[SOLVED] need help in my query

[pixeltrace]

hi,

 

i need help, i have a page wherein i need to display the items in jos_providers table

but i always get a "no data found" message.

initially i have 3 tables

jos_classifieds_providers

jos_classifieds_providers_categories

jos_classifieds_categories.

 

my "id" in jos_classifieds_providers is equal to the "provider" column in my

jos_classifieds_providers_categories and each column has its respective "category" value which is another column

in jos_classifieds_providers_categories

 

now each category item in jos_classifieds_providers_categories

has its respective name in jos_classifieds_categories where the "parent" column is

equal to the "category" column in jos_classifieds_providers_categories

 

i am not sure if my query is working since its not displaying any results

below is my current code

<? 
include 'db_connect.php';

$view = (isset($_GET['view'])) ? $_GET['view'] : "main";

if ($view == "main"){
echo " 
<a href='test.php?id=328&view=sub'>Furniture, Fittings and Decor</a><br>
<a href='test.php?id=354&view=sub'>Interior Design</a><br>
<a href='test.php?id=355&view=sub'>Cleaning and Repairs</a><br>
<a href='test.php?id=367&view=sub'>Home Accessories and Appliances</a><br>
<a href='test.php?id=381&view=sub'>Home Alterations and Additions</a><br>
<a href='test.php?id=388&view=sub'>Architects, Engineering and Construction</a><br>
<a href='test.php?id=391&view=sub'>Home Entertainment</a><br>
<a href='test.php?id=395&view=sub'>Real Estate Agents</a><br>
<a href='test.php?id=397&view=sub'>Property Services</a><br>
<a href='test.php?id=401&view=sub'>Garden and Landscaping</a><br>
";
}
if ($view =="sub"){
$id = $_GET['id'];
		$uSql = "SELECT * FROM jos_classifieds_categories WHERE parent = '$id' AND published=1";
		$uResult = mysql_query($uSql, $connection);
		if(!$uResult){
		echo 'no data found';
		}else{
		while($uRow = mysql_fetch_array($uResult)){
		$name = $uRow["name"];

echo "<a href='test.php?id=$id&view=subview'>$name</a>" . "<br>";
}
}
}
?>

====================================================<br><br>
<? 
if ($view = "sub"){
$id = $_GET['id'];
		$uSql = "SELECT * FROM jos_classifieds_providers WHERE `id` IN (SELECT DISTINCT(provider) FROM `jos__classifieds_providers_categories` WHERE `category` IN (SELECT DISTINCT(id) FROM `jos__classifieds_categories` WHERE `parent` = $id) ";

		$uResult = mysql_query($uSql, $connection);
		if(!$uResult){
		echo 'no data found';
		}else{
		while($uRow = mysql_fetch_array($uResult)){
		$name = $uRow["name"];

echo "$name $address" . "<br>";
}
}
}elseif ($view = "subview"){
$id = $_GET['id'];
		$uSql = "SELECT * FROM jos_classifieds_providers WHERE `id` IN (SELECT DISTINCT(provider) FROM `jos__classifieds_providers_categories` WHERE `category` IN (SELECT DISTINCT(id) FROM `jos__classifieds_categories` WHERE `parent` = $id) ";


		$uResult = mysql_query($uSql, $connection);
		if(!$uResult){
		echo 'no data found';
		}else{
		while($uRow = mysql_fetch_array($uResult)){
		$name = $uRow["name"];
		$address = $uRow["address"];

echo "$name $address" . "<br>";
}
}
}
?>

 

and this is my test page

http://mango.resonance.com.sg/myoochi/findit/test.php

 

 

hope you could help me with this.

 

thanks!

[cooldude832]

Your error message suggest you don't have a field (index) named id check your tables.

[rameshfaj]

Can u post us the structure of the tables.

[pixeltrace]

hi,

 

its working now, i just had a type error.

 

 

thanks so much for the help!