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InStrRev Function

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


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

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.

The compare argument can have the following values:

Constant

Value

Description

vbBinaryCompare

0

Perform a binary comparison.

vbTextCompare

1

Perform a textual comparison.

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

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 Note:

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

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