Please help me with this code

[Jackthumper]

@$pet = mysql_fetch_array(mysql_query("SELECT * FROM pets));

 

[jim_keller]

It's probably best to describe your problem in more detail, but my first guess is that the @ sign shouldn't go before the variable name. if you really need to use it (you probably don't), put it before @mysql_fetch_array().

[Jackthumper]

it show my "Array".

[Pikachu2000]

Some context would be helpful.

[Jackthumper]

heres the page http://toxicpets.co.uk/search.php

it just dosent get the stuff from the database

[litebearer]

As a simple test, try changing

this

@$pet = mysql_fetch_array(mysql_query("SELECT * FROM pets));

to this

$query ="SELECT * FROM pets";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)) {
  echo $row[0] . " - " . $row[1] . "<br>";
}