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 Value
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.