|
i had readed this artical and applied it , but i still got the same problem what could it be ?
|
|
|
|
|
No idea. I've only seen the problem once and it was corrected by that article.
RageInTheMachine9532
"...a pungent, ghastly, stinky piece of cheese!" -- The Roaming Gnome
|
|
|
|
|
Hi,
I want to find out least value and max value from an array variable. If variable contains Zero then it should be omitted. I hav sorted variable but i am getting value Zero.
Plz help me out.
Thanks & Regards,
Chandan Kumar.
|
|
|
|
|
Post the code for your function and we'll take a look.
...Steve
"Give a man a fish and you've fed him for a day. Teach him how to fish and you've fed him for life." (Translation: I'll show you the way, but not write the code for you.) I read that somewhere once 
|
|
|
|
|
Thanks for replying steve!
I am working on MAP control. I wrote this code in MouseDown event to store X and Y value of 8 map points.
If iVar > 7 Then Exit Sub
ReDim Preserve arrXVal(intCounter)
ReDim Preserve arrYVal(intCounter)
arrXVal(intCounter) = MapX
arrYVal(intCounter) = MapY
intCounter = intCounter + 1
iVar = iVar + 1
What I want to do is I need to retreive Least and Max value of X and Y so that I can set boundry of map & take snapshot. But suppose if any of variable contains value zero then least value would be 0. As a least value I want value other then 0.
plz give me idea how to solve this.
Chandan
|
|
|
|
|
The restriction on the minimum values of X and Y (in this case non-zero) are handled by setting the min and max values to constants defined in the procedure. Note that if legitimate zero values are recorded, they will be changed to whatever MinXValue and MinXValue are set.
Private Sub MinXY(ByVal arrXVal() As Integer, ByVal arrYVal() As Integer)
'
' MinXY will determine the minimum and maximum X and Y values from the
' arrXVal and arrYVal arrays.
'
' The mimimum values must be greater than zero. The default mimimum and
' maximum values are determined by the constants at the start of this
' procedure.
'
Const MinXValue = 1 ' this is the minimum permitted value for X
Const MaxXValue = 9999 ' this is the maximum permitted value for X
Const MinYValue = 1
Const MaxYValue = 9999
Dim maxX As Integer = MinXValue
Dim minX As Integer = MaxXValue
Dim maxY As Integer = MinYValue
Dim minY As Integer = MaxYValue
For i As Integer = 0 To arrxval.GetUpperBound(0)
Select Case arrXVal(i)
Case MinXValue To MaxXValue
'
' If the value of arrXVal(i) is LESS than
' the currently stored mimimum X value in minX, then
' set minX to the value of arrXVal(i).
'
If arrXVal(i) < minX Then
minX = arrXVal(i)
End If
'
' If the value of arrXVal(i) is GREATER than
' the currently stored maximum X value in maxX, then
' set maxX to the value of arrXVal(i).
'
If arrXVal(i) > maxX Then
maxX = arrXVal(i)
End If
Case Else
'
' All other values of arrXVal() are ignored.
'
End Select
'
' Same process for maxY and minY values...
'
Select Case arrYVal(i)
Case MinYValue To MaxYValue
'
' If the value of arrYVal(i) is LESS than
' the currently stored mimimum X value in minX, then
' set minX to the value of arrYVal(i).
'
If arrYVal(i) < minY Then
minY = arrYVal(i)
End If
'
' If the value of arrYVal(i) is GREATER than
' the currently stored maximum X value in maxX, then
' set maxX to the value of arrYVal(i).
'
If arrYVal(i) > maxY Then
maxY = arrYVal(i)
End If
Case Else
'
' All other values of arrYVal() are ignored.
'
End Select
Next
'
' The variables minX, minY, maxX and maxY now contain the non-zero
' minimum and maximum values. You'll need to pass these back to the
' caller or via global variables (not recommended).
'
End Sub
...Steve
"Give a man a fish and you've fed him for a day. Teach him how to fish and you've fed him for life." (Translation: I'll show you the way, but not write the code for you.) I read that somewhere once
|
|
|
|
|
'****************************************
Try
'Do Something
Catch ex As Exception
Msgbox ex.Message
End Try
'****************************************
Try
'Do Something
Catch ex As Exception
Throw New Exception(ex.Message)
End Try
'****************************************
Which is better and show me the best and explain why
!Thanks!
!alien!
|
|
|
|
|
alien viper wrote: Which is better and show me the best and explain why
The correct answer sepends what you want to achieve.
There are some good articles on The Code Project about structured exception handling. Have a squizz and all shall be revealed.
...Steve
"Give a man a fish and you've fed him for a day. Teach him how to fish and you've fed him for life." (Translation: I'll show you the way, but not write the code for you.) I read that somewhere once
|
|
|
|
|
First of all, you are using try/catch. So i'm assuming you're using VB.net. If so, then do not use legacy VB functions like "msgbox". Use "MessageBox.Show(..)".
Secondly, your first example displays the message to the user and will continue processing statements after the "End Try". Your second example, re-throws the exceptions back to the caller.
I think both of the examples are bad practice for error handling.
