dave111 Posted August 16, 2006 Share Posted August 16, 2006 (edited) Warning: fwrite(): supplied argument is not a valid stream resource in /home/***/public_html/admin/froogle.php on line 306 Warning: fclose(): supplied argument is not a valid stream resource in /home/***/public_html/admin/froogle.php on line 307 File completed: frooglefilee.txt Warning: chmod(): No such file or directory in /home/***/public_html/admin/froogle.php on line 309 Connected to uploads.google.com, for user *** uploads.google.com: FTP upload has failed! Heres what the error means: 1) It cant write the file - This is either because the feeds folder it not there, is not set to 777, or the filename is blank. 2) It cant close the file - This is because it was never able to open the file in step 1 3) It cant change permissions of file because, once again, there is no file. 4) It cant upload the file because, you got it, there is no file. So check that the "feeds" folder is located in the main store folder (not inside the admin folder). And also check that the filename you set in the configuration area is not blank. Edited August 16, 2006 by dave111 Quote Link to comment Share on other sites More sharing options...
Guest Posted August 17, 2006 Share Posted August 17, 2006 So check that the "feeds" folder is located in the main store folder (not inside the admin folder). And also check that the filename you set in the configuration area is not blank. That all checks out fine: www/feeds - feeds folder is in the main catalog and permissions 777 filename in the configuration: Froogle FTP Filename frooglefilee.txt Many thanks Kate Quote Link to comment Share on other sites More sharing options...
oscommerceking Posted August 18, 2006 Share Posted August 18, 2006 File completed: froogle.txt Warning: ftp_login(): Login incorrect. in /home/dir/dir_html/catalog/admin/froogle.php on line 333 FTP connection has failed! Attempted to connect to uploads.google.com for user login Please help, any information on how to correct this error line 333 of my froogle.php was not changed from the original install. Thanks Jon Quote My Contributions: Add Search + Drop Down Anywhere, Eliminate Subcategories in Drop Down Link to comment Share on other sites More sharing options...
xearoth Posted August 19, 2006 Share Posted August 19, 2006 Here is my error, can anybody please help me? : SQL error 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 ') left join manufacturers on ( manufacturers.manufacturers_id| sql = SELECT concat( 'http://products.desatech.com/product_info.php?products_id=' ,products.products_id) AS product_url, products_model AS prodModel, products_weight, manufacturers.manufacturers_name AS mfgName, manufacturers.manufacturers_id, products.products_id AS id, products_description.products_name AS name, products_description.products_description AS description, products.products_quantity AS quantity, products.products_status AS prodStatus, FORMAT( IFNULL(specials.specials_new_products_price, products.products_price) * 1,2) AS price, CONCAT( 'http://products.desatech.com/images/' ,products.products_image) AS image_url, products_to_categories.categories_id AS prodCatID, categories.parent_id AS catParentID, categories_description.categories_name AS catName FROM categories, categories_description, products, products_description, products_to_categories) left join manufacturers on ( manufacturers.manufacturers_id = products.manufacturers_id ) left join specials on ( specials.products_id = products.products_id AND ( ( (specials.expires_date > CURRENT_DATE) OR (specials.expires_date = 0) ) AND ( specials.status = 1 ) ) ) WHERE products.products_id=products_description.products_id AND products.products_id=products_to_categories.products_id AND products_to_categories.categories_id=categories.categories_id AND categories.categories_id=categories_description.categories_id ORDER BY products.products_id ASC Quote Link to comment Share on other sites More sharing options...
redguy Posted August 22, 2006 Share Posted August 22, 2006 I am getting an error with Bizrate. For some reason my products with 4 digits are adding a "," ($1,600). Any thoughts. Thanks for your help. Quote Link to comment Share on other sites More sharing options...
redguy Posted August 22, 2006 Share Posted August 22, 2006 Me again. For the time being I am downloading the feed to my desktop and fixing the price column. The problem I am not experiencing is that the bizrate feed is only pulling down 6 columns on the 15 Bizrate/shopzilla requires. Anyone else have this issue? Quote Link to comment Share on other sites More sharing options...
oscommerceking Posted August 22, 2006 Share Posted August 22, 2006 It is apparent that the support for this thread has been discontinued, please direct further questions to the froogle thread. `Jon Quote My Contributions: Add Search + Drop Down Anywhere, Eliminate Subcategories in Drop Down Link to comment Share on other sites More sharing options...
