RE: Within-group aggregate query help please - customers and latest subscription row
2011/10/24 16:31 -0700, Daevid Vincent WHERE cs.customer_id = 7 GROUP BY customer_id Well, the latter line is now redundant. How will you make the '7' into a parameter? -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/mysql?unsub=arch...@jab.org
RE: Within-group aggregate query help please - customers and latest subscription row
Okay, it seems I am learning... slowly... So there needs to be a second WHERE in the sub-select... To get ONE customer's last subscription (0.038s): SELECT c.customer_id, c.email, c.name, c.username, s.subscription_id, s.`date` FROM customers AS c INNER JOIN customers_subscriptions AS s ON c.customer_id = s.customer_id INNER JOIN (SELECT MAX(`date`) AS LastDate, customer_id FROM customers_subscriptions AS cs WHERE cs.customer_id = 7 GROUP BY customer_id ) AS `x` ON s.customer_id = x.customer_id AND s.date = x.LastDate WHERE c.customer_id = 7; To get ALL customers and their last subscription row (1m:28s) SELECT c.customer_id, c.email, c.name, c.username, s.subscription_id, s.`date` FROM customers AS c INNER JOIN customers_subscriptions AS s ON c.customer_id = s.customer_id INNER JOIN (SELECT MAX(`date`) AS LastDate, customer_id FROM customers_subscriptions AS cs GROUP BY customer_id ) AS `x` ON s.customer_id = x.customer_id AND s.date = x.LastDate ORDER BY customer_id LIMIT 10; Thanks to "you know who you are" for pointing me in the right direction. Hopefully this helps someone else. d. > -Original Message- > From: Daevid Vincent [mailto:dae...@daevid.com] > Sent: Monday, October 24, 2011 4:06 PM > To: mysql@lists.mysql.com > Subject: RE: Within-group aggregate query help please - customers and latest > subscription row > > A kind (and shy) soul replied to me off list and suggested this solution, > however, > this takes 28 seconds (that's for a single customer_id, so this is not going > to scale). > Got any other suggestions? :-) > > SELECT > c.customer_id, > c.email, > c.name, > c.username, > s.subscription_id, > s.`date` > FROM > customers AS c > INNER JOIN customers_subscriptions AS s > ON c.customer_id = s.customer_id > INNER JOIN > (SELECT > MAX(`date`) AS LastDate, > customer_id > FROM > customers_subscriptions AS cs > GROUP BY customer_id) AS `x` > ON s.customer_id = x.customer_id > AND s.date = x.LastDate > WHERE c.customer_id = 7; > > There are 781,270 customers (nearly 1 million) and 1,018,092 > customer_subscriptions. > > Our tables have many indexes on pretty much every column and for sure the > ones we use here. > > EXPLAIN says: > > id select_type table typepossible_keys key > key_len refrows Extra > -- --- -- -- --- > --- -- --- --- > 1 PRIMARY c const PRIMARY PRIMARY 4 > const 1 > 1 PRIMARY s ref date,customer_id customer_id 4 > const 2 > 1 PRIMARYALL (NULL)(NULL) > (NULL) (NULL) 781265 Using where > 2 DERIVED cs ALL (NULL)(NULL) > (NULL) (NULL) 1018092 Using temporary; Using filesort > > > -Original Message- > > From: Daevid Vincent [mailto:dae...@daevid.com] > > Sent: Monday, October 24, 2011 1:46 PM > > To: mysql@lists.mysql.com > > Subject: Within-group aggregate query help please - customers and latest > > subscription row > > > > I know this is a common problem, and I've been struggling with it for a > full > > day now but I can't get it. > > > > I also tried a few sites for examples: > > http://www.artfulsoftware.com/infotree/queries.php#101 > > > http://forums.devarticles.com/general-sql-development-47/select-max-datetime > > -problem-10210.html > > > > Anyways, pretty standard situation: > > > > CREATE TABLE `customers` ( > > `customer_id` int(10) unsigned NOT NULL auto_increment, > > `email` varchar(64) NOT NULL default '', > > `name` varchar(128) NOT NULL default '', > > `username` varchar(32) NOT NULL, > > ... > > ); > > > > CREATE TABLE `customers_subscriptions` ( > > `subscriptio
RE: Within-group aggregate query help please - customers and latest subscription row
