|
You could reference vjslib which should be listed under the .NET tab of the Add Reference dialog box.
The should then explore the java.util.zip namespace.
The only thing is, I as not sure it is available in the .NET Framework 1.x, but is available in .NET Framework 2
Steve Jowett
-------------------------
It is offen dangerous to try and see someone else's point of view, without proper training. Douglas Adams (Mostly Harmless)
|
|
|
|
|
I found the following article on CodeProject which may be of help http://www.codeproject.com/KB/recipes/cpzipstrip2.aspx[^]. It's in C# but should give you some pointers.
Steve Jowett
-------------------------
It is offen dangerous to try and see someone else's point of view, without proper training. Douglas Adams (Mostly Harmless)
|
|
|
|
|
i need to center align text when printing.( using Printer.print )
Thanks & Regards
|
|
|
|
|
I don't think that it is possible when printing, you have to use printer.currentX to position your cursor and do not forget to take in consideration the lenght of the sentence you write using textwidth
Hope it was helpful
Shay Noy
|
|
|
|
|
I would like to create a open file dialog that can display multiple images in 1 single click. However, i only manage to select an image in 1 time. Could anyone please teach me how to select multiple image per open file dialog? Thanks so much!
Dim OpenFile As OpenFileDialog = New OpenFileDialog
Dim ImgName() As String
Dim LoopCtrl As Integer
OpenFile.Filter = "JPG Images|*.JPG|Bitmaps|*.BMP|GIF Images|*.GIF|All Images|*.BMP;*.GIF;*.JPG"
OpenFile.FilterIndex = 10
OpenFile.Multiselect = True
If OpenFile.ShowDialog = Windows.Forms.DialogResult.OK Then
ImgName = OpenFile.FileNames
For LoopCtrl = 0 To ImgName.Length - 1
objImage(LoopCtrl) = Image.FromFile(ImgName(LoopCtrl))
Next
OriginalPic.Image = objImage(0)
OriginalPic1.Image = objImage(1)
End If
This is how i write the code. Could anyone please kindly teach me how to select multiple image per open file dialog? Thanks so much!
|
|
|
|
|
The code which you have written is correct.
This will allow you to open a FileOpen dialog box. In this dialog box you may select multiple image files either by holding down control key, while selecting files or you may simply click on any empty area of dialog box and drag your mouse to select multiple files.
In case if you are facing any problem in selecting multiple files, let us know.
-Ajay.
-------------------------
www.componentone.com
-------------------------
|
|
|
|
|
First of all, thanks for reply. I was holding down the control key while selecting file, but it appeared that the file name changed and only 1 selected instead of both filenames appeared. Any solution to it? Thanks!
|
|
|
|
|
I used same code which has been given by and it is showing two different images on two different picture boxes.
However I would like to inform you about one thing,when we select multiple files using Cntrl key, sequence of file names get changed.
Therefore if you first click on ‘img1.jpg’ then img2.jpg’, it will become ‘img2.jpg’ ‘img1.jpg’.
Therefore make sure that this fact is not creating any confusion.
You may also get into debug mode and check value of ImgName to see how many file names are listed there.
At my end I am getting correct number and list of files.
-Ajay.
-------------------------
www.componentone.com
-------------------------
|
|
|
|
|
Thanks..now i can select multiple file but it appeared some error lines. For the below coding
For LoopCtrl = 0 To ImgName.Length
objImage(LoopCtrl) = Image.FromFile(ImgName(LoopCtrl))
Next LoopCtrl
It come out error "Object reference not set to an instance of an object". How i can solve it?
|
|
|
|
|
Make it -
For LoopCtrl = 0 To ImgName.Length-1
objImage(LoopCtrl) = Image.FromFile(ImgName(LoopCtrl))
Next LoopCtrl
-----------
If you are selecting two files, ImgName.Length will be 2, however index for two elements would be 0 and 1.
Therefore we need to run loop from 0 to ImgName.Length-1.
-Ajay.
-------------------------
www.componentone.com
-------------------------
|
|
|
|
|
Sorry for the trouble, it seems the same problem occur although i changed it as u have stated. Is there any other error?
Dim OpenFile As OpenFileDialog = New OpenFileDialog
Dim ImgName() As String
Dim LoopCtrl As Integer
OpenFile.Filter = "JPG Images|*.JPG|Bitmaps|*.BMP|GIF Images|*.GIF|All Images|*.BMP;*.GIF;*.JPG"
OpenFile.FilterIndex = 10
OpenFile.Multiselect = True
If OpenFile.ShowDialog = Windows.Forms.DialogResult.OK Then
ImgName = OpenFile.FileNames
For LoopCtrl = 0 To ImgName.Length - 1
objImage(LoopCtrl) = Image.FromFile(ImgName(LoopCtrl))
Next LoopCtrl
OriginalPic.Image = objImage(0)
OriginalPic1.Image = objImage(1)
End If
|
|
|
|
|
What are OriginalPic and OriginalPic1?
I also don’t see any declaration of objimage array in your code.
Therefore I would like to how it is being declared and what is size of this array.
In my code its declared as -
Dim objimage(10) As Image
-Ajay.
-------------------------
www.componentone.com
-------------------------
|
|
|
|
|
Finally get it right. What a mistake i make! Thanks for help!!
|
|
|
|
|
Hello every one..
I am having a VB 2005 application.. Can i use a nested try and catch...
i.e can i have another try can catch block inside a catch block..