Guest Posted December 12, 2007 Share Posted December 12, 2007 Try this instead of the FTP section (from the other thread, but can't hurt to have it here too): $timeout = 60; $size = filesize($OutFile); $ftp_server = 'ftp://' . $ftp_server . ':21/' . $destination_file; $fp = fopen($OutFile, "r"); $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $ftp_server); curl_setopt($ch, CURLOPT_INFILE, $fp); curl_setopt($ch, CURLOPT_INFILESIZE, $size); curl_setopt($ch, CURLOPT_VERBOSE, 1); curl_setopt($ch, CURLOPT_UPLOAD, 1); curl_setopt($ch, CURLOPT_USERPWD, $ftp_user_name . ':' . $ftp_user_pass); curl_setopt($ch, CURLOPT_TIMEOUT, (int)$timeout); $ftp = curl_exec ($ch); if (curl_errno($ch)) { $error_from_curl = sprintf('Error [%d]: %s', curl_errno($ch), curl_error($ch)); // print $error_from_curl; } // print "<pre>"; // print_r(curl_getinfo($ch)); // print "\n\n cURL error number:" .curl_errno($ch); // print "\n\n cURL error:" . curl_error($ch); // print "</pre>"; curl_close ($ch); fclose($fp); Quote Link to comment Share on other sites More sharing options...
xlameee Posted January 3, 2008 Share Posted January 3, 2008 Here is my error, can anybody please help me? : SQL error 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 ') left join manufacturers on ( manufacturers.manufacturers_id| sql = SELECT concat( 'http://products.desatech.com/product_info.php?products_id=' ,products.products_id) AS product_url, products_model AS prodModel, products_weight, manufacturers.manufacturers_name AS mfgName, manufacturers.manufacturers_id, products.products_id AS id, products_description.products_name AS name, products_description.products_description AS description, products.products_quantity AS quantity, products.products_status AS prodStatus, FORMAT( IFNULL(specials.specials_new_products_price, products.products_price) * 1,2) AS price, CONCAT( 'http://products.desatech.com/images/' ,products.products_image) AS image_url, products_to_categories.categories_id AS prodCatID, categories.parent_id AS catParentID, categories_description.categories_name AS catName FROM categories, categories_description, products, products_description, products_to_categories) left join manufacturers on ( manufacturers.manufacturers_id = products.manufacturers_id ) left join specials on ( specials.products_id = products.products_id AND ( ( (specials.expires_date > CURRENT_DATE) OR (specials.expires_date = 0) ) AND ( specials.status = 1 ) ) ) WHERE products.products_id=products_description.products_id AND products.products_id=products_to_categories.products_id AND products_to_categories.categories_id=categories.categories_id AND categories.categories_id=categories_description.categories_id ORDER BY products.products_id ASC I've got the same problem is anybody can fix that please : SQL error Unknown column 'products.manufacturers_id' in 'on clause'| sql = SELECT concat( 'http://my_domain.com/product_info.php?products_id=' ,products.products_id) AS product_url, products_model , products_weight, products_quantity, manufacturers.manufacturers_name AS manufacturer, products.products_id AS id, products_description.products_name AS name, products_description.products_description AS description, FORMAT(products.products_price,2) AS price, CONCAT( 'http://my_domain.com/images/' ,products.products_image) AS image_url, products_to_categories.categories_id AS prodCatID, categories.parent_id AS catParentID, categories_description.categories_name AS catName FROM categories, categories_description, products, products_description, products_to_categories left join manufacturers on ( manufacturers.manufacturers_id = products.manufacturers_id ) WHERE products.products_id=products_description.products_id AND products.products_id=products_to_categories.products_id AND products_to_categories.categories_id=categories.categories_id AND categories.categories_id=categories_description.categories_id AND products.products_status != 0 ORDER BY products.products_id ASC That is crazy Thank you Nikolay Quote Link to comment Share on other sites More sharing options...
Guest Posted January 6, 2008 Share Posted January 6, 2008 I've got the same problem is anybody can fix that please : SQL error Unknown column 'products.manufacturers_id' in 'on clause'| sql = SELECT concat( 'http://my_domain.com/product_info.php?products_id=' ,products.products_id) AS product_url, products_model , products_weight, products_quantity, manufacturers.manufacturers_name AS manufacturer, products.products_id AS id, products_description.products_name AS name, products_description.products_description AS description, FORMAT(products.products_price,2) AS price, CONCAT( 'http://my_domain.com/images/' ,products.products_image) AS image_url, products_to_categories.categories_id AS prodCatID, categories.parent_id AS catParentID, categories_description.categories_name AS catName FROM categories, categories_description, products, products_description, products_to_categories left join manufacturers on ( manufacturers.manufacturers_id = products.manufacturers_id ) WHERE products.products_id=products_description.products_id AND products.products_id=products_to_categories.products_id AND products_to_categories.categories_id=categories.categories_id AND categories.categories_id=categories_description.categories_id AND products.products_status != 0 ORDER BY products.products_id ASC That is crazy Thank you Nikolay Diddo: : SQL error Unknown column 'products.manufacturers_id' in 'on clause'| sql = SELECT concat( 'http://www.petspl.us/product_info.php?products_id=' ,products.products_id) AS product_url, products_model AS prodModel, products_weight, manufacturers.manufacturers_name AS mfgName, manufacturers.manufacturers_id, products.products_id AS id, products_description.products_name AS name, products_description.products_description AS description, products.products_quantity AS quantity, products.products_status AS prodStatus, FORMAT( IFNULL(specials.specials_new_products_price, products.products_price) * 1,2) AS price, CONCAT( 'http://www.petspl.us/images/' ,products.products_image) AS image_url, products_to_categories.categories_id AS prodCatID, categories.parent_id AS catParentID, categories_description.categories_name AS catName FROM categories, categories_description, products, products_description, products_to_categories left join manufacturers on ( manufacturers.manufacturers_id = products.manufacturers_id ) left join specials on ( specials.products_id = products.products_id AND ( ( (specials.expires_date > CURRENT_DATE) OR (specials.expires_date = 0) ) AND ( specials.status = 1 ) ) ) WHERE products.products_id=products_description.products_id AND products.products_id=products_to_categories.products_id AND products_to_categories.categories_id=categories.categories_id AND categories.categories_id=categories_description.categories_id ORDER BY products.products_id ASC Quote Link to comment Share on other sites More sharing options...
