Wyo VB Lecture Notes

Objective #1: Use String variables.

• A string is a sequence of characters (letters, digits, and symbols) that may or not spell an English word. A string can be made up of digits. For example "1234" is a string. Strings very often spell phrases that can contain a blank space, such as "John Doe" (where there is a blank space between the 'n' and the 'D'.)
  
  
• The keyword String is a data type that can be used to store a string in a variable. The declaration statement
  
  Dim strName As String = "John"
  
  declares the string variable named strName and initializes it to "John".
  
  
• The " symbol is considered a string delimiter. In the string value
  " Basic"
  the space before the 'B' is called a leading space. In the string literal, "Programming ", the space AFTER the last 'g' is called a trailing space.
  
  
• You can assign a string value to a String variable with an assignment statement as in

      strName = "John Doe"
  
  This assigns the string value "John Doe" to the variable strName.

• The null string (aka empty string) is a special string value that contains no letters or characters. It is represented by typing two double quotes. The following declaration statement initializes strName to the null string:
  
  Dim strName As String = ""
  
  
• Concatenation is the operation in which you join together two strings. You must use the concatenation operator which is the plus symbol ( + ). In older versions of VB, programmers tended to use the ampersand symbol (&) as a concatenation operator but I prefer that you use the + operator. Look closely at the following example:

  strNameA = "John"
  strNameB = "John "
  strLastName = "Doe"
  strExample1 = strNameA + strLastName
  strExample2 = strNameB + strLastName

  strExample1 is currently equal to "JohnDoe" and strExample2 is "John Doe" because of the included blank space in strNameB.

  Here are more examples involving string concatenation:

"computer" + "programming"  becomes   "computerprogramming"
(VB does NOT automatically embed a space between the words)

"Wyomissing " + "Area"     displays as    "Wyomissing Area"        
(notice the space after the 'g' in "Wyomissing")

"54" + "32"  displays as      "5432"   
(both operands are strings and NOT numeric values)

"54" + 32  displays as      86  
(when at least one of the two operands is a number, the + symbol adds rather than concatenates)

54 + 32  displays as      86   
(both operands are numeric values so mathematical addition occurs rather than concatenation)

You can also use the += operator to concatenate two strings as in

strName = "John"
strLastName = "Doe"
strName += strLastName

strName is now "JohnDoe" (with no space in between)

This last statement is equivalent to

strName = strName + strLastName

• The Length method can be used to count the number of characters in a string. For example, 11 will show up in the message box from the following statement
  
  Dim strWord As String = "Programming"
  MessageBox.Show(strWord.Length())
  
  If strWord2 = " Programming is fun", then strWord2.Length() is 19 because of the three spaces that are included in the strWord2. Notice that there is a leading space before the 'P' and two spaces that separate the words. Spaces ARE considered to be characters and, therefore, they are counted as part of the length of a string.
  
  
• You can use the ToString method to turn a variable that represents a numeric value into a string. For example,
  
  Dim intNumber As Integer = 5
  strNum = intNumber.ToString()
  
  would turn the Integer variable intNumber that is equal to 5 into a string which is represented by "5" . In former versions of VB, the Str function was used to do the same thing as in
  
  strNum = Str(intNumber)
  
  though I prefer that you use the ToString method rather than the Str method when turning a numeric variable into a string.
  
  
• A programmer can use the Val function to turn a string into a numeric value like the statement
  
  intZip = Val("19610")
  
  or
  
  Dim strZip As String = "19610"
  intZip = Val(strZip)
  
  The CInt method can also be used to change a string into an integer as in
  
  intZip = CInt("19610")

• Strings can be compared using the string comparison operators =, <, <=, >, >=, and <>. Visual Basic compares strings alphabetically with these operators. In order to compare strings, VB simply examines the Unicode values of the corresponding characters of each string starting with the leftmost characters. Whichever first letter of the two strings has the lower Unicode value causes that string to be considered less than the other string. In comparing "Apple" to "Banana", VB first compares the Unicode value of the 'A' in "Apple" to the 'B' in "Banana. The Unicode value of 'A' is 65 which is less than the Unicode value of 'B' which is 66. Therefore, "Apple" is considered to be less than "Banana." If the two strings have the same first character, then Basic compares the second characters with each other.
  
  Examples:

  ("A" < "B")      simplifies to True since the Unicode value of A is 65 and the Unicode value of B is 66.
  ("D" < "d")      simplifies to True since the Unicode value of uppercase D is 68 and the Unicode value of d is 100.
  ("Apple" < "Banana")      simplifies to True since the Unicode value of A is 65 and the Unicode value of B is 66.
  ("ABC" = ("A" & "BC"))     simplifies to True since the "ABC" is equal to "ABC" after the string concatenation.
  ("Apple" < "Aphrodite")      simplifies to False since the Unicode value of the second p in Apple is 112 and the Unicode value of h is 104.
  ("Zeus" < "aphrodite")   simplifies to True since the Unicode value of uppercase Z is 90 and the Unicode value of lowercase a is 97.
  