I will tell what want to achieve
My application has a distributed DB on 1-4 DB servers (Configurable)
I am reading the Primary Server IP from a text file and then read the remaining servers IP from the setting table in Primary server...
In a single system setup what i am doing is accessing the DB on the network path...
But when the network is down the i will like to access the same DB as local host
The structure is
'get network IP
Tyr
conn.open
catch
'get alternate server IP
conn.open
end try
My intention is to do error handling of second attempt to connect also. So i wanted to put a try and catch block in catch block
Is there any other or better way
Regards
Nishkarsh
|
|
|
|
|
Yes you can but why not put the connection creation in a seperate method and return the connction or nothing, then test the return value and try again if it is nothing
Dim oCon as SQLConnection = nothing
do while ocon is nothing or i > counter
ocon = create connection(sip)
if not ocon is nothing then
exit for
end if
loop
Never underestimate the power of human stupidity
RAH
|
|
|
|
|
I am a newbie to VB platform... i wanted to write a program that can compare two text files and display the common strings along with the number of times the strings are found and the line number !
I heard can use string parsing to do the job.Can anyone help me with some codes ,really appreatiate your help.
|
|
|
|
|
mahan wrote: I heard can use string parsing to do the job
No other way to do it, really.
A full on differencing engine is a lot of work. However, File.ReadAllLines will give you a file as a string array, you can easily do that for two files and compare line by line.
Christian Graus
Please read this if you don't understand the answer I've given you
"also I don't think "TranslateOneToTwoBillion OneHundredAndFortySevenMillion FourHundredAndEightyThreeThousand SixHundredAndFortySeven()" is a very good choice for a function name" - SpacixOne ( offering help to someone who really needed it ) ( spaces added for the benefit of people running at < 1280x1024 )
|
|
|
|
|
Function for Comparing Two Files
Source from .net snippet.
Private Function CompareFiles(ByVal file1 As String, ByVal file2 As String) As Boolean
'Set to true if the files are equal; false otherwise
Dim filesAreEqual As Boolean = False
With My.Computer.FileSystem
' Ensure that the files are the same length before comparing them line by line
If .GetFileInfo(file1).Length = .GetFileInfo(file2).Length Then
Using file1Reader As New FileStream(file1, FileMode.Open), _
file2Reader As New FileStream(file2, FileMode.Open)
Dim byte1 As Integer = file1Reader.ReadByte()
Dim byte2 As Integer = file2Reader.ReadByte()
' If byte1 or byte2 is a negative value, we have reached the end of the file
While byte1 > 0 And byte2 > 0
If (byte1 <> byte2) Then
filesAreEqual = False
Exit While
Else
filesAreEqual = True
End If
'Read the next byte
byte1 = file1Reader.ReadByte()
byte2 = file2Reader.ReadByte()
End While
End Using
End If
End With
Return filesAreEqual
End Function
Rajesh B --> A Poor Workman Blames His Tools <--
|
|
|
|
|
1 - this won't compare words
2 - this won't let him do what he asked for, which was to mark words and lines that are the same
Christian Graus
Please read this if you don't understand the answer I've given you
"also I don't think "TranslateOneToTwoBillion OneHundredAndFortySevenMillion FourHundredAndEightyThreeThousand SixHundredAndFortySeven()" is a very good choice for a function name" - SpacixOne ( offering help to someone who really needed it ) ( spaces added for the benefit of people running at < 1280x1024 )
|
|
|
|
|
i have a listbox that contain ip address of the client, if my server can accept multiple client, how can i send data to the selected client.. so that it don't send the data to all client.. any guide..
|
|
|
|
|
Make the listbox multiselect, then send the data to all selected IPs ?
Christian Graus
Please read this if you don't understand the answer I've given you
"also I don't think "TranslateOneToTwoBillion OneHundredAndFortySevenMillion FourHundredAndEightyThreeThousand SixHundredAndFortySeven()" is a very good choice for a function name" - SpacixOne ( offering help to someone who really needed it ) ( spaces added for the benefit of people running at < 1280x1024 )
|
|
|
|
|
hi,
i'm now working in windows application vb.net, i have create one server and the problem is my server application only accept oly one client in one time.. when i start new client, the server accept the new client and the old client getting ignore by the server, is there any guide or step to make my server to receive multiple client in one time..
please help me...
|
|
|
|
|
Hard to say without seeing your code. I mean, it's obvious that it's written now in a way that connecting to more than one client is not possible, but why this is, I can't really say without seeing the code.
Christian Graus
Please read this if you don't understand the answer I've given you
"also I don't think "TranslateOneToTwoBillion OneHundredAndFortySevenMillion FourHundredAndEightyThreeThousand SixHundredAndFortySeven()" is a very good choice for a function name" - SpacixOne ( offering help to someone who really needed it ) ( spaces added for the benefit of people running at < 1280x1024 )
|
|
|
|
|
Thanks for your attention.. it's my bad.. anyway i'm really appreciate that.. thanks again..
|
|
|
|
|
I was asking for more info so I could help, not telling you off. Why apologise and then start a new thread ?
Christian Graus
Please read this if you don't understand the answer I've given you
"also I don't think "TranslateOneToTwoBillion OneHundredAndFortySevenMillion FourHundredAndEightyThreeThousand SixHundredAndFortySeven()" is a very good choice for a function name" - SpacixOne ( offering help to someone who really needed it ) ( spaces added for the benefit of people running at < 1280x1024 )
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
