Printing a certain field out of a DB isn't working :-(

[zoobooboozoo]

I've read the newbies sticky, yet it doesn't work. here's the code:

<?php
  $mysql_link=mysql_connect('localhost','root','')
 or die("ERROR: cannot connect to MySQL server.<br>");
  echo "connected successfully to MySQL server.<br>";

if (mysql_select_db('lespaul',$mysql_link))
   echo "DB selected.<br>";
else
   echo "Could not select the DB.<br>";

echo "Model 1111<br>";
$out=mysql_db_query("lespaul", "SELECT name,id FROM models WHERE id>100", $mysql_link);
echo $_POST['out'];

?>

Output:

connected successfully to MySQL server.
DB selected.
Model 1111
PHP Notice: Undefined index: out in C:\Program Files\EasyPHP1-8\www\main.php on line 13 

[Houdini]

Could it be this? echo $_POST['out'];
Maybe you could try echo $out;

[thorpe]

Youve already selected the database so there is no need to use mysql_db_query. To print the results of a query you'd need something like.

$sql = "SELECT name,id FROM models WHERE id>100";
if ($result = mysql_query($sql)) {
  while ($row = mysql_fetch_assoc($result)) {
    print $row['name'];
  }
} else {
  echo mysql_error();
}

I dont understand why you have echo $_POST['out'] there at all.

[zoobooboozoo]

It works guys!
thanks!

@:
may you please explain to me what that peace of code actually do?
I would like to lrean...

c-ya & thx ! 🙂

[zoobooboozoo]

as I said it works, but I want to understand what akk this functions does, can anyone explain it to me plz?

[laserlight]

// Construct the SQL statement
$sql = "SELECT name,id FROM models WHERE id>100";

// Query the database with the SQL statement
// Assign the result of query to $result
if ($result = mysql_query($sql)) {
  // Loop through result set returned
  // The current result row is assigned to $row
  while ($row = mysql_fetch_assoc($result)) {
    // Print the result column named 'name'
    print $row['name'];
  }
} else {
  // Query failed, print the error message
  echo mysql_error();
}