  Another Example:
  
  strMyName = "Minich"
  strYourName = "Anderson"
  
  If (strMyName > strYourName) Then
     MessageBox.Show("Ha Ha, I am greater than you alphabetically")
  End If

• You can also use the Compare method to compare two strings to see if one string is alphabetically less than, equal to, or greater than another string. The Compare method returns a positive number if its first parameter is alphabetically greater than the second parameter. It returns a negative number if its first parameter is less than the second parameter . It returns a zero if the two parameters are equal.
  
  Dim strYourName As String = "Zeus"
  Dim strMyName As String = "Aphrodite"
  
  If (String.Compare(strMyName, strYourName) < 0) Then
     MessageBox.Show("My name is alphabetically less than your name")
  ElseIf (String.Compare(strMyName, strYourName) > 0) Then
     MessageBox.Show("My name is alphabetically less than your name")
  ElseIf (String.Compare(strMyName, strYourName) = 0) Then
     MessageBox.Show("My name is exactly the same as your name")
  End If
  
  Notice that the Compare method is a static method from the String class like the Sqrt and Pow methods from the Math class.

• The ToUpper method is used to change lowercase letters into uppercase letters. The ToLower function is used to change uppercase letters into lowercase letters. These functions can be used to make a program case-insensitive. This would allow the user to type text input without regard to the "uppercaseness" or "lowercaseness" of what he/she is typing.
  
  Examples:

  strName = "FrEd"
  MessageBox.Show(strName.ToUpper())      "FRED" will show up in the message box
  MessageBox.Show(strName.ToLower())      "fred" will show up in the message box

• In general, you cannot assume that the user is going to enter data into your program in a consistent (or even logical) manner. Therefore, one person may enter his name as "John Doe" whereas another may enter it as "bill doe" (forgetting to capitalize proper nouns). This seems to occur more than ever since computer users are so sloppy typing email messages in lowercase letters. A skillful programmer can avoid any complications from bad user input by using the ToUpper and ToLower methods.
  
  
• Similarly, users may enter trailing or leading spaces by accident when inputting data. For example, someone may enter his name as " John Doe" (notice the leading space before the 'J') and not "John Doe". You can use the Trim function to avoid any complications. The Trim function removes all of the leading and trailing spaces from a string if any exist.
  
  
  Examples:

  Trim("    Fred") = "Fred"
  Trim("   Fred      ") = "Fred"
  Trim("Fred") = "Fred"

  
  Note that if you were compare two strings where one has a leading space then that string would always appear less than the other even if the first non-white space character was alphabetically less than its counterpart.
  
  That is, " Zeus" < "Arthur" since the Unicode value of the leading space in " Zeus" 32 is less than the Unicode value of 'A' which is 65.
  

• A substring is part of a larger string. For example, the string "miss" is a substring of "Wyomissing".
  
  
• The Substring method works in this way
  
  Dim strName As String = "George Washington"
  MessageBox.Show(strName.Substring(9, 4))   ' shin
  MessageBox.Show(strName.Substring(0, 6))   ' George
  MessageBox.Show(strName.Substring(7))   ' Washington
  
  where the first parameter is the starting point of the substring (counting the first character as zero) and the second parameter is the number of characters to extract from and including the starting point. If no second parameter is used then the rest of the string is extracted. Blank spaces count as characters. If the second parameter is too large (i.e. greater than the available number of remaining characters) an error occurs.
  
  
• The IndexOf method helps you find a substring within another string. In other words it finds one string within another string.
  
  Dim strName As String = "George Washington"
  MessageBox.Show(strName.IndexOf("g"))
  
  would display the value 4 in the message box since the first occurrence of a lowercase g is in position 4 of strName. The IndexOf method does not find the second or later occurrence of a substring. A -1 value is returned by the IndexOf method if it does not find an occurrence of the string parameter in the larger string. Here are some more examples with strName = "George Washington"
  
  strName.IndexOf("shin")    simplifies to 9 since the position of the "s" in "shin" is position 9. The blank space in strName counts as a position of course.
  
  strName.IndexOf("wash") simplifies to -1 since "wash" with a lowercase "w" is not found anywhere in strName
  
  strLetter = "r"
  strName.IndexOf(strLetter)    simplifies to 3 since the lowercase "r" is found in position 3 of strName.
  
  strVowels = "aeiou"
  strName = "George Washington"
  strVowels.IndexOf(strName.Substring(1, 1))     simplifies to 1 since the "e" in strName is found in position 1 of strVowels