[SOLVED] Warning: mysql_fetch_array():

[forumnz]

Why am I getting this?

It says this:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/local/www/vhosts/site/httpdocs/Browse/Listing/listinginfo.php on line 5

 

With this as the code:

<?php
include('connectdb.php');

$result = mysql_query("SELECT * FROM `fflists` WHERE `id`={$id} AND `active`='1' LIMIT 1");
$num = mysql_num_rows($result);

while($row = mysql_fetch_array($result))
  {
  $business_name = $row['business_name'];
  $listid = $row['id'];
  $dcr = $row['dcr'];
  $add1 = $row['add1'];
  }
?>

 

Thanks!

[marcus]

mysql error most likely

 

<?php
include ('connectdb.php');

$sql = "SELECT * FROM `fflists` WHERE `id`='" . $id .
    "' AND `active`='1' LIMIT 1";
$result = mysql_query($sql) or die(mysql_error());
$num = mysql_num_rows($result);

while ($row = mysql_fetch_array($result)) {
    $business_name = $row['business_name'];
    $listid = $row['id'];
    $dcr = $row['dcr'];
    $add1 = $row['add1'];
}
?>

[forumnz]

Comes up with:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'AND `active`='1' LIMIT 1' at line 1

[darkfreaks]

<?php
$sql = 'SELECT * FROM `fflists` WHERE `id`='$id ' AND `active`='1' LIMIT 1';?>

[marcus]

That's going to produce an error darkfreaks.

 

$sql = 'SELECT * FROM `fflists` WHERE `id`='.$id .' AND `active`="1" LIMIT 1';

[revraz]

$sql = "SELECT * FROM `fflists` WHERE `id`='$id' AND `active`= 1 LIMIT 1";?>

[phpSensei]

Where is $id ?

[darkfreaks]

<?php
sql = "SELECT * FROM `fflists` WHERE `id`='"$id"'
    AND `active`= '1' LIMIT 1";?>

[marcus]

Still an error for your darkfreaks

[darkfreaks]

corrected it 

[phpSensei]

<?php
sql = "SELECT * FROM `fflists` WHERE `id`='"$id"'
    AND `active`= '1' LIMIT 1";?>

 

I ment where is the value of it.