2 replies to this topic

[Ahmos]

hi all
am beginner in php and need your help
in the code i have error

Quote

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\db\db.php on line 11

the code

<? 

$conn = mysql_pconnect('localhost','root','123456');

if(!$conn ){echo mysql_error($conn);}

$db=mysql_select_db('info',$conn);


$sql="SELECT user_name FROM users";

$result=mysql_query($sql,$conn);




while($row= mysql_fetch_array( $result ))

{

	echo $row["user_name"];

	echo "<br>";

	echo $row["user_id"];echo "<br>";

}


?>

note my database name is info and table name is users and assume password 123456

[Alexander]

Your query is returning an error, and thus becomes "false" and cannot be extracted with mysql_fetch_array.

You must check mysql_error() if $result is false.

i.e.

$result=mysql_query($sql,$conn); [COLOR=#000000][COLOR=#007700]
[/COLOR][/COLOR]
if($result == false) {
  echo mysql_error();
  exit;
}

There are two issues with your code:

1: You should use mysql_fetch_assoc to return an associative array, rather than an array of numerical index
2: You do not ask for user_id in your query, so you cannot use it in your while loop. You must also select it, "SELECT user_id,user_name FROM ..."

[Ahmos]

thanks Alexander ,
i will search for mysql_fetch_assoc:blink:
and i will test it :)