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Error when extracting data from MySQL .. ?

Posted: Thu Jul 15, 2010 9:44 am
by dominod
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).

Code: Select all

<?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++;
  }



?>
And the error msg I get is:

Code: Select all

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
Can anyone help me? :(

Re: Error when extracting data from MySQL .. ?

Posted: Thu Jul 15, 2010 10:11 am
by dominod
Ok nevermind that problem.. 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:

Code: Select all

<?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 />";

?>

Re: Error when extracting data from MySQL .. ?

Posted: Thu Jul 15, 2010 10:34 am
by dominod
Nevermind, DigitalPoint helped me out :)