A kind (and shy) soul replied to me off list and suggested this solution, however, this takes 28 seconds (that's for a single customer_id, so this is not going to scale). Got any other suggestions? :-) SELECT c.customer_id, c.email, c.name, c.username, s.subscription_id, s.`date` FROM customers AS c INNER JOIN customers_subscriptions AS s ON c.customer_id = s.customer_id INNER JOIN (SELECT MAX(`date`) AS LastDate, customer_id FROM customers_subscriptions AS cs GROUP BY customer_id) AS `x` ON s.customer_id = x.customer_id AND s.date = x.LastDate WHERE c.customer_id = 7; There are 781,270 customers (nearly 1 million) and 1,018,092 customer_subscriptions. Our tables have many indexes on pretty much every column and for sure the ones we use here. EXPLAIN says: id select_type table typepossible_keys key key_len refrows Extra -- --- -- -- --- --- -- --- --- 1 PRIMARY c const PRIMARY PRIMARY 4 const 1 1 PRIMARY s ref date,customer_id customer_id 4 const 2 1 PRIMARYALL (NULL)(NULL) (NULL) (NULL) 781265 Using where 2 DERIVED cs ALL (NULL)(NULL) (NULL) (NULL) 1018092 Using temporary; Using filesort > -Original Message- > From: Daevid Vincent [mailto:dae...@daevid.com] > Sent: Monday, October 24, 2011 1:46 PM > To: mysql@lists.mysql.com > Subject: Within-group aggregate query help please - customers and latest > subscription row > > I know this is a common problem, and I've been struggling with it for a full > day now but I can't get it. > > I also tried a few sites for examples: > http://www.artfulsoftware.com/infotree/queries.php#101 > http://forums.devarticles.com/general-sql-development-47/select-max-datetime > -problem-10210.html > > Anyways, pretty standard situation: > > CREATE TABLE `customers` ( > `customer_id` int(10) unsigned NOT NULL auto_increment, > `email` varchar(64) NOT NULL default '', > `name` varchar(128) NOT NULL default '', > `username` varchar(32) NOT NULL, > ... > ); > > CREATE TABLE `customers_subscriptions` ( > `subscription_id` bigint(12) unsigned NOT NULL default '0', > `customer_id` int(10) unsigned NOT NULL default '0', > `date` date NOT NULL default '-00-00', > ... > ); > > I want to show a table where I list out the ID, email, username, and LAST > SUBSCRIPTION. > > I need this data in TWO ways: > > The FIRST way, is with a query JOINing the two tables so that I can easily > display that HTML table mentioned. That is ALL customers and the latest > subscription they have. > > The SECOND way is when I drill into the customer, I already know the > customer_id and so don't need to JOIN with that table, I just want to get > the proper row from the customers_subscriptions table itself. > > SELECT * FROM `customers_subscriptions` WHERE customer_id = 7 ORDER BY > `date` DESC; > > subscription_id processor customer_id date > --- - --- -- > 134126370 chargem 7 2005-08-04 > 1035167192 billme 7 2004-02-08 > > SELECT MAX(`date`) FROM `customers_subscriptions` WHERE customer_id = 7 > GROUP BY customer_id; > > gives me 2005-08-04 obviously, but as you all know, mySQL completely takes a > crap on your face when you try what would seem to be the right query: > > SELECT subscription_id, MAX(`date`) FROM `customers_subscriptions` WHERE > customer_id = 7 GROUP BY customer_id; > > subscription_id MAX(`date`) > --- --- > 1035167192 2005-08-04 > > Notice how I have the correct DATE, but the wrong subscription_id. > > In the example web sites above, they seem to deal more with finding the > MAX(subscription_id), which in my case will not work. > > I need the max DATE and the corresponding row (with matching > subscription_id). > > Thanks, > > d -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/mysql?unsub=arch...@jab.org
Within-group aggregate query help please - customers and latest subscription row
