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hi
sorry i have no idea about the timer1_click event but if you want tell me what does it do and for what control it is initiated. maybe i will be able to help
best regards
jamil abou khalil
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 Well, I don't know about the click event too  . Ok, I talk about the only one event of the timer control timer1_tick. Sorry again ). Can you help me?
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hi
i am not understanding what u want. first u said that u want to hide some controls when buttons are clicked or visa versa. That is solved by the click event of the button. and then u want to use the timer1_tick event which is fired as i saw on a given interval of time. i can't see what is the common between these information that you gave.
best regards
jamil abou khalil
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I have a panel that changes it's size gradually by using the timer_tick event of the timer control. I want when I press a menu button this panel to show by encreasing it's size and show some of the controls. After that when I press the second menu button the panel hides (as it gradually decreases it's size to panel1.height = 0 and in the very next moment to increase again by showing the other controls). I thought I could take the text property of the buttons (menuItems) in a switch-case construction placed in the timer_tick event but it didn't work. I don't know if there is a simle way. It's a little complicated. Hope it is clear now what I want to do
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hi
u should go to the first soulution that i gave to u; that is; using the button_click event. double click on the menu item and that event will write a function for you and then write a what u need inside it. i can't see another solution for that.
best regards
jamil abou khalil
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Create a global class level variable. On click of menu set the variable to clicked menu name. In the timer_tick event retrive the current menu from the variable.
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Hi, thanks both of you. Now I tryed to solve the problem with a global class variable and it works. It just needs a little fixing. I'll cope with that. thank u guys for the help.
Best Regards..
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by default, the datagrid add a pending new row after i have add my own row, the pending row has a * in front of the first column, is there anyway i can disable this pending new row? or is there anyway i could disable the user to add a new row on the datagrid itself?since the dataentry is done by another routine
Below is the screenshot
http://img.photobucket.com/albums/v608/mjay83/datagrid.jpg[^]
-- modified at 3:07 Thursday 2nd November, 2006
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hi
you can simply set the allowUserToAddRow property of the data grid to false.
this will eliminate the row of the * that u don't like and will allow the user of that interface to only view the table and the data in it.
hope it hleps
if you need more help plz contact me on MSN: jamilaboukhalil@hotmail.com
Jamil Abou Khalil
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Hi Jamil, thanks for reply
but i cant find the property u mentioned
i am using VS 2003, please advise 
Thanks a again
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hi
sorry i don't know about vs2003 but i am sure that this property is one of the properties of the datagrid of vs2005. I am sorry about that.
p.s: u should upgrade to vs2005 it is more relaiable and more friendly and have more functionalities.
best regards
jamil abou khalil
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Thanks for reply again
It is ok, i dun think my company gonna purchase license for VS 2005 though
Thanks anyway;)
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hi again,
i think i have the solution for you. in the columns property of the datagarid you can navigate throught the columns that appear in the datagrid view. and for each column there is some properties on the right hand side of the window that appear when you click on the columns property. there u can set the column to be frozen. this means that the column is read only.
tell me if this works
best regards
jamil abou khalil
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try to use the read only property of the grid
When you get mad...THINK twice that the only advice
Tamimi - Code
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Hi Tamimi,
The Read only mode wont work for me becoz there is some column i would like the user can edit it on the fly..
thanks for reply anyway
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trust me, there is no way to do this...........
i tried to do this several times with no luck
When you get mad...THINK twice that the only advice
Tamimi - Code
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i am using a treeView control in which i want to display image against its every node. the images are placed in project directiory's \bin\release folder. now the tree is popupulated when the form is loaded i.e the statements of populating the tree are after calling
InitializeComponent();
in the constructor of new form. at first time when i come 2 this form (calling it against the NEXT button of a prevous form) then no problems occur and the images are loaded without any complication but when i go back to a previous form and come back to this form through same path and using same NEXT button then fileNotFound exeception occurs at this line
Image img = Image.FromFile("pic1.jpg");
but previously images were displayed successfully...!
i am facing same problem when i rightclick and create a contextMenu against a node. images are displayed against each optin of contextMenu successfully at first but when i click some option of contextMenu then work n then rightclick same node then the same exeception fileNotFound occurs(while populating ContextMenu) at same sort of lines...
further more if i am debugging the project then these fileNotFoundException occurs even for the first time when i populate the tree.but when i run the project no error comes for the first time and the images are loaded successfully....but only for first time..!
i canot understand where the problem lies??
May God help u solv ur problems as well,thx in advance!!
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I program a programing which can enforce another program. How can I do? Please help me ,thank so much!
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you can use the system.diagnostic.process.start(fullpath) and put the full path of the second program as an argument.
this will start the second program
Jamil Abou Khalil
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thank for your help.But I don't understand clearly, can you give me source code for reference.Thank you!
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For example, I design a button which when I push it, a program (eg : sound recorder)run but run background not display.
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if you want to run any system recognized Program then proceed as
Process.Start("wordpad.exe");
else if you want to run your Program then proceed as
Process.Start(ExePath+"\\wordpad.exe");
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Use : System.Diagnostic.Process.Start(@"C:\My Temp\MyApplication.exe");
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