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#11
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Hold on, my one answer was inaccurate! If all you're doing on the second
command button is reading the data from the file, open it for Input, not Output! Open "C:\Accounts\Users\" & username & ".txt" For Input As #intFile You'd use Open "C:\Accounts\Users\" & username & ".txt" For Output Shared As #intFile if you were doing both reads and writes in the same routine. -- Doug Steele, Microsoft Access MVP http://www.AccessMVP.com/djsteele Co-author: "Access 2010 Solutions", published by Wiley (no e-mails, please!) "Douglas J. Steele" wrote in message ... "Bob H" wrote in message ... Update: I have used this line to create a text file with given information: Open "C:\Accounts\Users\" + username + ".txt" For Output As #1 Print #1, Text23 Print #1, Text25 I guess I should have highlighted in my previous reply that you should never refer to file handles by hard-coded numbers. While it may work fine for you, if other applications are reading or writing to files at the same time, you can run into problems. You should ALWAYS use the FreeFile function, even if you're positive no other applications will ever be running concurrent with yours. Dim intFile As Integer intFile = FreeFile() Open "C:\Accounts\Users\" & username & ".txt" For Output As #intFile Print #intFile, Text23 Print #intFile, Text25 You should also use & for concatenating text, not +. But now on another cmd button I want access to read that said information in that text file, and grant access. With this line: Open "C:\Accounts\Users\" + username + ".txt" For Output As #2 Input #2, openfile If username.Text = openfile Then Input #2, datafile Try using Open "C:\Accounts\Users\" & username & ".txt" For Output Shared As #2 Access throws up a runtime error 54 'Bad File mode' at Input #2 , openfile openfile Also the information in the previously created text file has been deleted. As you've found, opening a file For Output deletes the previous file if it exists. use For Append. That will create the file if it doesn't already exist, and append to the file if it does. -- Doug Steele, Microsoft Access MVP http://www.AccessMVP.com/djsteele Co-author: "Access 2010 Solutions", published by Wiley (no e-mails, please!) |
#12
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Just for the sake of completeness, Daniel's reply is correct. Somehow the
required spaces didn't appear in the post. username = Me!Text23 & vbNullString If Len(username) = 0 Then MsgBox "Enter a Username !" Else If Len(Me!Text25 & vbNullString) = 0 Then -- Doug Steele, Microsoft Access MVP http://www.AccessMVP.com/DJSteele/AccessIndex.html Co-author: "Access 2010 Solutions", published by Wiley (no private e-mails, please) "Bob H" wrote in message ... Thanks for the coding but now I get runtime error 2465 at the line: username = Me!Text23& vbNullString. Thanks On 04/06/2010 20:21, Douglas J. Steele wrote: There is no App object in Access. Private Sub CreateAcc_Click() Dim intFile As Integer Dim username As String username = Me!Text23& vbNullString If Len(username) = 0 Then MsgBox "Enter a Username !" Else If Len(Me!Text25& vbNullString) = 0 Then MsgBox "Enter a Password !" Else intFile = FreeFile() Open CurrentProject.Path& "\Accounts\Users\"& username& ".txt" For Output As #intFile Print #intFile, Me!Text23 Print #intFile, Me!Text25 Close #intFile End If End If End Sub |
#13
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On 05/06/2010 20:44, Douglas J. Steele wrote:
Hold on, my one answer was inaccurate! If all you're doing on the second command button is reading the data from the file, open it for Input, not Output! Open "C:\Accounts\Users\" & username & ".txt" For Input As #intFile You'd use Open "C:\Accounts\Users\" & username & ".txt" For Output Shared As #intFile if you were doing both reads and writes in the same routine. Ok, this is the code I am using to login to an account, using account creation files. Basically I want Access 2007 to read a previously created file, which has the username and password in it. Once it has been read, then access is granted to the whatever. Private Sub cmdLogin_Click() Dim openfile As String Dim datafile As String Dim intFile As Integer openfile = Text1 datafile = Text2 If Text1 = "" Then MsgBox "Please Enter a Username" Else If Text2 = "" Then MsgBox " Please Enter a password" Else intFile = FreeFile() Open "C:\Accounts\Users\" & Text1 & ".txt" For Input As #intFile Input #2, openfile I am getting a runtime error here as before If Text1 = openfile Then Input #2, datafile If Text2 = datafile Then MsgBox "Granted Access" Else MsgBox "Denied Access" End If End If Close #2 End If End If End Sub I know there is something wrong with it at that line but I don't know what else to change it to. Thanks |
#14
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You've still got the #2 everywhere in your code! That should be #intFile.
-- Doug Steele, Microsoft Access MVP http://www.AccessMVP.com/djsteele Co-author: "Access 2010 Solutions", published by Wiley (no e-mails, please!) "Bob H" wrote in message ... On 05/06/2010 20:44, Douglas J. Steele wrote: Hold on, my one answer was inaccurate! If all you're doing on the second command button is reading the data from the file, open it for Input, not Output! Open "C:\Accounts\Users\" & username & ".txt" For Input As #intFile You'd use Open "C:\Accounts\Users\" & username & ".txt" For Output Shared As #intFile if you were doing both reads and writes in the same routine. Ok, this is the code I am using to login to an account, using account creation files. Basically I want Access 2007 to read a previously created file, which has the username and password in it. Once it has been read, then access is granted to the whatever. Private Sub cmdLogin_Click() Dim openfile As String Dim datafile As String Dim intFile As Integer openfile = Text1 datafile = Text2 If Text1 = "" Then MsgBox "Please Enter a Username" Else If Text2 = "" Then MsgBox " Please Enter a password" Else intFile = FreeFile() Open "C:\Accounts\Users\" & Text1 & ".txt" For Input As #intFile Input #2, openfile I am getting a runtime error here as before If Text1 = openfile Then Input #2, datafile If Text2 = datafile Then MsgBox "Granted Access" Else MsgBox "Denied Access" End If End If Close #2 End If End If End Sub I know there is something wrong with it at that line but I don't know what else to change it to. Thanks |
#15
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On 06/06/2010 14:49, Douglas J. Steele wrote:
You've still got the #2 everywhere in your code! That should be #intFile. Thanks for your time and help, as that has done the job for me now. |
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