I know this is a common problem, and I've been struggling with it for a full day now but I can't get it. I also tried a few sites for examples: http://www.artfulsoftware.com/infotree/queries.php#101 http://forums.devarticles.com/general-sql-development-47/select-max-datetime -problem-10210.html Anyways, pretty standard situation: CREATE TABLE `customers` ( `customer_id` int(10) unsigned NOT NULL auto_increment, `email` varchar(64) NOT NULL default '', `name` varchar(128) NOT NULL default '', `username` varchar(32) NOT NULL, ... ); CREATE TABLE `customers_subscriptions` ( `subscription_id` bigint(12) unsigned NOT NULL default '0', `customer_id` int(10) unsigned NOT NULL default '0', `date` date NOT NULL default '-00-00', ... ); I want to show a table where I list out the ID, email, username, and LAST SUBSCRIPTION. I need this data in TWO ways: The FIRST way, is with a query JOINing the two tables so that I can easily display that HTML table mentioned. That is ALL customers and the latest subscription they have. The SECOND way is when I drill into the customer, I already know the customer_id and so don't need to JOIN with that table, I just want to get the proper row from the customers_subscriptions table itself. SELECT * FROM `customers_subscriptions` WHERE customer_id = 7 ORDER BY `date` DESC; subscription_id processor customer_id date --- - --- -- 134126370 chargem 7 2005-08-04 1035167192 billme 7 2004-02-08 SELECT MAX(`date`) FROM `customers_subscriptions` WHERE customer_id = 7 GROUP BY customer_id; gives me 2005-08-04 obviously, but as you all know, mySQL completely takes a crap on your face when you try what would seem to be the right query: SELECT subscription_id, MAX(`date`) FROM `customers_subscriptions` WHERE customer_id = 7 GROUP BY customer_id; subscription_id MAX(`date`) --- --- 1035167192 2005-08-04 Notice how I have the correct DATE, but the wrong subscription_id. In the example web sites above, they seem to deal more with finding the MAX(subscription_id), which in my case will not work. I need the max DATE and the corresponding row (with matching subscription_id). Thanks, d -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/mysql?unsub=arch...@jab.org
Fwd: Query help please
I wondered if anyone can help me ? Do you need any further information ? Cheers Neil -- Forwarded message -- From: Tompkins Neil Date: Thu, Sep 23, 2010 at 9:49 AM Subject: Query help please To: "[MySQL]" Hi all, I've the following query : SELECT fixtures_results.seasons_id , home_teams_id AS teams_id , 1 AS home ,0 AS away , (SELECT SUM(goals) FROM players_appearances WHERE fixtures_results.fixtures_results_id = players_appearances.fixtures_results_id AND players_appearances.teams_id = home_teams_id) AS home_goals_aa, IF(home_goals > away_goals, 1, 0) AS won_home , IF(home_goals = away_goals, 1, 0) AS drawn_home , IF(home_goals < away_goals, 1, 0) AS lost_home , home_goals AS scored_home , away_goals AS conceded_home , 0 AS won_away , 0 AS drawn_away , 0 AS lost_away , 0 AS scored_away , 0 AS conceded_away FROM fixtures_results WHERE fixtures_results.competitions_id = 1 AND fixtures_results.seasons_id = 1 AND fixtures_results.status = 'approved' Basically I have a table called player_appearances which contains a SUM of goals for each fixture for the home and away team. How can I use this SUM called "home_goals_aa", in my logic like IF(home_goals > away_goals, 1, 0) AS won_home , IF(home_goals = away_goals, 1, 0) AS drawn_home , IF(home_goals < away_goals, 1, 0) AS lost_home , Cheers Neil
Query help please
Hi all, I've the following query : SELECT fixtures_results.seasons_id , home_teams_id AS teams_id , 1 AS home ,0 AS away , (SELECT SUM(goals) FROM players_appearances WHERE fixtures_results.fixtures_results_id = players_appearances.fixtures_results_id AND players_appearances.teams_id = home_teams_id) AS home_goals_aa, IF(home_goals > away_goals, 1, 0) AS won_home , IF(home_goals = away_goals, 1, 0) AS drawn_home , IF(home_goals < away_goals, 1, 0) AS lost_home , home_goals AS scored_home , away_goals AS conceded_home , 0 AS won_away , 0 AS drawn_away , 0 AS lost_away , 0 AS scored_away , 0 AS conceded_away FROM fixtures_results WHERE fixtures_results.competitions_id = 1 AND fixtures_results.seasons_id = 1 AND fixtures_results.status = 'approved' Basically I have a table called player_appearances which contains a SUM of goals for each fixture for the home and away team. How can I use this SUM called "home_goals_aa", in my logic like IF(home_goals > away_goals, 1, 0) AS won_home , IF(home_goals = away_goals, 1, 0) AS drawn_home , IF(home_goals < away_goals, 1, 0) AS lost_home , Cheers Neil
Re: Query help, please..
