php + mysql problema

Taigi esme tokia, kad is formos paema duomenis, juos sumeta i duombaze, o vienus duomenis sumestu i txt failiuka. taigi viskas vyksta iki metimo i txt faila. $tk neigauna jokios reiksmes, ir tesiog info kelia i .txt faila. kame bedos cia gali buti?ar negalima tuo paciu prisijungimu tik padejus i duombaze is tos pacios eilutes paimt daleiskim id?

<?php

$con = mysql_connect('localhost', 'v7670_bb', 'bets');

if (!$con)

{

die('Neimanoma prisijungti: ' . mysql_error());

}

mysql_select_db("v7670_bb", $con);

 

$name=mysql_real_escape_string($_POST['name']);

$surname=mysql_real_escape_string($_POST['surname']);

$age=mysql_real_escape_string($_POST['age']);

$pilietybe=mysql_real_escape_string($_POST['pilietybe']);

$country=mysql_real_escape_string($_POST['country']);

$city=mysql_real_escape_string($_POST['city']);

$phone=mysql_real_escape_string($_POST['phone']);

$email=mysql_real_escape_string($_POST['email']);

$patirtis=mysql_real_escape_string($_POST['patirtis']);

$imones=mysql_real_escape_string($_POST['imones']);

$kalbos = serialize($_POST['kalbos']);

 

$sql="INSERT INTO info (name,surname,age,pilietybe,country,city,phone,email,kalbos,imones, patirtis) VALUES ('$name','$surname','$age','$pilietybe','$country','$city','$phone','$email','$kalbos','$imones', '$patirtis')";

if (!mysql_query($sql,$con)) {

die('Klaida: ' . mysql_error());

}

 

$sql2 = "SELECT id FROM info WHERE name = $name";

$tk = mysql_query($sql2,$con);

 

/* $tk neigauna jokios reiksmes. kodel? */

 

 

 

if ($_POST['phone_num_0'])

{

$continue = FALSE;

$i = 0;

while ($continue == FALSE)

{

if (isset($_POST['phone_'.$i]))

{

$myFile = $tk.'.txt';

$fh = fopen($myFile, 'a');

$stringData=$_POST['phone_'.$i] . ' - ' . $_POST['phone_num_'.$i];

fputs($fh, $stringData."\r\n");

$phone = $_POST['phone_'.$i];

$phone_num = $_POST['phone_num_'.$i];

fclose($fh);

}

else

{

$continue = TRUE;

}

$i++;

}

}

 

 

 

echo "Duomenys sėkmingai patalpinti duombazėje.";

 

mysql_close($con);

 

?>

mysql_fetch_array nepamaetei? ar aš čia kažko nesuprantu ? :/

mysql_fetch_array nepamaetei? ar aš čia kažko nesuprantu ? :/

$SQL2 = "SELECT * FROM info WHERE name = $name";

$result = mysql_query($SQL2,$con);

while ($db_field = mysql_fetch_array($result)) {

$tk = $db_field['id'];

}

sita vietoj:

$sql2 = "SELECT id FROM info WHERE name = $name";

$tk = mysql_query($sql2,$con);

?

vistiek neveikia :rolleyes:

Nepamirsai kabuciu? t.y. name='$name'

Nepamirsai kabuciu? t.y. name='$name'

My bad. Dekui.