Hi All, I am relativly new to all this and have been wrestling with this problem for the last few hours... I decided it was time to ask for a little help, so HEEEEELP!!

When I try to run this code;

<?php
$category_txt = $_GET['category'];
$_SESSION['category'] = $category_txt;
if ($category_txt=="bar") {$category_disp="Bars";}
if ($category_txt=="hotel") {$category_disp="Hotels";}
if ($category_txt=="restaurant") {$category_disp="Restaurants";}
if ($category_txt=="tours") {$category_disp="Tours";}
$category_yes = "1";
?>

I get an error that says:
"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 '=''' at line 1"

Any help would be greatly appreciated

Chheers,

Dave

There isn't a SQL query in the code you posted.

The error must be coming from somewhere else.

hi david can you please post the sql query since the error is with the query so that you can be helped.

WOW!! Thanks for the awesome response time....could it be here?

<?php
$result = mysql_query("SELECT * FROM advertiser WHERE $category_txt='$category_yes'") or die(mysql_error());
while($row = mysql_fetch_array($result)){
$id = $row['id'];
$name_txt = $row['name'];
$image_txt=$row['image'];
$description_txt=$row['description'];

I ran this in my database and didn't get an error

Are $category_txt and $category_yes being set?

cross check the field names of your advertiser table whether you have a field named '$category_txt' i mean with the ampersand preceding the name. the ampersand might be causing the error.

also check that $category_txt is not a typo. maybe you meant to type category_txt ??? Meaning that you did not want to use a PHP variable there but instead use a db field ??

Hi All,
Thanks for your help... all I had to do was add ?category="whatever" at the end of the URL and it works

Thanks again

<?php
$category_txt = $_GET['category'];
$_SESSION['category'] = $category_txt;

if ($category_txt == "bar")
{
$category_disp="Bars";
}

if ($category_txt=="hotel")
{
$category_disp="Hotels";
}

if ($category_txt=="restaurant")
{
$category_disp="Restaurants";
}

if ($category_txt=="tours")
{
$category_disp="Tours";
}
$category_yes = "1";
?>

According to me till now in that script, there is no error I face,then please mention here full code script,then , I'll definitely solve them.