Anders, >I also want to find out the user's position relative to others depending on the result. For a given pUserID, something like this? SELECT userid,result,rank FROM ( SELECT o1.userid,o1.result,COUNT(o2.result) AS rank FROM object o1 JOIN object o2 ON o1.result < o2.result OR (o1.result=o2.result AND o1.userid=o2.userid) GROUP BY o1.userid,o1.result ) WHERE userid = pUserID; PB - Anders Norrbring wrote: I'm looking at a situation I haven't run into before, and I'm a bit puzzled by it. I have this table structure: Table USERS: userid, class Table OBJECT: userid, class, result Now I want to query the database for a certain user's result in a specified class, which is very, very easy. No problems. But, I also want to find out the user's position relative to others depending on the result. So, if the specified user's result is the 9:th best of all of the users, I want to have a reply from the DB query that say he has position number 9. I really can't figure out how to do that... Somehow I have to make MySQL calculate the position based on the value in the result column. -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: Query help, please..
On Dec 11, 2007, at 10:46 AM, Rob Wultsch wrote: On Dec 11, 2007 8:38 AM, Anders Norrbring <[EMAIL PROTECTED]> wrote: I'm looking at a situation I haven't run into before, and I'm a bit puzzled by it. I have this table structure: Table USERS: userid, class Table OBJECT: userid, class, result Now I want to query the database for a certain user's result in a specified class, which is very, very easy. No problems. But, I also want to find out the user's position relative to others depending on the result. So, if the specified user's result is the 9:th best of all of the users, I want to have a reply from the DB query that say he has position number 9. I really can't figure out how to do that... Somehow I have to make MySQL calculate the position based on the value in the result column. Take a look at http://arjen-lentz.livejournal.com/55083.html . Very similar ideas in play, though you also have a join. The basic idea is that you do a count on the number of users that have a lower score. Is there any reason you wouldn't want to count the people in front of you and add 1 to get your place in line? It seems like depending on where you are, that may be a shorter number to count :) But I don't know anything about how to do stuff off of separate tables yet still trying to grasp that :) -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED] -- Jason Pruim Raoset Inc. Technology Manager MQC Specialist 3251 132nd ave Holland, MI, 49424 www.raoset.com [EMAIL PROTECTED] -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: Query help, please..
On Dec 11, 2007 8:38 AM, Anders Norrbring <[EMAIL PROTECTED]> wrote: > I'm looking at a situation I haven't run into before, and I'm a bit > puzzled by it. > > I have this table structure: > > Table USERS: userid, class > Table OBJECT: userid, class, result > > Now I want to query the database for a certain user's result in a > specified class, which is very, very easy. No problems. > > But, I also want to find out the user's position relative to others > depending on the result. > > So, if the specified user's result is the 9:th best of all of the users, > I want to have a reply from the DB query that say he has position number 9. > > I really can't figure out how to do that... Somehow I have to make MySQL > calculate the position based on the value in the result column. Take a look at http://arjen-lentz.livejournal.com/55083.html . Very similar ideas in play, though you also have a join. The basic idea is that you do a count on the number of users that have a lower score. -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Query help, please..
I'm looking at a situation I haven't run into before, and I'm a bit puzzled by it. I have this table structure: Table USERS: userid, class Table OBJECT: userid, class, result Now I want to query the database for a certain user's result in a specified class, which is very, very easy. No problems. But, I also want to find out the user's position relative to others depending on the result. So, if the specified user's result is the 9:th best of all of the users, I want to have a reply from the DB query that say he has position number 9. I really can't figure out how to do that... Somehow I have to make MySQL calculate the position based on the value in the result column. -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: DELETE query help please?
This should work DELETE Item FROM Item,ItemTag WHERE Item.ProductID =ItemTag.ItemID AND ItemTag.TagID = '168' Mark Kelly wrote: Hi I want to delete from the 'Item' table all the items identified by the folowing query: SELECT Item.ProductID FROM Item, ItemTag WHERE ItemTag.TagID = '168' AND ItemTag.ItemID = Item.ProductID; but I'm not sure how to go about it. Can anyone help? Thanks Mark -- Yoge, AdventNet, Inc. 925-965-6528 [EMAIL PROTECTED] site24x7.com -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: DELETE query help please?
