asocierea unui camp dintr-o alta tabela
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
Cod PHP:
<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..
Cod PHP:
<?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..
Cod PHP:
<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.
