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.
.gif) Note: |
|---|
The syntax for the InStrRev function is not the same as the syntax for the InStr function. |
Version 2
Reference