The following query should work if I understand what you're attempting correctly. Use at your own risk though ;) DELETE FROM Item USING Item, ItemTag WHERE ItemTag.ItemID = Item.ProductID AND ItemTag.TagID = '168'; -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: DELETE query help please?
Hi. On Thursday 05 July 2007 17:35, you wrote: > > I want to delete from the 'Item' table > > all the items identified by the folowing query: > > If you have MySQL 5+, you can do it using a sub-query: > > DELETE FROM > Item > WHERE > ProductID IN ( > SELECT > Item.ProductID > FROM > Item, ItemTag > WHERE > ItemTag.TagID = '168' > AND > ItemTag.ItemID = Item.ProductID > ); I'm on 4.1 due to host restrictions, but for both versions the manual page on subquery syntax says "Another restriction is that currently you cannot modify a table and select from the same table in a subquery. This applies to statements such as DELETE, INSERT, REPLACE, UPDATE" which seems to me would stop your suggestion working. Am I misunderstanding (perfectly possible, as I'm a long way from expert)? Thanks for the quick reply anyway, Mark -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
RE: DELETE query help please?
> I want to delete from the 'Item' table > all the items identified by the folowing query: If you have MySQL 5+, you can do it using a sub-query: DELETE FROM Item WHERE ProductID IN ( SELECT Item.ProductID FROM Item, ItemTag WHERE ItemTag.TagID = '168' AND ItemTag.ItemID = Item.ProductID ); thnx, Chris -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
DELETE query help please?
Hi I want to delete from the 'Item' table all the items identified by the folowing query: SELECT Item.ProductID FROM Item, ItemTag WHERE ItemTag.TagID = '168' AND ItemTag.ItemID = Item.ProductID; but I'm not sure how to go about it. Can anyone help? Thanks Mark -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: Bash script array from MySQL query - HELP Please!!!
I would look at the 15th URL to see if there are specials in there that are
breaking the hash somehow.
On 5/22/07, Ben Benson <[EMAIL PROTECTED]> wrote:
I'm having problems getting a monitoring script to run.
I've put the troublesome bit in a separate script just to test, and it
goes
like this:
declare -a HNdeclares the array "HN"
HN=(`echo "SELECT url FROM hosts" | mysql --user=netmon --password=n3tm0n
--skip-column-names check_http`) runs the query and assigns each record to
a
new element in the array
echo ${#HN} echo's number of elements in array
for ((i=0;i<${#HN};i++)); do
echo ${HN[${i}]} echo value of each element.
done
Seems simple enough yeah?! Well if I run echo "SELECT url FROM hosts" |
mysql --user=user --password=pass --skip-column-names check_http at the
command line, i get all of the records - 32. If I run the script above, it
simply refuses to put more than 14 elements in the array.
Then, to confuse it even more, if I sort the query, it gives a different
amount depending on what its sorted by!! For example, if I sort it by
'url'
it seems to generate 569 elements!
Can anyone please spot the undoubtedly obvious error I've made here?! I've
been scratching my head for days, to no avail!
Many thanks in advance,
Ben Benson
--
We are all slave to our own paradigm. -- Joshua Williams
If the letters PhD appear after a person's name, that person will remain
outdoors even after it's started raining. -- Jeff Kay
Bash script array from MySQL query - HELP Please!!!
I'm having problems getting a monitoring script to run.
I've put the troublesome bit in a separate script just to test, and it goes
like this:
declare -a HNdeclares the array "HN"
HN=(`echo "SELECT url FROM hosts" | mysql --user=netmon --password=n3tm0n
--skip-column-names check_http`) runs the query and assigns each record to a
new element in the array
echo ${#HN} echo's number of elements in array
for ((i=0;i<${#HN};i++)); do
echo ${HN[${i}]} echo value of each element.
done
Seems simple enough yeah?! Well if I run echo "SELECT url FROM hosts" |
mysql --user=user --password=pass --skip-column-names check_http at the
command line, i get all of the records - 32. If I run the script above, it
simply refuses to put more than 14 elements in the array.
Then, to confuse it even more, if I sort the query, it gives a different
amount depending on what its sorted by!! For example, if I sort it by 'url'
it seems to generate 569 elements!
