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Hi All
I am getting run time error
Run-time error '424' Object required
I am try to find excel worksheet password through this code
Set xl = CreateObject("Excel.application")
xl.Visible = False
ExcelFile = "D:\Users\Test\Desktop" + "\test1.xls"
On Error GoTo ErrorHandler
For i = 1 To 6
ExcelNword = Conversion.CStr(i)
ExcelPassword = "ABC" + ExcelNword
Set xlBook = xlApp.Workbooks.Open(FileName:=ExcelFile, _
Password:=ExcelPassword)
ErrorHandler:
MsgBox Err.Description, , "Error"
Next i
Can any one help me
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At the beginning of your code you have set the variable xl to point to the application, but later on you use a variable called xlApp .
It's time for a new signature.
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Thanks for reply
i change xlApp to xl.But still i have same problem.
Please help me
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Have you tried to declare the xlBook as a workbook object?
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yes
Please see my full code
Dim xl As New Excel.Application
Dim xlsheet As Excel.Worksheet
Dim xlwbook As Excel.Workbook
Dim ExcelFile As String
Dim ExcelPassword As String
Dim ExcelNword As String
Dim i As Integer
Private Sub Command1_Click()
For i = 1 To 6
Set xl = CreateObject("Excel.application")
xl.Visible = False
ExcelFile = "D:\Users\Test\Desktop" + "\test1.xls"
On Error GoTo ErrorHandler
ExcelNword = Conversion.CStr(i)
ExcelPassword = "ABC" + ExcelNword
Set xlBook = xl.Workbooks.Open(FileName:=ExcelFile, _
Password:=ExcelPassword)
ErrorHandler:
MsgBox Err.Description, , "Error"
xl.Quit
Next i
End Sub
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I am not sure if the location of your xl.quit is destroying the XLApp object.
move it to outside the For Next Loop.
I think the routine will work for the 1st loop, i.e, when i=1, but will fail on successive iterations of the loop.
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i did it but no effect.
Still i have same problem
Please help me
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In that case then if i were you i would stick a breakpoint in at the start and then step through the code line by line until you find exactly where the code is failing.
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Ok i check it through break point.You are right first i check then i post question in forum.
I got where is error.
When i click on Open button then Cursor goes into step by step when cursor came into this line
Set xlBook = xl.Workbooks.Open(FileName:=ExcelFile, _
Password:=ExcelPassword)
Then Message show password is wrong.As we know i use for loop when cursor goes into second time in line
Set xlBook = xl.Workbooks.Open(FileName:=ExcelFile, _
Password:=ExcelPassword)
Then run time error message show.
Please help me
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[Edit2: Original did work, sorry]
Sub Button1_Click()
Dim xl As New Excel.Application
Dim xlsheet As Excel.Worksheet
Dim xlwbook As Excel.Workbook
Dim ExcelFile As String
Dim ExcelPassword As String
Dim i As Integer
ExcelFile = "C:\Users\Dave Auld\Desktop\password_test.xlsx"
Set xl = CreateObject("Excel.application")
xl.Visible = False
On Error Resume Next
For i = 1 To 6
ExcelPassword = "ABC" + CStr(i)
Set xlBook = xl.Workbooks.Open(Filename:=ExcelFile, Password:=ExcelPassword)
If Err.Number > 0 Then
MsgBox "Error: " + Err.Number + ": " + Err.Description, , "Error"
Err = 0
Else
MsgBox "File Opened with Password " + ExcelPassword
Exit For
End If
Next i
xl.Quit
End Sub
Dave
Don't forget to rate messages!Find Me On: Web| Facebook| Twitter| LinkedInWaving? dave.m.auld[at]googlewave.com
modified on Monday, May 17, 2010 6:51 AM
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Thanks you solve my problem.
Sorry for late response
Once again Thanks.
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Your welcome, glad you got it working now.
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What environment are you using? If VB.NET, then you should have a lot of compiler errors anyway...
- You declare a workbook named "xlwbook" and refer to it as "xlbook"
- Declare your i integer inside the procedure
- You declared xl twice (7 instances of Excel running!). If you Dim xl as New Excel.Application that's enough. You shouldn't create a new instance of the Excel object every time your loop runs.
- In .NET you wouldn't need a conversion class. ExcelNword = CStr(i) is enough.
