$link inside a funtion

[devil_online]

why the variable $link doesn't work inside a funtion?
this is the $link

$link = @mysql_connect( "localhost", $users, $pass ); 

And there is the code that don't work

 mysql_query ( $query, $link) 

What should I do?

thanks

[edwardsbc]

Is $link actually a valid resource? Did you test it to make sure your connection to the database worked? (If it failed you wouldn't have noticed because you suppressed errors with the "@" sign.)

Try something like this:

if(!$link = @mysql_connect( "localhost", $users, $pass )) {
     die('Database connection failed.');
}

[devil_online]

yes it work
it's about the $link can't be in the funtion

thanks

[Plasma]

What you posted is valid php - your going to need to post the entire code in order to explain what problem you are getting.

[devil_online]

Here it goes:

 
function insertname(){ 

$firstnames = array("Jonh", "Paul", "Bill");
$lastnames = array("Simon", "Luck","Guy");
$name = $firstnames[array_rand($firstnames)]." ".$lastnames[array_rand($lastnames)];
$sql = mysql_query("SELECT names from home 
WHERE names = '$name' 
LIMIT 1") or die(mysql_error());
$resource = mysql_num_rows($sql);
print $name;
$query = "INSERT INTO home(names)
    values('$name')";
mysql_query ( $query, $link)

}

thanks

[thorpe]

Just leave $link out, its an optional argument and isn't required.

mysql_query($query);

If you really want it in there you will need to pass it into the function, but as I said, its not needed.

[edwardsbc]

Ah, $link is probably not available in the function's scope. Pass it in:

function insertname($link) {

And then it will probably work. It would be a good idea to verify that it is a valid resource reference within the functions before trying to use it.

[devil_online]

it still gives this error:

Parse error: syntax error, unexpected '}' in C:\(..)\tez1.php on line 20

what i'm doing wrong?

thanks

[thorpe]

Your missing a ; after your query.

[devil_online]

thanks...what a mistake
however now it says:

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\(...)\tez1.php on line 18

why?