sql query cant find p.product_id - even though it does exist

[Guest]

http://www.magshop.co.za/index.php?cPath=2

 

hi everyone....

 

having a bit of a problem. i built this site on my local pc, mysql version 3.6, and i had no problems - no errors.

 

then we uploaded to test online, mysql version 5.

 

and now its giving me this error:

 

1054 - Unknown column 'p.products_id' in 'on clause'

select count(p.products_id) as total from products_description pd, products p left join manufacturers m on p.manufacturers_id = m.manufacturers_id, products_to_categories p2c left join specials s on p.products_id = s.products_id where p.products_status = '1' and p.products_id = p2c.products_id and pd.products_id = p2c.products_id and pd.language_id = '1' and p2c.categories_id = '2'

[TEP STOP]

 

and here is the actual query on my products_listing_col.php page:

 

$listing_sql = "select " . $select_column_list . " p.products_id, p.manufacturers_id, p.products_price, p.products_tax_class_id, IF(s.status, s.specials_new_products_price, NULL) as specials_new_products_price, IF(s.status, s.specials_new_products_price, p.products_price) as final_price from " . TABLE_PRODUCTS . " p, " . TABLE_PRODUCTS_DESCRIPTION . " pd, " . TABLE_MANUFACTURERS . " m, " . TABLE_PRODUCTS_TO_CATEGORIES . " p2c left join " . TABLE_SPECIALS . " s on p.products_id = s.products_id where p.products_status = '1' and p.manufacturers_id = m.manufacturers_id and m.manufacturers_id = '" . (int)$HTTP_GET_VARS['manufacturers_id'] . "' and p.products_id = p2c.products_id and pd.products_id = p2c.products_id and pd.language_id = '" . (int)$languages_id . "' and p2c.categories_id = '" . (int)$HTTP_GET_VARS['filter_id'] . "'";

 

we tried to pull the online database to my local site & files, and we get thesame error.

 

now this indicates to me that between the 2 mysql versions, there's a problem, its interpreting the query differently.

 

can anyone please help me to get this to work?

[Jack_mcs]

Yes, it a compatibility problem. You can find the fix here.

 

Jack

[Guest]

its not that difficult i think, but i did find a solotion online.

 

google.com to the rescue ^_^

 

http://mysqldatabaseadministration.blogspo...e-transfer.html