Error when extracting data from MySQL .. ?

Hi

I am having troubles extracting data from my MySQL database.. The strange thing is that this code has worked before.. I just copied and pasted (changing the nessessary names of cource).

<?php
ini_set("display_errors", true);
error_reporting(-1);

include("connect.php");

$extract = mysql_query("SELECT * FROM suggest");
$numrows = mysql_num_rows($extract);

while ($row = mysql_fetch_array($extract))
{
$url[$i]=$row['url'];
$i++;
}

$id=1;
while($id < count($name))
  {
  echo "The URL is " . $url[$id] . "<br />";
  $id++;
  }



?>

PHP:

And the error msg I get is:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/haukaas/public_html/new/suggest-search-engine/admin/index.php on line 8

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/haukaas/public_html/new/suggest-search-engine/admin/index.php on line 10

Notice: Undefined variable: name in /home/haukaas/public_html/new/suggest-search-engine/admin/index.php on line 17

PHP:

Can anyone help me?

 

Its basically saying your theres a problem with your SQL. Do you have the correct table name specified? (suggest?)

You might want to try adding or die(mysql_error()); onto the end of the SQL statements for more information into the actual problem.

Also.

while($id < count($name))

PHP:

$name is not defined..... it looks like that should be

while($id < count($url))

PHP:

 

You might not be connecting to mysql properly, didn't do mysql_seletct_db() or table doesn't exist.

 

<?php
    ini_set("display_errors", true);
    error_reporting(-1);

    include("connect.php");

    // you have not tested what has been returned,
    // don't assume that the functon has worked
    $extract = mysql_query('SELECT * FROM suggest') || die('Error: ' . mysql_error());
    $numrows = mysql_num_rows($extract);

    // always declare variables
    $i = 0;
    $url = array();
    while ($row = mysql_fetch_array($extract) {
        $url[$i] = $row['url'];
        $i++;
    }

    
    // no idea what you're trying to do here. What is $name?
    $id = 1;
    while ($id < count($name)) {
      echo "The URL is " . $url[$id] . "<br />";
      $id++;
    }
?>

PHP:

 

Ok first of all I was trying to connect to the wrong database (Stupid me! Yes I know!)

I also made some changes to the code and now all the error msg is gone and it is diplaying the last entry in the mysql db.

How can I make it display all entries?

Here is the code:

<?php
ini_set("display_errors", true);
error_reporting(-1);

include("connect.php");

$extract = mysql_query("SELECT * FROM suggest");
$numrows = mysql_num_rows($extract);

while ($row = mysql_fetch_array($extract))
{
$url=$row['url'];
}


  echo "The URL is " . $url . "<br />";

?>

PHP:

 

Because you are replacing the $url in the while loop. put the echo in the while loop also.

 

Yep.

<?php
    ini_set("display_errors", true);
    error_reporting(-1);

    include("connect.php");

    $extract = mysql_query("SELECT * FROM suggest");
    $numrows = mysql_num_rows($extract);

    while ($row = mysql_fetch_array($extract)) {
        echo 'The URL is ' . $row['url'] . '<br />'; 
    }
?>

PHP:

 

Of cource!

Im a newbie so this isnt logic for me yet ...

Thanks alot for the help guys