Loading ...Description:
Returns the position of the first occurrence of one string within another
Syntax:
InStr([start,]string1,string2[,compare])
| Parameters | Description |
|---|---|
| start | Optional. Default 1 Numeric expression that sets the starting position for each search |
| string1 | Required. String expression being searched |
| string2 | Required. String expression searched for. |
| compare | Optional. Numeric value indicating the kind of comparison to use when evaluating substrings |
Compare Argument:
| Constants | Value | Description |
|---|---|---|
| vbUseCompareOption | -1 | Performs a comparison using the setting of the Option Compare statement |
| vbBinaryCompare | 0 | Default. Perform a binary comparison |
| vbTextCompare | 1 | Perform a textual comparison |
| vbDatabaseCompare | 2 | Microsoft Access only. Performs a comparison based on information in your database |
Return Data Type:
Long
Usage / Example Code:
Dim SearchStr as String, SearchChar as String
SearchStr = "Banyan Seed"
| Statement | Result / Value | Remark / Note |
|---|---|---|
| instr(SearchStr,”b”) | 0 | |
| instr(SearchStr,”B”) | 1 | |
| instr(1,SearchStr,”b”,vbTextCompare) | 1 | is equal to instr(1,SearchStr,”b”,vbTextCompare) |
| instr(SearchStr,”b”,vbTextCompare) | Error | Run-time error ’13′: Type mismatch |
| instr(SearchStr,”e”) | 9 | |
| instr(10,SearchStr,”e”) | 10 | Second occurance |
| instr(“”, “e”) | 0 | |
| instr(4, SearchStr, “”) | 4 | value of start argument |
| instr(Null, “e”) | Null | |
| instr(SearchStr,Null) | Null | |
| xxx | xxx | |
| xxx | xxx |
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