The following table lists the functions that Visual Basic provides in theMicrosoft.VisualBasic.Strings class to search and manipulate strings. They can be regarded as Visual Basic intrinsic functions; that is, you do not have to call them as explicit members of a class, as the examples show. Additional methods, and in some cases complementary methods, are available in theSystem.String class.

You can use theOption Compare statement to set whether strings are compared using a case-insensitive text sort order determined by your system's locale (Text) or by the internal binary representations of the characters (Binary). The default text comparison method isBinary.

Example: UCase

This example uses theUCase function to return an uppercase version of a string.

' String to convert.Dim lowerCase As String = "Hello World 1234"' Returns "HELLO WORLD 1234".Dim upperCase As String = UCase(lowerCase)

Example: LTrim

This example uses theLTrim function to strip leading spaces and theRTrim function to strip trailing spaces from a string variable. It uses theTrim function to strip both types of spaces.

' Initializes string.Dim testString As String = "  <-Trim->  "Dim trimString As String' Returns "<-Trim->  ".trimString = LTrim(testString)' Returns "  <-Trim->".trimString = RTrim(testString)' Returns "<-Trim->".trimString = LTrim(RTrim(testString))' Using the Trim function alone achieves the same result.' Returns "<-Trim->".trimString = Trim(testString)

Example: Mid

This example uses theMid function to return a specified number of characters from a string.

' Creates text string.Dim testString As String = "Mid Function Demo"' Returns "Mid".Dim firstWord As String = Mid(testString, 1, 3)' Returns "Demo".Dim lastWord As String = Mid(testString, 14, 4)' Returns "Function Demo".Dim midWords As String = Mid(testString, 5)

Example: Len

This example usesLen to return the number of characters in a string.

' Initializes variable.Dim testString As String = "Hello World"' Returns 11.Dim testLen As Integer = Len(testString)

Example: InStr

This example uses theInStr function to return the position of the first occurrence of one string within another.

' String to search in.Dim searchString As String = "XXpXXpXXPXXP"' Search for "P".Dim searchChar As String = "P"Dim testPos As Integer' A textual comparison starting at position 4. Returns 6.testPos = InStr(4, searchString, searchChar, CompareMethod.Text)' A binary comparison starting at position 1. Returns 9.testPos = InStr(1, SearchString, SearchChar, CompareMethod.Binary)' If Option Compare is not set, or set to Binary, return 9.' If Option Compare is set to Text, returns 3.testPos = InStr(searchString, searchChar)' Returns 0.testPos = InStr(1, searchString, "W")

Example: Format

This example shows various uses of theFormat function to format values using bothString formats and user-defined formats. For the date separator (/), time separator (:), and the AM/PM indicators (t andtt), the actual formatted output displayed by your system depends on the locale settings the code is using. When times and dates are displayed in the development environment, the short time format and short date format of the code locale are used.

Dim testDateTime As Date = #1/27/2001 5:04:23 PM#Dim testStr As String' Returns current system time in the system-defined long time format.testStr = Format(Now(), "Long Time")' Returns current system date in the system-defined long date format.testStr = Format(Now(), "Long Date")' Also returns current system date in the system-defined long date ' format, using the single letter code for the format.testStr = Format(Now(), "D")' Returns the value of testDateTime in user-defined date/time formats.' Returns "5:4:23".testStr = Format(testDateTime, "h:m:s")' Returns "05:04:23 PM".testStr = Format(testDateTime, "hh:mm:ss tt")' Returns "Saturday, Jan 27 2001".testStr = Format(testDateTime, "dddd, MMM d yyyy")' Returns "17:04:23".testStr = Format(testDateTime, "HH:mm:ss")' Returns "23".testStr = Format(23)' User-defined numeric formats.' Returns "5,459.40".testStr = Format(5459.4, "##,##0.00")' Returns "334.90".testStr = Format(334.9, "###0.00")' Returns "500.00%".testStr = Format(5, "0.00%")

See also

• Keywords
• Visual Basic Runtime Library Members
• String Manipulation Summary
• System.String class methods