H, Right click on solution explorer then select Add option in that New Item. It will open dialog box in that, from last 3 rd option is Application config file. Select that option it will add App.config to solution.

pallavi

Hi rahul, can u explain me u r table relationship? means is your first table have blogid as primary key and second table with blogid as forgien key? can u give me few table content of both tables (2, 3 records for both tables)?

pallavi

Hi jai, set column width and wrap property to true.

pallavi

Hi, you can add row in datatable which is bind to gridview so it will create new row in gridview .

pallavi

hi, can u try query like this .. and tell me what is out put now... select blog_new.blgid,blog_new.blogtitle,blog_new.createddate from blog_new inner join blog_post on blog_new.blgid = blog_post.blgid and blog_new.status= 1 order by blog_post.editeddate desc

pallavi

Hi Rahul, can u show me out put here .. as i dont know if u r using distinct blgid it is showing 3 records to u ? Do u have one record in first table and 3 records in second table with same ids ?

pallavi

Hi, you need to find that control on any event or what ? you can use 2 ways : 1. In code behind itself u can add attribute to control e.g you can use button click event btnAdd.Attributes["onclick"] = "javascript:DisableControl('"+ddlCombo.ClientID+"');"; and in javascript if(document.getElementById(ddvalue).getAttribute("selectedIndex") or u can directly pass selected index ito javascript 2. in your javascript document.getElementById('Combo1').getAttribute ("selectedIndex") will give u control's index but i m not sure.

pallavi

Hi Rahul, if you have same ids how will u get 3 records ..its one record na ..:) see in join, u join in which field is matter a lot .. so if u are joining on blogid it must be unique...and so it will have uique ids no matter title same or diffrent. if u have any query plz ask :)

pallavi

hi Rahul, you are using distinct blgid and join on blgid so even though thr are 3 same blogtitle but if they have diffrent id it will return 3 records

pallavi

hi durga, put OleDbCommand cmd = new OleDbCommand(strQuery, myConnection); cmd.ExecuteNonQuery(); code in foreach loop .. now u will get what u want and put where cluase in your query :-D

pallavi

Hi DKalepu, I think , string str = ((TextBox)dgRow.FindControl("txtPartNo")).Text.ToString(); this line your code giving you problem as you finding control but grid view is having rows of controls so it take last value and gives u .. try to get values for perticular row . try this , myGridView1.Rows[dgRow.RowIndex].FindControl("txtPartNo")).Text.ToString(); i am not sure ..try some R&D on this u will find answer

pallavi

thanx michael

pallavi

thanks Michael can i use exchange server ,lotus notes for reading mail's content in same way as we use smtp.. as i dont know much about exchange server and lotus notes

pallavi

I want to send mail and read mail for from and subject fields from SMTP,Exchange server and lotus note in application, anybody have any idea about this ....any site where i can find solution...i serached alot but not find any good solution...

pallavi

Hi Vasudevan, Thank you for your help ..:)

pallavi

Hi Sajan, first take one textbox , apply required field validator to it ... set its property(control to validate = textbox ,display property =none and error message ) then add validator callout extender to page set its TargetControlID property to control i.e textbox and u done.

pallavi

hi , i want to create newsletter in my webapplication ..can any one tell me idea behind this and site which have examples for creating newsletter .. Thanx in advance ,

pallavi

Hi Magi, consider one example. If my dropdown list have datasource which conatin employee id and employee name. your display field will be employee name and value field will be employee id. so if your datagrid is having datasource in which it contains employee id column. it will show respective name in dropdown list column...

pallavi

Hi Sebastian , I never used command builder ..i use sqlcommand object..check if it works for u ... SqlCommand updatecmd = new SqlCommand(strSql, sqlcon); SqlCommand deletecmd = new SqlCommand(strSql1, sqlcon); SqlCommand insertcmd = new SqlCommand(strSql2, sqlcon); SqlDataAdapter sda = new SqlDataAdapter(); sda.UpdateCommand = updatecmd; //here cmd generater for update sda.DeleteCommand = deletecmd; //here cmd generater for delete sda.InsertCommand = insertcmd ; //here cmd generater for insert sda.Update(dsStorageArea, "StorageArea");

pallavi

hi, how i am using session in my application .. its in c#.. Session["sername"]=TextBox1.Text //its right then on page 2 no need to define variable again Label3.Text = "Welcome " + Session["Username"].ToString(); check ,if this giving u value or not. :)

pallavi