xlameee Posted January 7, 2008 Share Posted January 7, 2008 Diddo: : SQL error Unknown column 'products.manufacturers_id' in 'on clause'| sql = SELECT concat( 'http://www.petspl.us/product_info.php?products_id=' ,products.products_id) AS product_url, products_model AS prodModel, products_weight, manufacturers.manufacturers_name AS mfgName, manufacturers.manufacturers_id, products.products_id AS id, products_description.products_name AS name, products_description.products_description AS description, products.products_quantity AS quantity, products.products_status AS prodStatus, FORMAT( IFNULL(specials.specials_new_products_price, products.products_price) * 1,2) AS price, CONCAT( 'http://www.petspl.us/images/' ,products.products_image) AS image_url, products_to_categories.categories_id AS prodCatID, categories.parent_id AS catParentID, categories_description.categories_name AS catName FROM categories, categories_description, products, products_description, products_to_categories left join manufacturers on ( manufacturers.manufacturers_id = products.manufacturers_id ) left join specials on ( specials.products_id = products.products_id AND ( ( (specials.expires_date > CURRENT_DATE) OR (specials.expires_date = 0) ) AND ( specials.status = 1 ) ) ) WHERE products.products_id=products_description.products_id AND products.products_id=products_to_categories.products_id AND products_to_categories.categories_id=categories.categories_id AND categories.categories_id=categories_description.categories_id ORDER BY products.products_id ASC and still nobody can fix that problem :'( :'( :'( if you tell me just where I have to looked I'll try to fix that anybody has any idea please help or tell me how to unistall this contribution plese Quote Link to comment Share on other sites More sharing options...
Guest Posted January 7, 2008 Share Posted January 7, 2008 Those are two different problems. The first one, with the syntax error, is perhaps due to an version difference in MySQL. I copied and pasted the query into my phpMyAdmin, on a server running MySQL 4.1.20, and it worked fine. What version are you running? Regarding the second, with the unknown column, my guess is that you are using table prefixes. By that I mean you may be using tables named: something_products something_categories something_manufacturers instead of: products categories manufacturers Is that the case? -jared Quote Link to comment Share on other sites More sharing options...
xlameee Posted January 25, 2008 Share Posted January 25, 2008 If you talking about PHPMyAdmin version is 2.10.0.2 but I'm not so good with PHP if anybody has any idea how to unistall the Froogle/Google Base, Yahoo, and Bizrate Feeds Please let me know I'm gonna get something different or somebody to know what contribution to use for OSComnerce v2.2 RC1 please let me know Thank you ion advance Nikolay Those are two different problems. The first one, with the syntax error, is perhaps due to an version difference in MySQL. I copied and pasted the query into my phpMyAdmin, on a server running MySQL 4.1.20, and it worked fine. What version are you running? Regarding the second, with the unknown column, my guess is that you are using table prefixes. By that I mean you may be using tables named: something_products something_categories something_manufacturers instead of: products categories manufacturers Is that the case? -jared Quote Link to comment Share on other sites More sharing options...
Guest Posted February 16, 2008 Share Posted February 16, 2008 Nicolay, On the first screen, phpMyAdmin will give you the MySQL server version as well as the phpMyAdmin version. What is the MySQL server version? -jared Quote Link to comment Share on other sites More sharing options...
sbat Posted March 3, 2008 Share Posted March 3, 2008 Hi everbody, im trying to run Run the feed.sql file in my SLQ, but have few questions. this is what Im inserting INSERT INTO `configuration` VALUES ('', 'Froogle FTP username', 'FROOGLE_FTP_USER', 'froogleusername', 'Froogle FTP username', 62, 1, '2006-06-18 17:49:23', '0000-00-00 00:00:00', NULL, NULL); What is the froogle FTP username? is this the same as the FTP username to get to my own website or another FTP username & password I need to create with/for froogle?? & once I find out what it means, where do I insert it in either in 'FROOGLE_FTP_USER' or 'froogleusername' OR 'Froogle FTP username' Thanks so much Quote Link to comment Share on other sites More sharing options...
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