InStrRev

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

Syntax: InStrRev( stringcheck, stringmatch [ , start [ , compare ] ] )

stringcheck: required String expression being searched.
stringmatch: 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. start must be -1, or 1 or greater: 0 or any other negative value raises error 5 (Invalid procedure call or argument). 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 below.

The compare argument can have the following values:

Constant	Value	Description
vbUseCompareOption	-1	Performs a comparison by using the setting of the Option Compare statement.
vbBinaryCompare	0	Performs a binary comparison.
vbTextCompare	1	Performs a textual comparison.

Return values: the first row that applies gives the result.

If	InStrRev returns
stringcheck is zero-length	0
start > Len(stringcheck)	0
stringmatch is zero-length	start, or Len(stringcheck) when start is -1
stringmatch is not found	0
stringmatch is found within stringcheck	Position at which match is found

If stringcheck or stringmatch is Null, error 94 (Invalid use of Null) occurs. Unlike InStr, InStrRev does not return Null.

Note

The syntax for the InStrRev function is not the same as the syntax for the InStr function — note the swapped order of the search arguments.

InStrRev will not find an instance of stringmatch unless the position of the end character of stringmatch is less than or equal to start.

Example

This example uses InStrRev to find the last occurrence of a substring.

' The last dot.
Debug.Print InStrRev("a.b.c", ".")       ' 4
' The last dot at or before position 3.
Debug.Print InStrRev("a.b.c", ".", 3)    ' 2
' Not found.
Debug.Print InStrRev("a.b.c", "x")       ' 0

This example shows the end-character rule and the rows of the return-value table that are easy to get wrong.

' "bc" at position 5 ends at 6, after start, so the match at 2 is found.
Debug.Print InStrRev("abcabc", "bc", 5)   ' 2
Debug.Print InStrRev("abcabc", "bc", 6)   ' 5
Debug.Print InStrRev("abc", "c", 4)       ' 0
Debug.Print InStrRev("abc", "")           ' 3
Debug.Print InStrRev("abc", "", 4)        ' 0

On Error Resume Next
Debug.Print InStrRev(Null, "a")
Debug.Print Err.Number                    ' 94

InStrRev

Example

See Also