Can anyone please spot the undoubtedly obvious error I've made here?! I've
been scratching my head for days, to no avail!
Many thanks in advance,
Ben Benson
RE: Problem Query - Help Please
> When I execute the following query I get duplicate > product_id's as shown > below: > > SELECT * FROM product, product_category_xref, category WHERE > product_parent_id='' > AND product.product_id=product_category_xref.product_id > AND category.category_id=product_category_xref.category_id > AND product.product_publish='Y' > AND product.product_special='Y' ORDER BY product_name DESC\G > > > Results ( As you can see product_id 4139 occurs twice and I desire the > product_id's to be unique in this query) > > I have also included the descriptions of the tables. > > I would appreciate someone assisting me with a query that > works correctly. Product 2139 has two different categories BOOKS and EDUCATION and therefore appears twice. category_id: 7920cfab5c630ca88ceabcfda6b3848d product_id: 4139 product_list: NULL category_id: 7920cfab5c630ca88ceabcfda6b3848d vendor_id: 1 category_name: BOOKS category_id: 4ee8c8513ee84c95c8eb7f24e63d7222 product_id: 4139 product_list: NULL category_id: 4ee8c8513ee84c95c8eb7f24e63d7222 vendor_id: 1 category_name: EDUCATION If you need to show all the categories you will probably need to retrieve thos separately and build a list programatically to give something like categories: EDUCATION, BOOKS -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Problem Query - Help Please
When I execute the following query I get duplicate product_id's as shown below: SELECT * FROM product, product_category_xref, category WHERE product_parent_id='' AND product.product_id=product_category_xref.product_id AND category.category_id=product_category_xref.category_id AND product.product_publish='Y' AND product.product_special='Y' ORDER BY product_name DESC\G Results ( As you can see product_id 4139 occurs twice and I desire the product_id's to be unique in this query) I have also included the descriptions of the tables. I would appreciate someone assisting me with a query that works correctly. *** 1. row *** product_id: 4199 vendor_id: 1 product_parent_id: 0 product_sku: ToBeAs product_s_desc: Coming Soon! Preorder Today! A series of essays on the influential thinkers and ideas in modern times. product_desc: Coming Soon! Preorder Today! By R.J. Rushdoony. This monumental work is a series of essays on the influential thinkers and ideas in modern times. The author begins with De Sade, who self-consciously broke with any Christian basis for morality and law. Enlightenment thinking began with nature as the only reality, and Christianity was reduced to one option among many. It was then, in turn, attacked as anti-democratic and anti-freedom for its dogmatic assertion of the supernatural. Literary figures such as Shelly, Byron, Whitman, and more are also examined, for the Enlightenment presented both the intellectual and the artist as replacement for the theologian and his church. Ideas, such as the spirit of the age, truth, reason, Romanticism, persona, and Gnosticism are related to the desire to negate God and Christian ethics. Reading this book will help you understand the need to avoid the syncretistic blending of humanistic philosophy with the Christian faith. Paperback, 230 pages, and indices. product_thumb_image: 62c16392f436313324d9922ecf2f5a30.jpg product_full_image: d99c1de85355c6bc853102a4d85065b3.jpg product_publish: Y product_weight: 0. product_weight_uom: pounds product_length: 0. product_width: 0. product_height: 0. product_lwh_uom: inches product_url: product_in_stock: 0 product_available_date: 0 product_special: y product_discount_id: 0 ship_code_id: NULL cdate: 1057785021 mdate: 1059273555 product_name: To Be As God: A Study of Modern Thought Since the Marquis De Sade product_discount_use: category_id: 7920cfab5c630ca88ceabcfda6b3848d product_id: 4199 product_list: NULL category_id: 7920cfab5c630ca88ceabcfda6b3848d vendor_id: 1 category_name: BOOKS category_description: category_thumb_image: NULL category_full_image: NULL category_publish: Y menu_image_1: menu_image_2: menu_image_3: page_image: cdate: 1028759226 mdate: 1028759226 category_flypage: category_browsepage: *** 2. row *** product_id: 4139 vendor_id: 1 product_parent_id: 0 product_sku: Victims product_s_desc: The decline of Americas public education - how and why. product_desc: By Samuel L. Blumenfeld. Americas most effective critic of public education shows us how Americas public schools were remade by educators who used curriculum to create citizens suitable for their own vision of a utopian socialist society. This collection of essays will show you how and why Americas public education declined. You will see the educator-engineered decline of reading skills. The author describes the causes for the decline and the way back to competent education methodologies that will result in a self-educated, competent, and freedom-loving populace. Paperback, 266 pages, and index. product_thumb_image: 63ad73b92ddd18d83eb6942914bcf277.jpg product_full_image: 1440c376576aba8783f183ff145c248b.jpg product_publish: Y product_weight: 0. product_weight_uom: pounds product_length: 0. product_width: 0. product_height: 0. product_lwh_uom: inches product_url: product_in_stock: 0 product_available_date: 0 product_special: y product_discount_id: 0 ship_code_id: NULL cdate: 1056405288 mdate: 1061947639 product_name: The Victims of Dick and Jane product_discount_use: category_id: 7920cfab5c630ca88ceabcfda6b3848d product_id: 4139 product_list: NULL category_id: 7920cfab5c630ca88ceabcfda6b3848d vendor_id: 1 category_name: BOOKS category_description: category_thumb_image: NULL category_full_image:
