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Visual Basic 2005Visual Basic 2005III. STRINGS, CHARACTERS AND REGULAR

EXPRESSIONS

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Strings, Characters and Regular Strings, Characters and Regular ExpressionsExpressionsObjectives

In this chapter you will learn:•To create and manipulate immutable character String objects of class String.•To create and manipulate mutable character String objects of class StringBuilder.•To manipulate character objects of structure Char.•To use regular expressions in conjunction with classes Regex and Match.

Visual Basic 20052

STRINGS, CHARACTERS AND REGULAR EXPRESSIONS

Fundamentals of Characters and Fundamentals of Characters and StringsStrings

Characters are the fundamental building blocks of Visual Basic source code

Every program is composed of characters that, when grouped together meaningfully, create a sequence that the compiler interprets as instructions

a program also can contain character literals, also called character constants

A character literal is a character that is represented internally as an integer value, called a character code

Visual Basic 2005

Strings, Characters and Regular Expressions

3STRINGS, CHARACTERS AND REGULAR EXPRESSIONS


Character literals are established according to the Unicode character set

A string is a series of characters treated as a single unit

A string is an object of class String in the System namespace

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We write string literals as sequences of characters in double quotation marks:

A declaration can assign a String literal to a String variable:

Dim color As String = “Blue”

“Gerick Andrei M. Salalima" “257-G Teresa St." “Sta. Mesa, Manila" "(632) 4689900"

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String ConstructorsString Constructors

Class String provides constructors for initializing Strings

' string initialization originalString = "Welcome to VB programming!" string1 = originalString string2 = New String(characterArray) string3 = New String(characterArray, 6, 3) string4 = New String("C"c, 5)

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Sample program

STRINGS, CHARACTERS AND REGULAR EXPRESSIONS 7

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‘Demonstrating String class constructorsModule StringConstructor Sub Main() Dim originalString, string1, string2, _ string3, string4 As String Dim characterArray() As Char = _ {"b"c, "i"c, "r"c, "t"c, "h"c, " "c, "d"c, "a"c, "y"c}

' string initialization originalString = "Welcome to VB programming!" string1 = originalString string2 = New String(characterArray) string3 = New String(characterArray, 6, 3) string4 = New String("C"c, 5) Console.WriteLine("string1 = " & """" & string1 & _ """" & vbCrLf & "string2 = " & """" & string2 & _ """" & vbCrLf & "string3 = " & """" & string3 & _ """" & vbCrLf & "string4 = " & """" & string4 & _ """" & vbCrLf) End Sub ' Main End Module ' StringConstructor


Output

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string1 = "Welcome to VB programming!" string2 = "birth day" string3 = "day" string4 = "CCCCC"

StringString Indexer, Indexer, LengthLength Property and Property and CopyToCopyTo Method Method

The String indexer is used to access individual characters in a String

The property Length returns the length of the String

The CopyTo method copies a specified number of characters from a String into a Char array

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StringString Indexer, Indexer, LengthLength Property and Property and CopyToCopyTo Method Method

Using the indexer, property Length and method CopyTo

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Dim string1 As String = “Hello there”Dim i As Integer

For i = string1.Length - 1 To 0 Step -1 Console.Write(string1(i)) Next i

Dim len As Integer

len = string1.Length

Dim characterArray() As Char

' copy characters from string1 into characterArraystring1.CopyTo(0, characterArray, 0, 5)

Comparing StringsComparing Strings

Comparing Strings with = and String Methods Equals and CompareTo


' test for equality with = If string1 = "hello” Then Console.WriteLine("string1 equals ""hello""") End If

' test for equality using Equals methodIf string1.Equals("hello") Then Console.WriteLine("string1 equals ""hello""") End If

' test CompareToIf string1.CompareTo("hello") = 0 Then Console.WriteLine("string1 equals ""hello""") Else Console.WriteLine("string1 not equal to ""hello""")End If

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Locating Characters and Substrings in Locating Characters and Substrings in StringStringss

In many applications, it is necessary to search for a character or set of characters in a String

For example, a programmer creating a word processor would want to provide capabilities for searching through documents


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Searching for characters and substrings in Strings


Dim letters As String = "abcdefghijklmabcdefghijklm“Dim searchLetters As Char () = {"c"c, "a"c, "$"c}Dim index As Integer

index = letters.IndexOf("c"c) ‘this returns 2. . .index = letters.IndexOf("a"c, 1) ‘this returns 13. . .index = letters.IndexOf("$"c, 3, 5) ‘this returns -1

Character to be located

Starting index where search begins

No of characters to seek

IndexOf retruns the index of the first occurrence of the

character in the string

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index = letters.LastIndexOf("c"c) ‘this returns 15. . .index = letters.LastIndexOf("a"c, 25) ‘this returns 13. . .index = letters.LastIndexOf("$"c, 15, 5) ‘this returns -1

Character to be located

Starting index where search begins

No of characters to seek

LastIndexOf retruns the index of the last occurrence of the character in the string

