Help me to fix the PHP code

Dear Alls,

I have just studied PHP program. Now I have a problem with my website.

"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in"

And here my code

$result = mysql_query("SELECT a.brand_id as brand_id
a.product_name as product_name
b.brand_name as brand_name
FROM product a, supplier b
Where (a.brand_id=b.brand_id) AND (brand_id=$_GET[id])");// and (channel=($age)");


The idea is that I will click on item from my website with brand_id=15, all products which have brand_id=15 will be displayed on the screen.

Please help me

Thks all & happy new year

Sam

 

this is the correct code $brand_id = $_GET[id];

u seem to have forgot the ";"

 

Add an ; after $_GET[id]

 

Thank you very much

 

<?php

$brand_id = $_GET['id'];
$result = mysql_query("SELECT * FROM supplier WHERE brand_id = $brand_id");

?>

Code (PHP):

OR

<?php

$brand_id = $_GET['id'];
$result = mysql_query("SELECT * FROM supplier WHERE brand_id = $brand_id AND channel = $age");

?>

Code (PHP):

 

New one (

"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in"

And here my code

$result = mysql_query("SELECT a.brand_id as brand_id
a.product_name as product_name
b.brand_name as brand_name
FROM product a, supplier b
Where (a.brand_id=b.brand_id) AND (brand_id=$_GET[id])");// and (channel=($age)");


The idea is that I will click on item from my website with brand_id=15, all products which have brand_id=15 will be displayed on the screen.

Please help me

Thks all & happy new year

Sam

 

You're not getting that error from the code you posted. Post the full code.

Brandon, both of your solutions would error.

 

Yea both ur codes dont have a ; and therefore will error

 

Fixed. Thanks for letting my know.

 

$result = mysql_query("
	SELECT a.brand_id, a.product_name, b.brand_id, b.brand_name
	FROM producta AS a
	LEFT JOIN productb AS b ON (a.brand_id = b.brand_id)
	WHERE brand_id = $brand_id
");
[CODE]

OR

[CODE=PHP]
$result = mysql_query("
	SELECT a.brand_id, a.product_name, b.brand_id, b.brand_name
	FROM producta AS a
	LEFT JOIN productb AS b ON (a.brand_id = b.brand_id)
	WHERE brand_id = $brand_id
	AND channel = $age
");
[CODE]

For the above to work, both table  producta and productb will have to contain the field brand_id with a matching value to join both together.

Code (PHP):

 

check your mysql... some field is named wrong.