izvade no mysql

[ziedinjsh]

Sveiki, man nepieciešama palīdzība šādā jautājumā:

 

es ievadu informāciju no feildiem:

 

mysql_query("INSERT INTO users (artist, email, location, genere, pass, pass2, producer_id)VALUES ('$_POST[artist]','$_POST','$_POST[location]','$_POST[genere]','$_POST[pass]','$_POST[pass2]','$_POST[producer_id]')");

 

bet kāviņu izvadīt? atpakaļ php kodā?

[yeahz]

$row = mysql_fetch_array(mysql_query("SELECT * FROM users WHERE id=$id"));

echo "<table>";
echo "<tr><td>artist</td><td>email</td><td>location</td><td>genere</td><td>pass</td><td>pass2</td><td>producer_id</td></tr>";
echo "<tr><td>$row[artist]</td><td>$row[email]</td><td>$row[location]</td><td>$row[genere]</td><td>$row[pass]</td><td>$row[pass2]</td><td>$row[producer_id]</td></tr>";
echo "</table>";

[ziedinjsh]

izveidojas brīdinājums:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in E:\WEB\htdocs\reg.php on line 58

artist email location genere pass pass2 producer_id

[marcis]

http://lv.php.net/mysql_query - piemēri arī tur ir.