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index = letters.IndexOfAny(searchLetters) ‘this returns 0. . .index = letters.IndexOfAny(searchLetters, 7) ‘this returns 13. . .index = letters.IndexOfAny(searchLetters, 7, 5) ‘this returns -1

IndexOfAny retruns the index of the first occurrence in the string of any character

in a specified array of unicode characters

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index = letters.LastIndexOfAny(searchLetters) ‘this returns 15. . .index = letters.LastIndexOfAny(searchLetters, 1) ‘this returns 0. . .index = letters.LastIndexOfAny(searchLetters, 25, 5)‘this returns -1

LastIndexOfAny retruns the index of the last occurrence in the string of any character

in a specified array of unicode characters

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Extracting Substrings From Extracting Substrings From StringStringss

Class String provides two Substring methods, which are used to create a new String by copying part of an existing String


Dim letters As String = "abcdefghijklmabcdefghijklm“Dim output As String = “”

output = letters.Substring(20) ‘output gets hijklm. . .output = letters.Substring(0, 6) ‘output gets abcdef. . .

Starting index No of characters to extract from starting index

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Concatenating Concatenating StringStringss

Like the & operator, the String class's Shared method Concat can be used to concatenate two Strings


Dim string1 As String = “Visual “Dim string2 As String = “Basic”Dim newString As String

newString = String.Concat(string1, string2) ‘newString gets Visual Basic. . .

Concat method

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Miscellaneous Miscellaneous StringString Methods Methods

Class String provides several methods that return modified copies of Strings

String methods Replace, ToLower, ToUpper and Trim:


Dim string1 As String = “cheers!“Dim string2 As String = “GOOD BYE ”Dim string3 As String = “ spaces “Dim output As String

output = string1.Replace(“e”c, “E”c) ‘output gets “chEErs!”. . .output = string1.ToUpper() ‘output gets “CHEERS!”. . .output = string2.ToLower() ‘output gets “good bye” . . .output = string3.Trim() ‘output gets “spaces”. . .

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Class Class StringBuilderStringBuilder

Objects of class String are constant strings, whereas object of class StringBuilder are mutable sequences of characters


' Demonstrating StringBuilder class constructors. Imports System.Text

Module StringBuilderConstructor Sub Main() Dim buffer1, buffer2, buffer3 As StringBuilder

buffer1 = New StringBuilder() buffer2 = New StringBuilder(10) buffer3 = New StringBuilder("hello") Console.WriteLine("buffer1 = """ & buffer1.ToString() & """") Console.WriteLine("buffer2 = """ & buffer2.ToString() & """") Console.WriteLine("buffer3 = """ & buffer3.ToString() & """") End Sub ' Main End Module ' StringBuilderConstructor

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LengthLength and and CapacityCapacity Properties, Properties, EnsureCapacityEnsureCapacity Method and Indexer of Class Method and Indexer of Class StringBuilderStringBuilder

Class StringBuilder provides the Length and Capacity properties to return the number of characters currently in a StringBuilder and the number of characters that a StringBuilder can store without allocating more memory, respectively


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LengthLength and and CapacityCapacity Properties, Properties, EnsureCapacityEnsureCapacity Method and Indexer of Class Method and Indexer of Class StringBuilderStringBuilder

StringBuilder size manipulation


Dim buffer As New StringBuilder("Hello, how are you?")

Console.WriteLine(buffer.ToString())Console.WriteLine(“Length = “ & buffer.Length)Console.WriteLine(“Capacity = “ & buffer.Capacity)

buffer.EnsureCapacity(75)Buffer.Length = 10

Console.WriteLine(buffer.ToString())

Hello, how are you?Length = 19Capacity = 32Hello, how

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AppendAppend and and AppendFormatAppendFormat Methods of Methods of Class Class StringBuilderStringBuilder

Class StringBuilder provides 19 overloaded Append methods for appending values of various types to the end of a StringBuilder's contents

There are versions of this method for each of the primitive types and for character arrays, Strings and Objects


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AppendAppend and and AppendFormatAppendFormat Methods of Methods of Class Class StringBuilderStringBuilder

Append methods of StringBuilder


Dim objectValue As Object = "hello“ Dim stringValue As String = "good bye“ Dim characterArray As Char() = {"a"c, "b"c, "c"c, "d"c, "e"c, "f"c} Dim booleanValue As Boolean = True Dim characterValue As Char = "Z"c Dim integerValue As Integer = 7 Dim longValue As Long = 1000000 Dim floatValue As Single = 2.5F ' F indicates that 2.5 is a float Dim doubleValue As Double = 33.333 Dim buffer As New StringBuilder()

' use method Append to append values to buffer buffer.Append(objectValue) buffer.Append(" ") buffer.Append(stringValue) buffer.Append(" ") buffer.Append(characterArray) buffer.Append(" “)buffer.Append(integerValue). . .

