insert(startIndex, insertString)
  • strResult receives the string returned from the method
  • origString is the string to be accept inserts
  • startIndex is the index position of the insertion.
  • insertString is the string to be inserted.

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String.Remove -- removes a specified instance of a string beginning at a specified index position in this instance. 

  • strResult = origString.Remove(startIndex, numChars)
  • strResult receives the string returned from the method
  • origString is the string to be accept inserts
  • startIndex is the index position of the insertion.
  • numChars is the number of characters to be deleted.

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String.Copy -- creates a new instance of String with the same value as a specified String.

  • strResult = String.Copy(origString)
  • strResult receives the string returned from the method
  • origString is the string to be copied
  • Dim str1 As String = "abc"
    Dim str2 As String = "xyz"
    Console.WriteLine("1) str1 = " & str1)  ' writes abc
    Console.WriteLine("2) str2 = " & str2)  ' writes xyz
    Console.WriteLine("Copy...")
    str2 = String.Copy(str1)
    Console.WriteLine("3) str1 = '{0}'", str1)  ' writes abc
    Console.WriteLine("4) str2 = '{0}'", str2)  ' writes abc

 

Arrays of Strings

String.Split -- Identifies the substrings in a string that are delimited by one or more characters specified in an array, then places the substrings into a String array.

  • varResult = String.Split (charSeparatorArray() [, intElemCnt])
  • varResult receives the array of strings returned from the method (a dynamic array is useful here)
  • String is the string list that is to be separated 
  • charSeparatorArray() is an An array of Unicode characters that delimit the substrings in this instance.
  • intElemCnt is an optional parameter that specifies the number of string array elements to create from strList.
  • wordArray = stringOfWords.Split ( " " )
     
  • Dim delimStr As String = " ,.:"
    Dim delimiter As Char() = delimStr.ToCharArray()
    Dim wordList As String = "one two,three:four."
    Dim split As String() = Nothing

    Console.WriteLine("The delimiters are -{0}-", delimStr)
    Dim ctr As Integer
    For ctr = 1 To 5
    split = wordList.Split(delimiter, ctr)
    Console.WriteLine(ControlChars.Cr + "count = {0,2} ..............", ctr)
    Dim stringElt As String
    For Each stringElt In split
    Console.WriteLine("-{0}-", stringElt)
    Next stringElt
    Next ctr
  • The delimiters are - ,.:-

    count = 1 ..............
    -one two,three:four.-

    count = 2 ..............
    -one-
    -two,three:four.-

    count = 3 ..............
    -one-
    -two-
    -three:four.-

    count = 4 ..............
    -one-
    -two-
    -three-
    -four.-

    count = 5 ..............
    -one-
    -two-
    -three-
    -four-
    --

Practice exercise #1:

The Split method was not available in previous versions of VB.  Write a routine called getWord to accomplish the same task that Split accomplishes, namely parse a long string into individual words using the other string functions such as IndexOf and Substring.  For additional practice, be sure that the words do not have any punctuation or whitespace characters attached, i.e., extract only the word.
Link

Practice exercise #2:

Write a function that returns a count of the number of words in a sentence like 

line = "        The old        dog, named       ""Peewee,"" ate a lot. Still, we all loved her...."

Tutorial

MS Link