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:
 Copy Code | |
|---|---|
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. |
Requirements
See Also