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InsertInsert, , RemoveRemove and and ReplaceReplace Methods of Methods of Class Class StringBuilderStringBuilder

Class StringBuilder provides 18 overloaded Insert methods to allow values of various types to be inserted at any position in a StringBuilder

There are versions of Insert for each of the primitive types and for character arrays, Strings and Objects


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InsertInsert, , RemoveRemove and and ReplaceReplace Methods of Methods of Class Class StringBuilderStringBuilder

StringBuilder text insertion and removal


Dim buffer As New StringBuilder("Hello, how are you?")

Console.WriteLine(buffer.ToString())buffer.Insert(18, " Gerick")Console.WriteLine(buffer.ToString())buffer.Replace(“Gerick", “Rainne")Console.WriteLine(buffer.ToString())buffer.Remove(18, 7)Console.WriteLine(buffer.ToString())

Hello, how are you?Hello, how are you Gerick?Hello, how are you Rainne?Hello, how are you?

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CharChar Methods Methods

Many of the primitive types that we have used are actually aliases for different structures

For instance, an Integer is defined by structure System.Int32, a Long by System.Int64 and so on

Char, is the structure for charactersMost Char methods are Shared


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Char's Shared character-testing and case-conversion methods


Dim character As Char = “A”cDim output As String

output = "is digit: " & Char.IsDigit(character) & vbCrLf output &= "is letter: " & Char.IsLetter(character) & vbCrLf output &= "is letter or digit: " & _ Char.IsLetterOrDigit(character) & vbCrLf output &= "is lowercase: " & Char.IsLower(character) & vbCrLf output &= "is uppercase: " & Char.IsUpper(character) & vbCrLf output &= "to uppercase: " & Char.ToUpper(character) & vbCrLf output &= "to lowercase: " & Char.ToLower(character) & vbCrLf output &= "is punctuation: " & _ Char.IsPunctuation(character) & vbCrLf output &= "is symbol: " & Char.IsSymbol(character)

Console.WriteLine(output)

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Char's Shared character-testing and case-conversion methods


is digit: Falseis letter: Trueis letter or digit: Trueis lowercase: Falseis uppercase: Trueto uppercase: Ato lowercase: ais punctuation: Falseis symbol: False

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Regular Expressions and Class Regular Expressions and Class RegExRegEx

Regular expressions are specially formatted Strings used to find (and possibly replace) patterns in text

They can be useful during information validation, to ensure that data is in a particular format

For example, a Philippine ZIP code must consist of four digits, and a last name must start with a capital letter


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Regular expression character classes:


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Character class Matches Character class Matches

\d Any digit \D Any non-digit

\w Any word character \W Any non-word character

\s Any whitespace \S Any non-whitespace



Regular expressions checking birthdays


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‘ Demonstrating Class Regex. Imports System.Text.RegularExpressions

Module RegexMatches Sub Main() ' create regular expression Dim expression As New Regex("J.*\d[0-35-9]-\d\d-\d\d") Dim string1 As String = _ "Jane's Birthday is 05-12-75" & vbCrLf & _ "Dave's Birthday is 11-04-68" & vbCrLf & _ "John's Birthday is 04-28-73" & vbCrLf & _ "Joe's Birthday is 12-17-77"

' match regular expression to string and ' print out all matches For Each myMatch As Match In expression.Matches(string1) Console.WriteLine(myMatch) Next myMatch End Sub ' Main End Module ' RegexMatches



Quantifiers used in regular expressions


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Quantifier Matches

* Matches zero or more occurrences of the preceding pattern.

+ Matches one or more occurrences of the preceding pattern.

? Matches zero or one occurrences of the preceding pattern.

{n} Matches exactly n occurrences of the preceding pattern.

{n,} Matches at least n occurrences of the preceding pattern.

{n,m} Matches between n and m (inclusive) occurrences of the preceding pattern.



Validating user information using regular expressions


Visual Basic 2005

Dim strLastName As String = "Salalima"

If Regex.Match(strLastName, "^[A-Z][a-zA-Z]*$").Success Then MessageBox.Show("Valid last name", "Message", _ MessageBoxButtons.OK, MessageBoxIcon.Information)Else MessageBox.Show("Invalid last name", "Message", _ MessageBoxButtons.OK, MessageBoxIcon.Error)End If





Visual Basic 2005

Dim strAddress As String = "257-G Teresa St."

If Regex.Match(strAddress, _ "^[0-9]+\s+([a-zA-Z]+|[a-zA-Z]+\s[a-zA-Z]+)$").Success Then MessageBox.Show("Valid address", "Message", _ MessageBoxButtons.OK, MessageBoxIcon.Information)Else MessageBox.Show("Invalid address", "Message", _ MessageBoxButtons.OK, MessageBoxIcon.Error)End If





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Dim strZip As String = "12345"

If Regex.Match(strZip, "^\d{5}$").Success Then MessageBox.Show("Valid zip code", "Message", _ MessageBoxButtons.OK, MessageBoxIcon.Information)Else MessageBox.Show("Invalid zip code", "Message", _ MessageBoxButtons.OK, MessageBoxIcon.Error)End If


ExerciseExerciseSTRINGS, CHARACTERS AND REGULAR EXPRESSIONS 37

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End of Strings, Characters and End of Strings, Characters and Regular ExpressionsRegular Expressions


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