nead coding for array

[atck]

can give me coding for array...

i use this code [code]while($row = mysql_fetch_array ($result))[/code]
but have error in my browser..

this is error...
[code]Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program files\easyphp1-8\www[/code]

can you help me...

[Stooney]

$result would have to come from a mysql query ex:

[code]$result=@mysql_query("select * from table");[/code]

then while($row = mysql_fetch_array ($result))    would be valid.

Also the query has to be successful. 

[atck]

can you continue my code...
and add your code in my coding.. plz..

[Stooney]

[url=http://www.phpfreaks.com/forums/index.php/board,8.0.html]http://www.phpfreaks.com/forums/index.php/board,8.0.html[/url]

[sylesia]

Not positive about the while loop part since you have to put in the info you want it to do, but you have to make sure you have the four lines before it to access the whole DB, and when you do, what you do with it

        $conn = mysql_connect("sqlLocation", "userName", "password");
mysql_select_db("DBName", $conn);

        $sql ="SELECT * FROM table";
        $result = mysql_query($sql, $conn);
        while($row = mysql_fetch_array ($result))

[atck]

why must put username and password?..

[sylesia]

its to access the DB.  if there is not a user name or pwd, it would look like
$conn = mysql_connect("sqlLocation", "", "");
so, you only put the username and password there if you actually have one for the DB, for example, if my username for one of my DB is me, and pwd is pwd, so mine looks like
$conn = mysql_connect("sqlLocation", "me", "pwd");