The VBA FileAttr function returns an integer, representing the mode (or system file handle) of a file that has been opened using the VBA Open statement.
The syntax of the function is:
Where the arguments are as follows:
| FileNumber | - | The file number associated to the file that you are querying. | ||||||
| [ReturnType] | - |
A number, representing the type of information that you want to be returned. This can be either:
|
If the [ReturnType] argument is set to 1 (or is omitted), the FileAttr function returns an integer, representing the mode of the specified file. This may be any of the following:
| Value | Mode |
|---|---|
| 1 | Input |
| 2 | Output |
| 4 | Random |
| 8 | Append |
| 32 | Binary |
In the example below, the VBA FileAttr function is used to return the mode of two open files.
|
' Find the modes of two open files.
Dim fmode1 As Integer Dim fmode2 As Integer
' Open the files data1.txt and data2.txt.
Open "C:\Users\John\Documents\data1.txt" For Input As #1 Open "C:\Users\John\Documents\data2.txt" For Output As #2
' Get the mode of the file data1.txt.
' Get the mode of the file data2.txt.fmode1 = FileAttr( 1 ) ' fmode1 is now equal to 1 (indicates file open for Input). fmode2 = FileAttr( 2 ) ' fmode2 is now equal to 2 (indicates file open for Output). |
Note that, in both of the above calls to the FileAttr function, the [ReturnType] argument is omitted and so the default value 1 is used (meaning the mode of the supplied file should be returned).
Therefore, after running the above VBA code:
If the FileAttr function is supplied with a FileNumber that does not exist, you will get the error:
If the [ReturnType] argument is set to 2, and the current system is not a 16-bit system, you will get the error: