InStrRev Function

Returns the position of an occurrence of one string within another, from the end of string.

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

Arguments

string1: Required. String expression being searched.
string2: Required. String expression being searched for.
start: Optional. Numeric expression that sets the starting position for each search. If omitted, -1 is used, which means that the search begins at the last character position. If start contains Null, an error occurs.
compare: Optional. Numeric value indicating the kind of comparison to use when evaluating substrings. If omitted, a binary comparison is performed. See Settings section for values.

Settings

The compare argument can have the following values:

Constant	Value	Description
vbBinaryCompare	0	Perform a binary comparison.
vbTextCompare	1	Perform a textual comparison.

Return Values

InStrRev returns the following values:

If	InStrRev returns
string1 is zero-length	0
string1 is Null	Null
string2 is zero-length	start
string2 is Null	Null
string2 is not found	0
string2 is found within string1	Position at which match is found
start > Len(string2)	0

Remarks

The following examples use the InStrRev function to search a string:

Dim SearchString, SearchChar, MyPos
SearchString ="XXpXXpXXPXXP"   ' String to search in.
SearchChar = "P"   ' Search for "P".
MyPos = InstrRev(SearchString, SearchChar, 10, 0)   ' A binary comparison starting at position 10. Returns 9.
MyPos = InstrRev(SearchString, SearchChar, -1, 1)   ' A textual comparison starting at the last position. Returns 12.
MyPos = InstrRev(SearchString, SearchChar, 8)   ' Comparison is binary by default (last argument is omitted). Returns 0.

Note The syntax for the InStrRev function is not the same as the syntax for the InStr function.

Requirements