asocierea unui camp dintr-o alta tabela

[impelo]

Salutare!
Incerc sa invat PHP-ul si cel mai bine este cand faci ceva util... In felul asta vreau sa invat si eu, insa m-am impotmolit la o treaba... Vreau sa atribui datele din prima tabela celei de-a doua si aici mi-am prins urechile.. Prima tabela #angajati arata astfel:
#angajati
id_angajat
ang_nume
ang_data_n
ang_cnp
ang_serie_bul
ang_nr_bul
ang_elib_de
ang_elib_la
ang_local
ang_jud
ang_str
ang_nr_str
ang_bl
ang_et
ang_ap

Am introdus datele, am adaugat vreo 20 de angajati, acum vreau sa creez adeverintele pentru ei. Cum fac asta? Iata codul pe care l-am creat

                    <form method="post" action="?x=new_adev_gen_add">
                        <fieldset> 
                            <label>Numele si prenumele</label>
                            <select>
                                <option name="angajat_id">Alege</option>
<?php
$host="localhost";
$username="";
$password="";
$db_name="";
$tbl_name="angajati";
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY ang_nume ASC";
$result=mysql_query($sql);
while($rows=mysql_fetch_array($result)){
?>
                                <option value="<?php echo $rows['id_angajat']; ?>"><?php echo $rows['ang_nume']; ?> | <?php echo $rows['ang_cnp']; ?></option>
<?php
}
mysql_close();
?>
                            </select>
                        </fieldset>
                        <fieldset style="width:48%; float:left; margin-right: 3%;"> 
                            <label>Numarul adeverintei</label>
                            <input type="text" style="width:92%;" name="adev_gen_nr">
                        </fieldset>
                        <fieldset style="width:48%; float:left;"> 
                            <label>Data adeverintei</label>
                            <input type="text" style="width:92%;" name="adev_gen_data">
                        </fieldset>
                        <fieldset> 
                            <label>A-i servi la</label>
                            <input type="text" name="adev_gen_serveste">
                        </fieldset>
                        <div class="clear"></div>
                    </form>

cum ar trebui sa creez cea de-a doua tabela?
in principiu, #adev_gen va avea
-id_adev_gen
-angajat_id (din #angajati)
-adev_gen_nr
-adev_gen_data
-adev_gen_serveste

corect? sper sa nu gresesc pana aici, ca asta am facut deja... introducerea datelor stiu s-o fac..

<?php
$host="localhost";
$username="";
$password="";
$db_name=""; 
$tbl_name="adev_gen";
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");
$angajat_id=$_POST['angajat_id '];
$adev_gen_nr=$_POST['adev_gen_nr'];
$adev_gen_data=$_POST['adev_gen_data'];
$adev_gen_serveste=$_POST['adev_gen_serveste'];
$sql="INSERT INTO $tbl_name(angajat_id , adev_gen_nr, adev_gen_data, adev_gen_serveste)VALUES('$angajat_id ', '$adev_gen_nr', '$adev_gen_data', '$adev_gen_serveste')";
$result=mysql_query($sql);
if($result){
echo '<h4 class="alert_success">Angajatul a fost inregistrat cu succes!</h4>';
}
else {
echo '<h4 class="alert_error">Angajatul nu a putut fi inregistrat. Va rugam sa mai incercati!</h4>';
}
mysql_close();
?>

.. apoi sa fac si afisarea... aici cum o fac? iata codul creat de mine..

            <table class="tablesorter" cellspacing="0"> 
            <thead> 
                <tr> 
                       <th></th> 
                    <th>Nume angajat</th> 
                    <th>Nr. adeverinta</th> 
                    <th>Data</th> 
                    <th>Serveste la</th> 
                    <th>Optiuni</th> 
                </tr> 
            </thead> 
            <tbody> 
            <?php
$host="";
$username="";
$password="";
$db_name="";
$tbl_name="adev_gen";
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);
while($rows=mysql_fetch_array($result)){
?>
                <tr> 
                       <td><input type="checkbox"></td> 
                    <td>?????????????????</td> 
                    <td><?php echo $rows['adev_gen_nr']; ?></td> 
                    <td><?php echo $rows['adev_gen_data']; ?></td> 
                    <td><?php echo $rows['adev_gen_serveste']; ?></td>
                    <td>
                      <input type="image" src="images/icn_profile.png" title="Vizualizeaza">
                      <input type="image" src="images/icn_edit.png" title="Editeaza">
                      <input type="image" src="images/icn_trash.png" title="Sterge">
                    </td> 
                </tr> 
                <?php
}
mysql_close();
?>
            </tbody> 
            </table>

Nu stiu cum sa fac sa afiseze numele corespunzator fiecarei adeverinte. Ma puteti ajuta?

Va multumesc anticipat.

[raxiro]

acel select trebuie sa fie facut un un join pe tabele de angajati ceva de genul:

SELECT * FROM angajati,adev_gen where abgajati.id_angajat=adev_gen.angajat_id

Asta e varianta simpla in care ai lista intreaga.

[impelo]

Da... eu in principiu o sa am afisarea la gramada (toate adeverintele adaugate) si afisarea separata (fiecare adeverinta in parte).
Daca ai putea fi mai explicit, nu am inteles foarte bine...cum fac afisarea numelui atribuit fiecarei adeverinte? si celelalte date ale angajatului? Deasemenea cum o sa fac afisarea separata/?
Multumesc mult pentru sfat, o sa mai caut si sper sa ajung la un numitor comun.

[raxiro]

Selectul din ultimul php unde faci afisarea in loc de
$sql="SELECT * FROM $tbl_name";

va fi :
$sql="SELECT * FROM angajati,adev_gen where abgajati.id_angajat=adev_gen.angajat_id";

Pe urma acolo unde ai pus:
<td>?????????????????</td>

va fi :
<td><?php echo $rows['ang_nume']; ?></td>

In acel rows vei avea si coloanele din adeverinta si cele din angajat.
Sper ca e mai clar acum

[impelo]

Nu-mi apare nimic, dar de cec?

...?!

Acum am inteles ce mi-ai aratat, un cod face cat o mie de vorbe. Merci mult. Sa vad cum fac cu afisarea asta...

--- Later Edit --- (ca mi-a fost lene sa apas "Edit")

Desi am pus codul cum trebuie, nu inteleg.. nu-mi afiseaza nimic.. Cred ca gresesc un caracter ceva, dar nu-mi dau seama unde..

<?php
    $host="localhost";
    $username="";
    $password="";
    $db_name="";
    mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
    mysql_select_db("$db_name")or die("cannot select DB");
    $sql="SELECT * FROM angajati,adev_gen where angajati.id_angajat=adev_gen.angajat_id"; 
    $result=mysql_query($sql);
    while($rows=mysql_fetch_array($result)){
?>
<tr> 
    <td><input type="checkbox"></td> 
    <td><?php echo $rows['ang_nume']; ?></td> 
    <td><?php echo $rows['adev_gen_nr']; ?></td> 
    <td><?php echo $rows['adev_gen_data']; ?></td> 
    <td><?php echo $rows['adev_gen_serveste']; ?></td>
</tr> 
<?php
    // close while loop 
    }
    // close connection 
    mysql_close();
?>

[raxiro]

Primul pas e sa rulezi selectul tau sa vezi daca intoarce ceva.
Daca dezvolti local poti folosi "Mysql query browser" daca nu, phpmysqladmin din cpanel.

E posibil ca sql-ul sa nu intoarca nimic pentru ca nu exista adeverinte pentru angajat.