RE: mySQL query help please
[snip] So if the end result is to have the records selected appended to the Customer_Equipment table, I have to make a temporary table based on the characteristics of the Customer_Equipment table... then insert from that temporary table. Correct? [/snip] BINGO! Goofy rules can be frustrating :) Jay sql, mysql, query (goofy spam rule) - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail <[EMAIL PROTECTED]> To unsubscribe, e-mail <[EMAIL PROTECTED]> Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
RE: mySQL query help please
Greetings Jay: RE: mySQL query help please I love rules that don't make sense ;-) So if the end result is to have the records selected appended to the Customer_Equipment table, I have to make a temporary table based on the characteristics of the Customer_Equipment table... then insert from that temporary table. Correct? Thank you. At 12:29 PM 6/24/2002 -0500, you wrote: >[snip] >INSERT INTO Customer_Equipment SELECT Customer.ID, Server_ID, >Configuration, Equipment_Type, Group_ID, Location, Rack_Location, >Network_Status_Name, Creator_ID, Primary_IP_Address FROM >Customer_Equipment, Customer WHERE SUBSTRING(Server_ID, 5,4) = >Customer.User_ID; > >The select statement works perfectly, but when I try to combine the select >statement with an INSERT INTO statement (I've done this in the past), I get >the error message: > > ERROR 1066: Not unique table/alias: 'Customer_Equipment' >[/snip] > >You cannot INSERT into a table that you have SELECT 'ed from; > >INSERT INTO Customer_Equipment SELECT ... >FROM Customer_Equipment, Customer Peter M. Perchansky, President/CEO Dynamic Net, Inc. Helping companies do business on the Net 420 Park Road; Suite 201 Wyomissing PA 19610 Non-Toll Free: 1-610-736-3795 Personal Email: [EMAIL PROTECTED] Company Email: [EMAIL PROTECTED] Web:http://www.dynamicnet.net/ http://www.manageddedicatedservers.com/ http://www.wemanageservers.com/ - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail <[EMAIL PROTECTED]> To unsubscribe, e-mail <[EMAIL PROTECTED]> Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
RE: mySQL query help please
[snip] INSERT INTO Customer_Equipment SELECT Customer.ID, Server_ID, Configuration, Equipment_Type, Group_ID, Location, Rack_Location, Network_Status_Name, Creator_ID, Primary_IP_Address FROM Customer_Equipment, Customer WHERE SUBSTRING(Server_ID, 5,4) = Customer.User_ID; The select statement works perfectly, but when I try to combine the select statement with an INSERT INTO statement (I've done this in the past), I get the error message: ERROR 1066: Not unique table/alias: 'Customer_Equipment' [/snip] You cannot INSERT into a table that you have SELECT 'ed from; INSERT INTO Customer_Equipment SELECT ... FROM Customer_Equipment, Customer HTH! Jay sql, mysql, query - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail <[EMAIL PROTECTED]> To unsubscribe, e-mail <[EMAIL PROTECTED]> Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
mySQL query help please
Greetings: RE: mySQL query help please INSERT INTO Customer_Equipment SELECT Customer.ID, Server_ID, Configuration, Equipment_Type, Group_ID, Location, Rack_Location, Network_Status_Name, Creator_ID, Primary_IP_Address FROM Customer_Equipment, Customer WHERE SUBSTRING(Server_ID, 5,4) = Customer.User_ID; The select statement works perfectly, but when I try to combine the select statement with an INSERT INTO statement (I've done this in the past), I get the error message: ERROR 1066: Not unique table/alias: 'Customer_Equipment' What does this error message mean, and how should I accomplish my insert into statement so I can use the results of the select statement above? Thank you. Peter M. Perchansky, President/CEO Dynamic Net, Inc. Helping companies do business on the Net 420 Park Road; Suite 201 Wyomissing PA 19610 Non-Toll Free: 1-610-736-3795 Personal Email: [EMAIL PROTECTED] Company Email: [EMAIL PROTECTED] Web:http://www.dynamicnet.net/ http://www.manageddedicatedservers.com/ http://www.wemanageservers.com/ - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail <[EMAIL PROTECTED]> To unsubscribe, e-mail <[EMAIL PROTECTED]> Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
RE: Query help please!
