select dropdown with two Sql query

[aspkiddy]

Hi,

I have some difficulty with my 'select'

In my select/menu, I display all the options come from a table (table_db_email) in my database

In this table (tb_code_prmtn11_email) I have two field :
fld_email_id
fld_name_email

It works here is a code

<select name="email_adress_menu" id="email_adress_menu"  class="valid"  onchange="submit()">
        <?php
            
            echo "<option selected=\"selected\" value=''>Choose your name</option>"; 
           

            $req_email_adress_menu =   " select DISTINCT id_email, fld_name_email, fld_adresse_email FROM $table_db_email ORDER BY fld_name_email ";
                      
            $rep_email_adress_menu =  mysql_query($req_email_adress_menu, $cnx) or die( mysql_error() ) ;
            
            
            
            while($show_email_adress_menu = mysql_fetch_assoc($rep_email_adress_menu)) {
               
                
                echo '<option value="'.$show_email_adress_menu['id_email'].'"';
                                    //if($primes==$show_email_adress_menu['fld_name_email']){echo " selected";} //display to select an option!!!!!!!!!!!!!!!!
                                    echo '>'.$show_email_adress_menu['fld_name_email'].'  - '.$show_email_adress_menu['fld_adresse_email'].'</option>';
                                    
                                    
                            
                

            }
        ?>
      </select>

I have some other information come from another table : tb_code_prmtn11 (containing information about people)
I have a few fields :
id_resultat
fld_name
fld_email_id ( FOREIGN KEY (`fld_email_id`) REFERENCES `tb_code_prmtn11_email` (`id_email`) ON DELETE NO ACTION ON UPDATE CASCADE
...

I want to put this menu on another Web page and this select must display all of the options as below( with) BY SELECTING THE OPTION THAT MATCHES THE INFORMATION FOUND IN THE SECOND TABLE : tb_code_prmtn11

How can I do this ?

I know how I can make to my request :

here is a code MySql :

SELECT td.id_resultat,td.fld_email_id,email.fld_name_email
FROM $table_db td 
									   
INNER JOIN $table_db_email email
ON td.fld_email_id = email.id_email 
			
WHERE td.id_resultat=$id

This code works also but I don't know how I can include this second query in my menu... Can you help me please

[Janos™]

Can you try to add some line of code in dropdownlist's selected index changed to get the job done as following.

protected void DropDownList1_SelectedIndexChanged(object sender, EventArgs e) 
    { 
        SqlDataSource1.InsertCommandType = SqlDataSourceCommandType.Text; 
        SqlDataSource1.InsertCommand = "Insert into [trails] (Location) values('" + DropDownList1.SelectedValue + "')"; 
        SqlDataSource1.Insert(); 
    } 
 
please add the folloing line at dropdownlist's tag in aspx page 
 
OnSelectedIndexChanged="DropDownList1_SelectedIndexChanged1" 

[aspkiddy]

Hi Janos,

Thanks for your help...


I made some mistake... Now, my menu works : It display all of the option : with this following code :

<select name="email_menu" id="email_menu" onchange="submit()">
        <?php
            
            echo "<option selected=\"selected\" value=''>Choose your name</option>"; 
           

            $req_email_menu =   " select DISTINCT id_email, fld_nom_email, fld_adresse_email FROM $table_db_email ORDER BY fld_nom_email ";
                      
            $rep_email_menu =  mysql_query($req_email_menu, $cnx) or die( mysql_error() ) ;
            
            
            
            while($show_contenu_email_menu = mysql_fetch_assoc($rep_email_menu)) {
               
                
                echo '<option value="'.$show_contenu_email_menu['id_email'].'"';
                
                
                

                        
                        //if($primes==$show_contenu_email_menu['fld_nom_email']){echo " selected";} // pour afficher la selectionne
                                    echo '>'.$show_contenu_email_menu['fld_nom_email'].'  - '.$show_contenu_email_menu['fld_adresse_email'].'</option>';
                                    
                                    
                            
                

            }
        ?>
      </select>

I get the following code in FireFox :

<select name="email_menu" id="email_menu" onchange="submit()">

        <option selected="selected" value=''>Choose your name</option><option value="tomo">TOTO MONO  - toto.mono@test.com</option><option value="kito">KIKI TOTO - kiki.toto@test.com</option></select>

Now I want display option selected with other option

Here is my new code :

<select name="email_menu" id="email_menu" onchange="submit()">
        <?php
            
            echo "<option selected=\"selected\" value=''>Choose your name</option>"; 
           

            $req_email_menu =   " select DISTINCT id_email, fld_nom_email, fld_adresse_email FROM $table_db_email ORDER BY fld_nom_email ";
                      
            $rep_email_menu =  mysql_query($req_email_menu, $cnx) or die( mysql_error() ) ;
            
            
            
            while($show_contenu_email_menu = mysql_fetch_assoc($rep_email_menu)) {
               
                
                echo '<option value="'.$show_contenu_email_menu['id_email'].'"';
                
                
                
////////***************************************** to display/show item selected
                
                $req_email_adress_menu_for_selected =   "SELECT td.id_resultat,td.fld_email_id,email.fld_nom_email
                                       FROM $table_db td 
                                       
                                       INNER JOIN $table_db_email email
                                       ON td.fld_email_id = email.id_email 
            
                                       WHERE td.id_resultat=$id ";
               $rep_email_adress_menu_for_selected =  mysql_query($req_email_adress_menu_for_selected, $cnx) or die( mysql_error() ) ;
                   while($show_email_adress_menu_for_selected = mysql_fetch_assoc($rep_email_adress_menu_for_selected)) 
                   {
                    if($emailselected==$show_email_adress_menu['fld_email_id']){echo " selected";} //display to select an option!!!!!!!!!!!!!!!!
                    }
                
/////--------------------------------
                        //if($primes==$show_contenu_email_menu['fld_nom_email']){echo " selected";} // pour afficher la selectionne
                                    echo '>'.$show_contenu_email_menu['fld_nom_email'].'  - '.$show_contenu_email_menu['fld_adresse_email'].'</option>';
                                    
                                    
                            
                

            }
        ?>
      </select>

My menu works but all options are selected, however, in reality, there is an option that is selected because in my database, there is one in each but all options are selected, however, in reality, there is an option that is selected because in my database, there is one in each recording

I get the following code in FireFox

<select name="email_menu" id="email_menu" onchange="submit()">

        <option selected="selected" value=''>Choose your name</option><option value="tomo"  selected>TOTO MONO  - toto.mono@test.com</option><option value="kito"  selected>KIKI TOTO - kiki.toto@test.com</option></select>