Alternate for Example 1:
Try<br />
'Do Something<br />
Catch ex As Exception<br />
Utilities.HandleError(ex)<br />
End Try
Advantages: You have a central error handling mecahnism. So if 2 weeks from today your boss tells you to start logging the errors to a text file, you wont have to change every "catch" statement but only the handler function. --> Refactoring
Alternate 1 for Example 2:
'Do Something
Advantages: In your example, you simply rethrow an exception without adding more information to it. By doing so, you actually lose critical error information -- such as inner exception and/or stack trace -- which is vital for finding the root of the problem. So simply remove try/catch block entirely.
Alternate 2 for Example 2:
Try<br />
'Do Something<br />
Catch ex As Exception<br />
Throw New Exception("Error opening your file", ex)<br />
End Try
Advantages: The advantage of this approach is that you mask the original exception message (which could be too technical for the user to understand) with your custom message without losing the original error. You're passing "ex" as an inner error so this way the caller can access both the user-friendly message and also the stacktrace of the original exception.
Hope this helps.
- Malhar
|
|
|
|
|
A class for convert HTML documet in Text (no HTML TAG bat only body text)
bay
Gaudenzio
|
|
|
|
|
be... sembra che funzioni
Public Function ConvertiHTML(ByVal html As String) As String
Dim out As String
Dim i As Integer
Dim j As Integer
Dim elenco As String()
Try
If html.Equals("") Then
Return ""
Exit Try
End If
elenco = html.Split(CChar(">"))
For i = 0 To elenco.Length - 1
j = elenco(i).LastIndexOf(CChar("<"))
If j > 0 Then
out = out & elenco(i).Substring(0, j)
End If
Next
elenco = Nothing
Return out
Catch ex As Exception
Throw ex
End Try
End Function
|
|
|
|
|
I want to corp an image iin VB 6.0 by giving the coordinates of the place I need to corp and want to save it to another place.
Any one knows how to do it.Please help
A,Miftha Idroos
Trainee Software Engineer.
HNB. Sri Lanka
|
|
|
|
|
Just .NET code but it may help:
<br />
Bitmap newone = new Bitmap(100,100);<br />
Graphics g4 = Graphics.FromImage(newone);<br />
System.Drawing.Rectangle dstRect = new System.Drawing.Rectangle(0, 0 ,100, 100);<br />
g4.DrawImage(screenGrab, dstRect, 30,40,100,100,System.Drawing.GraphicsUnit.Pixel);<br />
newone.Save("C:/thenewone.bmp");<br />
|
|
|
|
|
hi all,
i am using DBnull.value to insert Null value into the data base.
but it is giving me error as
"Object required". can anybody help me .....
thanx
raj
|
|
|
|
|
U confirm the table name i think the table which u r using is not in the db thats y it is giving such an error.
|
|
|
|
|
M Javed Khan wrote: U
M Javed Khan wrote: u r
M Javed Khan wrote: y
Not everyone on the forum is a native English speaker. Please, in the interests of clarity, type words fully rather than use abbreviations such as the above.
Heck! Even although English is my first language - I have difficulty in reading it.
ColinMackay.net
"Man who stand on hill with mouth open will wait long time for roast duck to drop in." -- Confucius
"If a man empties his purse into his head, no man can take it away from him, for an investment in knowledge pays the best interest." -- Joseph E. O'Donnell
-- modified at 5:06 Friday 3rd February, 2006
|
|
|
|
|
You have to show a more specific error message and the code that causes the error. From what you have shown it is not even possible to tell if it's a compiler error or a runtime error.
---
b { font-weight: normal; }
|
|
|
|
|
hi
you should check your stored procedures.
and use like this
SqlCommand1.Parameters("@Value").Value = DBNull.Value
!alien!
|
|
|
|
|
How do I prevent the user from typing in the datagrid cells. what property of the datagrid needs to be set?
With Best Regards,
Mayur
|
|
|
|
|
|
Thnx a lot Javed.
With Best Regards,
Mayur
|
|
|
|
|
 I've got a Pocket PC (iPaq 2215) running Windows Mobile 2003SE and I just downloaded VB 2005 Express. Will this develop Pocket PC apps? Anyone have an example Browser app for VB05X? I've also downloaded VC#2005X, VC++2005X and VJ#2005X, so if any of these would be better for Pocket PC app development, please let me know. I'm coming from a Delphi/MSAccess background.
Derek Benner
|
|
|
|
|
delphidab wrote: Will this develop Pocket PC apps?
Nope. The Express editions of Visual Studio don't support Mobile development projects. You'll have to get the Professional version of Visual Basic.NET or Visual Studio.NET.
RageInTheMachine9532
"...a pungent, ghastly, stinky piece of cheese!" -- The Roaming Gnome
|
|
|
|
|
hi
i am doing my final year project in vb.net (i am using 2003 version). my project is a learning package which wil demonstrate ospf and eigrp algorithms.
now i can create my routers and endusers by clicking on the buttons at runtime (they are represented by image boxes) and drag and drop them. now i need to set up the links among routers and endusers. ( i want to make them work as in packet tracer software)
i have got some code working for rubber banding but the problem is after i link say 2 routers, i cant get through programming which 2 routers are linked and if i drag one router to a new place on the from, the linked line doesnt go with it, i mean it stays as it was drawn at the first time.
can anyone please help me?
Mahdi H Miraz
Final Year Student
Computer Networks
University of Wales
NEWI Plas Coch Campus
|
|
|
|
|
Create a class that will represent and draw the connector line. It should hold the references of the two objects it needs to connect by a line.
RageInTheMachine9532
"...a pungent, ghastly, stinky piece of cheese!" -- The Roaming Gnome
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
