Contains |
Concat |
Empty |
IndexOf |
LastIndexOf |
Insert |
Join |
Length |
Trim |
Split |
Replace |
StartsWith |
Substring |
Char |
PadLeft |
PadRight |
System.String.Contains
Tells you if a string Contains another string:
Dim str1 As String = "Lions, tigers and bears"
Dim str2 As String = "tigers"
Dim str3 As String = "Elephants"
Dim hasIt as Boolean
In this example, the value of hasIt will be TRUE:.
hasIt = str1.Contains(str2)
In this example, the value of hasIt will be FALSE:
hasIt = str1.Contains(str3)
System.String.Concat
Concatenate up to four strings:
NewString = String.Concat (str1, str2)
NewString = String.Concat (str1, str2, str3, str4)
You can concatenate two, three or four strings.
You can also concatenate the strings in a string array.
NewString = String.Concat (strArray)
System.String.Empty
This function returns an empty string.
An empty string is a zero-length string, or "".
xStr = String.Empty
xstr is now "".
System.String.IndexOf
Find the position (or IndexOf), a character or string object in a string.
Dim str1 As String = "Lions, tigers and bears"
Dim str2 As string = "tigers"
Dim str3 As string = "bears"
Dim x As Integer = str1.IndexOf(str2)
The value of x will be 8.
Dim y As Integer = str1.IndexOf(str3)
The value of y will be 19.
If the searched for string is not found, the value returned is -1.
If the string being searched is empty, the value returned is 0.
System.String.LastIndexOf
Finds the last position (or IndexOf), a character or string object in a string.
This is similar to the right function in VB6.
The last position of a string will be the rightmost position.
Dim str1 As String = "No bananas, no bananas today."
Dim str2 As string = "bananas"
Dim x As Integer = str1.LastIndexOf(str2)
The value of x will be 16. This is the rightmost incidence of banana.
If the searched for string is not found, the value returned is -1.
If the string being searched is empty, the value returned is 0.
System.String.Insert(start index, value)
Inserts a string within another string at a specified starting point.
Dim str1 As String = "Lions tigers and bears"
Dim str2 As string = "and "
Dim str3 As string = "Oh My!"
Dim NewString as String = ""
NewString = str1.Insert(7, str2)
NewString will contain the string "Lions and tigers and bears"
NewString = str1.Insert(28,str3)
NewString will contain the string "Lions and tigers and bears Oh My!"
System.String.Join(separator, stringArray)
System.String.Join(separator, stringArray, startIndex, count)
Creates a single string from a string array.
Join also sends a string containing the Separator value for the new string.
Dim strArr() As String = {"Boston", "Chicago", "Detroit", "Memphis", "Denver"}
Dim sep As String = "|"
Dim newString As String
newString = String.Join(sep, strArr)
newString will contain "Boston|Chicago|Detroit|Memphis|Denver"
Join can also specify a starting index in the array and a count of elements to extract.
newString = String.Join(sep, strArr, 2, 3)
This example will join the array starting at the second element and will extract three array elements.
The result will be "Chicago|Detroit|Memphis"
System.String.Length
Find the number of characters in a string.
Dim str1 As String = "Today is Wednesday"
Dim x As Integer = str1.Length
The value of x will be 18.
System.String.Substring(start, length)
Extract characters from the starting position for the specified length.
Dim str1 As String = "Today is Wednesday"
Dim newstr as string = str1.Substring(0, 5)
This will return the string Today.
The start is 0 which is the first position.
5 characters are returned.>br/>
Dim str1 As String = "Today is Wednesday"
Dim newstr as string = str1.Substring(8, 9)
This will return the string Wednesday.
System.String.Trim
System.String.TrimStart
System.String.TrimEnd
Trim removes whitespace from the beginning and end of a string.
TrimStart removes whitespace from the beginning of a string.
TrimEnd removes whitespace from the end of a string.
Dim str1 As String = " Abe Lincoln "
Dim NewString As String = str1.Trim
NewString will be "Abe Lincoln".
Dim str1 As String = " Abe Lincoln "
Dim NewString As String = str1.TrimStart
NewString will be "Abe Lincoln ".
Dim str1 As String = " Abe Lincoln "
Dim NewString As String = str1.TrimEnd
NewString will be " Abe Lincoln".
System.String.Split
Takes a string and breaks it into a string array, based on the specified separator character.
Dim str1 As String = "Boston, Chicago, Detroit, Memphis, Denver"
Dim strArray() As String = str1.Split(",")
This example will create an array (strArray) with these elements:
strArray(0) = "Boston"
strArray(1) = "Chicago"
strArray(2) = "Detroit"
strArray(3) = "Memphis"
strArray(4) = "Denver"
The array was created by splitting the string into elements where the specified separator is found.
In this case, the designated separator character was a comma.
System.String.Replace
Replace a set of characters or a string, from another string.
Dim str1 As String = "Today is Wednesday"
Dim oldValue As String = "Wednesday"
Dim newValue As String = "Thursday"
NewString = str1.Replace(oldValue, newValue)
The value of NewString is "Today is Thursday".
System.String.StartsWith(string)
Determine if a string Starts With a specified character or string.
Dim str1 As String = "<h1> Heading </h1>"
Dim markup as Boolean
Dim value As String = "<"
markup = str1.StartsWith(value)
The value of markup will be TRUE in this example.
System.String.Char (index)
Get the character at the index location in the string.
This uses a zero-based index.
Dim str1 As String = "Happy New Year!"
Dim val As Char
val = str1(0)
The value of val will be H.
val = str1(4)
The value of val will be y.
val = str1(10)
The value of val will be Y.
System.String.PadLeft(width)
System.String.PadLeft(width, character)
System.String.PadRight(width)
System.String.PadRight(width, character)
PadLeft right-aligns the characters in a string and pads the leftside to the specified length.
PadRight left-aligns the characters in a string and pads the rightside to the specific length.
Both PadLeft and PadRight have an optional parameter to specify the character to fill the string.
If this is not specified, the padding will be blanks.
Dim str1 As String = "Happy Birthday"
NewString = str1.PadRight(20)
The value of NewString is "Happy Birthday ".
Dim str1 As String = "Happy Birthday"
NewString = str1.PadRight(5)
The value of NewString is "Happy Birthday"
In this example, the period character is specified.
This will create leading for a Table Of Contents.
Dim str1 As String = "Chapter 1"
NewString = str1.PadRight(15, ".")
The valueNewstring is "......Chapter.1"
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