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.
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
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

Unlike InStr, InStrRev does not return Null: a Null argument is an error.

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.

Common Errors

Error When
5 (&H5), Invalid procedure call or argument start is 0, or less than -1
compare is less than 0, or greater than 1 and not a valid locale identifier
94 (&H5E), Invalid use of Null any argument is Null

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

See Also