How to compare Values from static gridview in asp.net

Question

2.00/5 (1 vote)

See more:

I have inserted values in grid view and i dont want duplicate values in that gridview how it will be?

means if i again insert same values it should show error.

below is my insert code for gridview

C#

rejfields s1 = new rejfields();



s1.Application = ddplAR.SelectedItem.ToString();
s1.Section = ddplRS.SelectedItem.ToString();
data.Add(s1);
ViewState["_data"] = data;

GridView2.DataSource = data;
GridView2.DataBind();

Posted 8-Nov-12 3:25am

vivek886

Updated 8-Nov-12 3:29am

v3

Add a Solution

Comments

adriancs 8-Nov-12 9:30am

data.GetType().ToString()
what do you get?

adriancs 8-Nov-12 9:50am

check if the values exist in data or not.
if yes, throw error
if no, add it.

Sergey Alexandrovich Kryukov 8-Nov-12 10:48am

First of all, did you really make an instance of grid view static? And why?
--SA

vivek886 8-Nov-12 11:39am

I simplify my question.

From my webform i send values to Gridview.
then i send the values from grid to database.

When i send my values to gridview, i want if i have already added particular record to gridview then again it should not be added and remains unique in the gridview.

data.GetType().ToString() returns
System.Collections.Generic.List`1[rejfields]

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Sergey Alexandrovich Kryukov · Answer 1 · 2012-11-08T04:56:00

Solution 1

Please see my comment: there is no need to make control instances static. Do you really understand what is static? Perhaps you did not do it, but just use wrong or ambiguous terminology.

The answer depends on what you bind your grid view with. If this is a relational database table, think about having non-duplicated data in the table.

If situation is more complex, you might want to have an intermediate data source which guarantees uniqueness, such as some dictionary of records, System.Collections.Generic.Dictionary and bind you grid view with the instance of it — dictionary keys are unique.

—SA

Posted 8-Nov-12 4:56am

Sergey Alexandrovich Kryukov

Comments

vivek886 8-Nov-12 11:42am

i do understand what is static. Grid always have same record fixed, no connectivity to database. grid is binded with fixed values or record only. but if sometimes i need to add record once in a while it should check for duplicated record entry in gridview @ binding time.

vivek886 8-Nov-12 11:46am

the system we designed needs this kind of gridview @ someplace compulsary for administrative purpose only.

thanks Sergery for making me think twice again, and again simplify my question

fjdiewornncalwe 8-Nov-12 14:12pm

The DataGrid which holds the data that can be made static in certain situations, but the Control should NEVER be made static.

Sergey Alexandrovich Kryukov 8-Nov-12 14:32pm

Technically speaking, it can be done static and work, but I totally agree if you would NEVER recommend it.
--SA

fjdiewornncalwe 8-Nov-12 14:40pm

I hope I didn't imply that it wouldn't work, because it does and I've ~~seen~~ fixed a few projects where people did that. What I meant to reiterate and support from your answer is that it really is a bad practice because it is akin to using a UI control as a Business object. Cheers. (+5, BTW)

Sergey Alexandrovich Kryukov 8-Nov-12 17:07pm

Thank you, Marcus.
No, you did not imply that; I'm just trying to be more conservative. Using static instance of it is certainly a bad idea.
--SA

vivek886 9-Nov-12 13:09pm

not a bad idea, everyone has different way of designing a system, where needs are particular, and never have to work hypothetically.

Sergey Alexandrovich Kryukov 9-Nov-12 13:20pm

You are right in principle, but in this particular case -- do you think there is a good reason to use a static control member? I don't think so; I think this is apparently the bad practice. I could agree with you on this particular case only if you provided some compelling reason to use such static object, but I doubt you would be able to do so.
--SA

vivek886 · Answer 2 · 2012-11-09T07:05:00

C#

rejfields s1 = new rejfields();

           if(ddplRS.SelectedValue == "0")
           {
               lblDataMsg.Text = "Please select Reject Section.";
           }
           else if (GridView2.Rows.Count.Equals(0))
           {
               s1.Section = ddplRS.SelectedItem.ToString();
               s1.Description = ddplRS0.SelectedItem.ToString();
               data.Add(s1);
               ViewState["_data"] = data;

               GridView2.DataSource = data;
               GridView2.DataBind();
               lblDataMsg.Text = "";


           }
           else
           {
               bool found = false;
               for (int i = 0; i < GridView2.Rows.Count; i++)
               {
                       if (GridView2.Rows[i].Cells[0].Text == ddplRS.SelectedItem.Text)
                       {
                           lblDataMsg.Text = String.Format("Section {0} already added.", GridView2.Rows[i].Cells[0].Text);
                           found = true;
                       }
               }
               if (!found)
               {
                   s1.Section = ddplRS.SelectedItem.ToString();
                   s1.Description = ddplRS0.SelectedItem.ToString();
                   data.Add(s1);
                   ViewState["_data"] = data;

                   GridView2.DataSource = data;
                   GridView2.DataBind();
                   lblDataMsg.Text = "";
               }

           }
       }