display search result on a form

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I'm looking for help on how to display results when searching for clients informations with their “surname” on my application
I want to display a new form with multiple results in a List View (clientid, surname, othername) and allow the user to select the one they are looking for

codes in class

public bool searchpersonDetails(string personname)
	    {
           if (surname == string.Empty) 
                {
                    Fom1 frm = new Fom1(personname);
		    frm.Show();
		    personname = surname.ToString();

	        }

            SqlCommand cmd = new SqlCommand();
	    string sqlQuery = null;

            sqlQuery = "select client_id, surname, othername from tblspersonaldetails where surname like '%" + personname + "%';";

            cmd.Connection = conn;
            cmd.CommandText = sqlQuery;
	    cmd.CommandType = System.Data.CommandType.Text;

            SqlDataReader dr = null;
	    dr = cmd.ExecuteReader();
	if (dr.Read()) 
            {
                
		client_id = dr["client_id"].ToString();
                surname = dr["surname"].ToString();
                othername = dr["othername"].ToString();

                return true;
            }

            else
            {
                return false;
            }

        }

    }
}

codes behind button click

C#

private void button1_Click(object sender, EventArgs e)
        {
            try
            {

                getid getidentity = new getid();

                if (getidentity.searchpersonDetails(txtSurname.Text))
                {

                    var
                    _with9 = this;

                    _with9.txtSurname.Text = getidentity.Sname.ToString();
                }

                else
                {
                    MessageBox.Show("Record not found");
                }
            }


            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }

            finally
            {
                // Close data reader object and database connection


                if (conn.State == ConnectionState.Open)
                    conn.Close();
            }

        }
    }
}

codes behind form

public partial class Fom1 : Form
    {

            SqlConnection conn = new SqlConnection();

            public Fom1(string personname)
            {
                InitializeComponent();

                // Add any initialization after the InitializeComponent() call.
                SqlCommand cmd = new SqlCommand();


                string sqlQuery = null;
                //this criteria As String = personName & "%"
                sqlQuery = "select * from tblspersonaldetails where surname like '" + personname + "%" + "' order by surname,othername asc";
                cmd.Connection = conn;
                cmd.CommandText = sqlQuery;
                cmd.CommandType = System.Data.CommandType.Text;
                SqlDataReader dr = null;
                dr = cmd.ExecuteReader();
                while (dr.Read())
                {
                    ListViewItem lst = default(ListViewItem);

                    lst = this.listView1.Items.Add(dr["client_id"].ToString());
                    lst.SubItems.Add(dr["surname"].ToString());
                    lst.SubItems.Add(dr["othername"].ToString());
                }
                cmd.Dispose();

            }

BUT IT IS STILL NOT DISPLAYING ON THE FORM PLease HELP ME OUT

Posted 29-Dec-11 4:06am

mikeoabban

Updated 8-Jan-12 5:14am

v6

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Comments

Wendelius 29-Dec-11 10:09am

What part is the problem, creating the query, using the controls etc?

mikeoabban 30-Dec-11 14:31pm

i just dont understand why the result is not displaying at all on the form

Uday P.Singh 29-Dec-11 10:26am

where are you getting stuck?

mikeoabban 30-Dec-11 14:33pm

the result isnot displaying clientid, surname, othername on the form

RaviRanjanKr 29-Dec-11 13:19pm

[Edited]Code is wrapped in pre tag[/Edited]

3 solutions

Solution 3

Based on your latest comment, there seems to be no code that's actually putting the results in the form. Should you have something like:
...

C#

TextBoxSurName.Text = dr["surname"].ToString();
...

Posted 30-Dec-11 8:52am

Wendelius

Comments

mikeoabban 4-Jan-12 13:07pm

am having this in the class surname = dr["surname"].ToString();
this behind the button _with9.txtSurname.Text = getidentity.Sname.ToString();

Pls where should i put this TextBoxSurName.Text = dr["surname"].ToString();
and how. am confused

Wendelius 4-Jan-12 13:11pm

What I meant is that I don't see the code actually putting the content of dr["surname"] to a control in the window. It looks like you assign the value to a variable but is it also set to a control somewhere?

mikeoabban 4-Jan-12 14:10pm

Ok that is i have write a code to put the content of dr["surname"] to the new form (which is Fom1 )
if so then i will do my best to write the codes though this is my first c#
program

mikeoabban 4-Jan-12 13:41pm

i sometimes get this error message
ExecuteReader requires an open and available Connection. The connection's current state is closed

Wendelius 4-Jan-12 13:54pm

Then you try to use the reader but yo uhave closed the connection. There's no conn.Open() visible in your code so probably you open/close the connection somewhere else. I'd suggest putting a breakpoint on the line where you close the connectionand see when it hit's that breakpoint.

Another possibility is to change the design so that you always open the connection, execute reader, close connection. This way the connectionwould be open only when something is done against the database.

Wendelius 4-Jan-12 13:57pm

Ok, then you simply have closed the connection too early.

I'd suggest putting a breakpoint on the line where you close the connection and to see when the code hits that breakpoint.

Another option is to change the design so that the connection is open only when something is executed against the database. So you would have a connection open, executereader, connection close in every place.

mikeoabban 4-Jan-12 14:16pm

ok i will go give it a try.
i will update you n the outcome
thanks alot

Wendelius 4-Jan-12 14:19pm

You're welcome :)

mikeoabban 8-Jan-12 11:18am

hi, was able to display the data in a data grid view (thnaks to your help)
am trying to display it in a LIST VIEW. can you please help me out

Wendelius 8-Jan-12 11:36am

If you're using a list view, then I think the esiest way is to create a loop for your result set and add a ListViewItem[^] in the list view. After that add the remaining columns on each row to SubItems[^]

mikeoabban 8-Jan-12 12:49pm

i can't believe it, i just displayed the data in a list view.
but after displaying a message pop-up "object reference not set to an
instance of an object" what should i do

Wendelius 8-Jan-12 12:52pm

Use the debugger and go through the code line by line. Before executing each line, investigate the values of the variables. Some of the variables you're trying to use is null

mikeoabban 11-Jan-12 17:41pm

thanks Mika, i now display the list view with the needed data.
if i want the data in the list view to respond to a 'double click'
how do i go about it

Solution 2

use code like

C#

public datatable search(string sirname) //in your class
{
   datatable dt=new datatable();
   new sqldataadaptor("query",con).fill(dt);
   return dt;
}

on btn click

C#

datatable dt=classobject.search(txtsirname); //input by user
 datagrid.datasource=dt; //bind datagrid here

Posted 29-Dec-11 8:05am

devbtl

Updated 29-Dec-11 8:20am

v2

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OriginalGriff · Accepted Answer · 2011-12-29T04:26:00

This isn't difficult, but it is hard to give you code because we don't know how you do things. But the process itself is pretty simple:

1) Retrieve the info from the database.
2) Package it into a suitable structure - a List<T>, or a DataTable (if you use the later, it can be combined with (1) above very easily)
3) Create your new form and add your structure to teh default constructor. So if for example you do use a DataTable, the default constructor gets a single parameter - the DataTable.
4) Add a public property (a getter only) which returns the selected info - id, DataRow, or element of List that you handed in the constructor
5) Add the info to your grid and supply an OK and Cancel button.

Done!
To use it, create an instance of the new form, passing it the info. Call the ShowDialog method of the form instance, and check for OK return. If you get it, get the selected info from the property.