1. String Manipulation and More Controls

2. Objectives Determine the number of characters contained in a string
Remove characters from a string
Replace one or more characters in a string
Insert characters within a string
Search a string for one or more characters
Access characters contained in a string
Compare strings
Include radio buttons in an interface
Include check boxes in an interface
1/3/2019
VB 2005

3. Manipulating Strings in Visual Basic
Most applications need to manipulate string data in some fashion
String properties and methods are used to manipulate string data
1/3/2019
VB 2005

4. Determining the Number of Characters Contained in a String
Length Property
Syntax
String.length
Example
NumChar=ziptextbox.text.length
1/3/2019
VB 2005

5. Removing Characters from a String
TrimStart method: removes one or more characters from the beginning of a string
Syntax:
String.TrimStart([trimChars])
TrimEnd method: removes one or more characters from the end of a string
String.TrimEnd([trimChars])
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VB 2005

6. Removing Characters from a String (continued)
Trim method: removes one or more characters from both the beginning and end of a string
Syntax
String.TrimEnd([trimChars])
Each of these methods returns a string with the appropriate characters removed
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VB 2005

7. Removing Characters from a String (continued)
trimChars argument:
Comma-separated list of characters to be removed
If omitted, spaces will be removed
Default value is the space character: “ ”
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VB 2005

8. The Remove Method Index: Remove method:
Removes characters from a string
Can remove one or more characters located anywhere in the string
Returns a string with the appropriate characters removed
Syntax:
string.rRemove(startIndex, count)
Index:
The position of a character in a string
Is zero-relative (starts with 0 as first position)
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VB 2005

9. Replacing Characters in a String
Replace method: replaces a sequence of characters in a string with another sequence of characters
Syntax:
String.Replace (oldValue, newValue)
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VB 2005

10. The Mid Statement Mid statement:
Replaces a specified number of characters in a string with characters from another string
targetString argument: the string in which characters are to be replaced
replacementString argument: the replacement characters
start argument: the starting position for the replacement
count argument: number of characters to replace
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VB 2005

11. The Mid Statement (continued)
Character position of characters in the string starts with 1 (not the same as index)
Count argument is optional
Syntax:
Mid(targetString,start [,count])=ReplacementString
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VB 2005

12. Inserting Characters in a String
PadLeft method: inserts characters at the beginning of a string
Syntax
String.PadLeft(Length[, Character])
PadRight method: inserts characters at the end of a string
String.PadlRight(Length[, Character])
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VB 2005

13. Inserting Characters in a String (continued)
length argument: represents the total length of the desired resulting string
character argument:
The character used to pad the string
Default value is the space character
1/3/2019
VB 2005