hello again 
this is my new problem: I am trying to write a script in order to compare data between two tables of a db.
[php]
$sql = “SELECT * FROM table1
WHERE table1.id NOT IN
(SELECT table2.id FROM table2 WHERE table1.id=table2.id)”;
$res = mysql_query($sql)or die(“there is no hope any more!!”);
[/php]
The query is ok BUT it dosn’t work…i get the message “there is no hope”!!! 
any ideas?
try
[php]
$sql = “SELECT * FROM table1
WHERE table1.id NOT IN
(SELECT table2.id FROM table2 WHERE table1.id=table2.id)”;
$res = mysql_query($sql)or die(“there is still hope:
”.myscql_error());
[/php]
and fix the query
hi and thanx for the reply
i still get the same message,nothing changed 
did you replace with my codes?
you SHOULD not get the same message it should be different because i changed.
you getting this message “there is no hope any more!!”
with my code you get "there is still hope: "
and on a new line a more descriptive error.
can you post what you getting on the screen.
hi
i get the message : “there is still some hope : No database selected”
this is the code:
[php]
<?php $mysqli=mysql_connect("localhost","username","password","data base"); if (mysqli_connect_errno()){ printf("connection failed"); } else { $sql = "SELECT * FROM table1 WHERE table1.id NOT IN (SELECT table2.id FROM table2 WHERE table1.id=table2.id)"; $res = mysql_query($sql)or die("there is still hope:[/php]
what am i doing wrong?
you not selecting the database.
so before $sql query insert this:
[php]
mysql_select_db(“Database_name”) or die(mysql_error());
[/php]
Great now it works fine…
but why should i select the database? since i have a already set the connection shouldn’t it work properly without this line?