sql - How to properly use a DataGridView in C# -

well, very, broad question. i've done lot of research, , i'm still confused.

so, problem have little c# program connected sql database. through c# interface can insert data sql database (i know because checked in mysqlworkbench, , data there), want see it, edit, update , erase needed; , need use datagridview, so, complicated, i've seen lot of info, , confuses me lot more.

i'm gonna almost-complete copypaste of c# program:

private void form1_shown(object sender, eventargs e)     {         conexion.open();         textbox2.focus();         try         {             dataset ds = new dataset();             mysqldataadapter da = new mysqldataadapter("select cveestado, nombre tbestados", conexion);             da.fill(ds, "filldropdown");             combobox1.displaymember = "nombre";             combobox1.valuemember = "cveestado";             combobox1.datasource = ds.tables["filldropdown"];             conexion.close();         }         catch (exception ex)         {             messagebox.show(ex.message, application.productname, messageboxbuttons.ok, messageboxicon.error);         }      }      private void button1_click(object sender, eventargs e)     {         guardar();     }      private void guardar()     {         if (textbox2.text == "")         {             messagebox.show("ingrese un nombre", "error");             textbox2.focus();         }         else if (combobox1.selecteditem.tostring() == "")         {             messagebox.show("elija un estado", "error");             combobox1.focus();         }         else         {             conexion.open();             try             {                 cmd.commandtext = "insert tbmunicipios (nombre, cveestado) values ('" + textbox2.text + "', '" + combobox1.selectedvalue.tostring() + "')";                 cmd.executenonquery();                 cmd.clone();                 messagebox.show("datos guardados", "mensaje");                 conexion.close();                 textbox2.text = "";                 combobox1.text = "";                 textbox2.focus();             }             catch (exception ex)             {                 messagebox.show(ex.message, application.productname, messageboxbuttons.ok, messageboxicon.error);             }         }     } 

with above display "nombre" {name}, obtain "clave" {id}, in combobox; , want same datagridview; repeat, i´ve seen lot of information, confused me lot more.

and, clear, i'm gonna copy-paste sql code too:

create table tbestados (     cveestado int not null,     nombre varchar (45) not null,     constraint pkcveestado primary key (cveestado) )engine=innodb; create table tbmunicipios (     cvemunicipio int not null auto_increment,     nombre varchar (45) not null,     cveestado int not null,     constraint pkcvemunicipio primary key (cvemunicipio),     constraint fkcveedo foreign key (cveestado) references tbestados (cveestado) )engine=innodb; 

thanks in advance answers :d

bro recommend use datatable instead of dataset , drag datagridview on ui

private void form1_shown(object sender, eventargs e) {     conexion.open();     textbox2.focus();     try     {         datatable dt = new datatable();         mysqldataadapter da = new mysqldataadapter("select cveestado, nombre tbestados", conexion);         da.fill(dt);         //here comes datagridview         datagridview1.datasource = dt;         conexion.close();     }     catch (exception ex)     {         messagebox.show(ex.message, application.productname, messageboxbuttons.ok, messageboxicon.error);     }  } 

and best practice doing call method in form_load event