Using Listview to Show Information Saved in the Database

Question

I'm new to Visual Studio 2010 C# and MySQL. I am creating an application where there is a part in there that will show all the information in the database(MySQL) in a listview in c#. I already created the adding part of the data in the database. I have some codes in here but it doesn't work, no information is shown in my listview.

This is the code:

 listViewCompany.Items.Clear(); string cmd = "select company_name, company_other_names, company_contactperson, company_contactperson_position from company"; DBConn db = new DBConn(); DataTable tbl = db.retrieveRecord(cmd); int x = 0; foreach (DataRow row in tbl.Rows) { ListViewItem lv = new ListViewItem(row[0].ToString()); lv.SubItems.Add(row[1].ToString()); lv.SubItems.Add(row[2].ToString()); lv.SubItems.Add(row[3].ToString()); listViewCompany.Items.Add(lv); }

@sean, I meant whether you checked if any data came back, which means that the row count of the datatable should be non-zero. Try putting a breakpoint at your original code and do a quick watch at tbl.Rows. Check its .Count property if there are any rows returned. — Alex R.
– Alex R., Commented Jul 19, 2011 at 6:25

sean · Accepted Answer · 2011-07-19 09:50:37Z

This is what I did and it's working.

 string query = "SELECT * FROM company where company_name Like '" + textBoxSearchCompany.Text + "%'"; listViewCompany.Items.Clear(); DBConn db = new DBConn(); DataTable tbl = db.retrieveRecord(query); int x = 0; foreach (DataRow row in tbl.Rows) { ListViewItem lv = new ListViewItem(row[1].ToString()); lv.SubItems.Add(row[2].ToString()); lv.SubItems.Add(row[28].ToString()); lv.SubItems.Add(row[29].ToString()); listViewCompany.Items.Add(lv); }

You really don't want to do "SELECT *" and then index into all those results. The first time you change your table structure, that code will break. Much better to SELECT just the columns you want.

Ramgy Borja · Accepted Answer · 2017-07-31 06:03:34Z

Make it in safe way

DataTable tbl = new DataTable(); using (var con = new MySqlConnection { ConnectionString = conn.config }) { using (var command = new MySqlCommand { Connection = con }) { if (con.State == ConnectionState.Open) con.Close(); con.Open(); command.CommandText = @"SELECT * FROM company where company_name Like Concat(@search,'%')"; command.Parameters.AddWithValue("@search", textBoxSearchCompany.Text); tbl.Load(command.ExecuteReader()); foreach(DataRow row in tbl.Rows) { ListViewItem lv = new ListViewItem(row[1].ToString()); lv.SubItems.Add(row[2].ToString()); lv.SubItems.Add(row[28].ToString()); lv.SubItems.Add(row[29].ToString()); listView1.Items.Add(lv); } } }

Javed Iqbal · Accepted Answer · 2019-01-04 08:36:28Z

0

Set listview property View Details

listView.View = View.Details;

answered Jan 4, 2019 at 8:36

Javed Iqbal

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