- You should also rather use good old Try... Catch instead of OnError
I gave it a test in VB.NET with then enclosed changed code. Good luck with your further experiments
Michael
Imports Excel = Microsoft.Office.Interop.Excel
Public Class Form1
Dim xl As New Excel.Application
Dim xlwbook As Excel.Workbook
Dim ExcelFile As String
Dim ExcelPassword As String
Dim ExcelNword As String
Private Sub btnStart_Click() Handles btnStart.Click
xl.Visible = False
ExcelFile = "D:\" + "text.xls"
For i = 1 To 6
Try
ExcelNword = CStr(i)
ExcelPassword = "ABC" + ExcelNword
xlwbook = xl.Workbooks.Open(Filename:=ExcelFile, Password:=ExcelPassword)
MsgBox(String.Format("Workbook could be opened with password {0}", ExcelPassword))
Catch ex As Exception
If Err.Number = 1004 Then
MsgBox(String.Format("'{0}' is the wrong password!", ExcelPassword), , "Error")
Else
MsgBox(String.Format("Fehler {0}: {1}", Err.Number.ToString, Err.Description), , "Error")
End If
End Try
Next i
xl.Quit()
xlwbook = Nothing
xl = Nothing
End Sub
End Class
modified on Monday, May 17, 2010 7:06 AM
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I'd say the problem in this code is that you don't increment the name of the file. You set it to "D:\Users\Test\Desktop" + "\test1.xls"
So, it would work the first time, but on the second iteration, it would be comparing the password for workbook 2 with workbook 1.
Did you mean to write:
ExcelFile = "D:\Users\Test\Desktop\test" & i &".xls"
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I don't think he was wanting to increment the file name, but test a sequential number of passwords against the file to find the correct one.
Going by his other post in response to my code listing, the problem is now fixed, so he is a happy bunny.
Cheers,
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Well if that's what he was trying to do, it sure wasn't clear. Why would you open it and if it didn't open, just display the error message? Why display a message at all if it didn't open? You only need to know the correct password, not the incorrect ones. That was very strange code to try to break a password.
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agree, but he's happy!
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Hey Guys,
to first: Sorry for my bad english !!
I have found nothing suitable, despite gegoogle. But maybe someone knows of you the problem.
I have run a VB.Net 3.5 - App with Interbase 6.1. As ADO driver do I use "Firebird ADO.NET Data provider for. NET and Mono 1.7" from
Sean Leyne. (This driver also works with Interbase.)
Now I wanted to switch to Firebird 2.1. The change in the database has
even if folded. All domains, triggers, etc. run from DB.
With my SQL-Console (IB-Expert) can also easily edit the database.
The Connection String is also true:
<Password = masterkey; Database = C: \ XYZ \ DB.FDB; Dialect = 3; User Id = SYSDBA; Data Source = localhost; charset = ISO8859_1>
<br />
...<br />
Public myDataConn As New FbConnection<br />
Public myDataConnString As New FbConnectionStringBuilder<br />
myDataConn.ConnectionString = "Password = masterkey; Database = C:\XYZ\DB.FDB; Dialect = 3; User ID = SYSDBA; Data Source = localhost; charset = ISO8859_1"<br />
myDataConn.Open()<br />
...<br />
But I get the Open() an error message that I do not
can understand .
Error message
"Offset and length of the array are out of range, or
the number is greater than the number of elements from index to end of
Source collection. "(Error Target: Void BlockCopy (System.Array, Int32,
System.Array, Int32, Int32)
With Interbase 6.1 is full working and no problems.
Does anyone know of the problem or you have a tip for the solution?
Thank you in advance for your help
Frank
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You shouldn't post in more than one forum, it is bad form.
You will be answered quicker and better in Quick Answers as it is a specific question.
Thanks.
------------------------------------
I will never again mention that I was the poster of the One Millionth Lounge Post, nor that it was complete drivel. Dalek Dave
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if im being honest never used firebird before but i always double check my connection strings on this site connection strings i've looked and they have a firebird section too
Marc Clifton wrote: That has nothing to do with VB. - Oh crap. I just defended VB!
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Hi,everyone!I need some help about "find" in excel with vb.net.I want to find a text like "name" in each sheet's range(j5:l5),if it found "name" in such range,then exports data from range(J5:Q12) to other sheet.I don't know whether the function "find" could do this,or I need wrote function myself?
Please give me suggestion and help me !THX!
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Try this[^] link. Might help.
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Good Link
------------------------------------
I will never again mention that I was the poster of the One Millionth Lounge Post, nor that it was complete drivel. Dalek Dave
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hey guys..i want to make a vb.net program to let users control a machine arm to draw on a paper based on what he draw in the program form.
basically i want to design a control form with small dots on it(grid)
and every dot represent a coordiantion point of my machine
and a drawing function provide to user to draw anything on the dots,after that the machine can move to specified location based on the dots being link/draw by user.
so first..how can i make small dots on a vb.net form?
second...how can i provide a control button for user to draw line or input text on that form?do i need shape powerpack or something??
third...how can i detect the dots(which represent the coordinates) that had been link/draw by users?
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If you are a beginner then this as a starting project is way too much.
I would suggest you start by searching the CodeProject articles relating to vb.net graphics winforms.
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