Description

The InStrRev() function returns the position of the first occurrence of one string within another. The search begins from the end of string, but the position returned counts from the beginning of the string.

The InStrRev() function can return the following values:

• If string1 is "" - InStrRev returns 0
• If string1 is Null - InStrRev returns Null
• If string2 is "" - InStrRev returns start
• If string2 is Null - InStrRev returns Null
• If string2 is not found - InStrRev returns 0
• If string2 is found within string1 - InStrRev returns the position at which match is found
• If start > Len(string1) - InStrRev returns 0

Tip: Also look at the InStr function

This function has the form:

InStrRev(string1,string2 [,start [,compare]])

Parameter	Description
string1	(required) The string to be searched
string2	(required) The string expression to search for
start	(optional) Specifies the starting position For Each search. The search begins at the last character position by default (-1)
compare	(optional) Specifies the string comparison to use. (default = 0) This parameter can have one of the following values: 0 = vbBinaryCompare - Perform a binary comparison 1 = vbTextCompare - Perform a textual comparison

Examples

The following examples show the basic use of this function:

Example 1

txt="This is a beautiful day!"
Msgbox(InStrRev(txt,"beautiful"))

This produces the following result:

Example 2

Finding the letter "i", using different starting positions:

txt="This is a beautiful day!"
Msgbox(InStrRev(txt,"i",-1))
Msgbox(InStrRev(txt,"i",7))

This produces the following result (each line in a separate pop-up message box):

Example 3

Finding the letter "T", with textual, and binary, comparison:

txt="This is a beautiful day!"
Msgbox(InStrRev(txt,"T",-1,1))
Msgbox(InStrRev(txt,"T",-1,0))

This produces the following result (each line in a separate pop-up message box):

Miscellaneous Information

Supported in Version:	2.0