One more thing... What are you using to pull the data? PHP? Perl? Etc? -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]] Sent: Tuesday, March 05, 2002 12:05 PM To: [EMAIL PROTECTED] Subject: Query help please! I need help writing query that would give me parent categories of catID from categories table. For example, if catID=030 then it should give me: Birthday | Special Birthday | Special Birthday If catID=028 Birthday | General Birthday | NULL >desc categories +---+--++ | catID | parentID | catName| +---+--++ | 001 | 000 | Birthday | | 002 | 000 | Get Well | | 003 | 000 | Special Occasions | | 038 | 029 | 40th | | 037 | 029 | 30th | | 036 | 029 | 21st | | 035 | 029 | 16th | | 029 | 001 | Special Birthday | | 028 | 001 | General Birthday | | 030 | 029 | Inspirational | | 045 | 001 | Children's Birthday| | 046 | 045 | 1st| +---+--++ Pinkesh - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail <[EMAIL PROTECTED]> To unsubscribe, e-mail <[EMAIL PROTECTED]> Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail <[EMAIL PROTECTED]> To unsubscribe, e-mail <[EMAIL PROTECTED]> Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
RE: Query help please!
Do you have another table with all the birthdates in it? -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]] Sent: Tuesday, March 05, 2002 12:05 PM To: [EMAIL PROTECTED] Subject: Query help please! I need help writing query that would give me parent categories of catID from categories table. For example, if catID=030 then it should give me: Birthday | Special Birthday | Special Birthday If catID=028 Birthday | General Birthday | NULL >desc categories +---+--++ | catID | parentID | catName| +---+--++ | 001 | 000 | Birthday | | 002 | 000 | Get Well | | 003 | 000 | Special Occasions | | 038 | 029 | 40th | | 037 | 029 | 30th | | 036 | 029 | 21st | | 035 | 029 | 16th | | 029 | 001 | Special Birthday | | 028 | 001 | General Birthday | | 030 | 029 | Inspirational | | 045 | 001 | Children's Birthday| | 046 | 045 | 1st| +---+--++ Pinkesh - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail <[EMAIL PROTECTED]> To unsubscribe, e-mail <[EMAIL PROTECTED]> Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail <[EMAIL PROTECTED]> To unsubscribe, e-mail <[EMAIL PROTECTED]> Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
Query help please!
I need help writing query that would give me parent categories of catID from categories table. For example, if catID=030 then it should give me: Birthday | Special Birthday | Special Birthday If catID=028 Birthday | General Birthday | NULL >desc categories +---+--++ | catID | parentID | catName| +---+--++ | 001 | 000 | Birthday | | 002 | 000 | Get Well | | 003 | 000 | Special Occasions | | 038 | 029 | 40th | | 037 | 029 | 30th | | 036 | 029 | 21st | | 035 | 029 | 16th | | 029 | 001 | Special Birthday | | 028 | 001 | General Birthday | | 030 | 029 | Inspirational | | 045 | 001 | Children's Birthday| | 046 | 045 | 1st| +---+--++ Pinkesh - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail <[EMAIL PROTECTED]> To unsubscribe, e-mail <[EMAIL PROTECTED]> Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
Query Help Please
Hi all, I was hoping that I could get some help with this query. TABLE: Items FIELDS: ProductDescription, ItemPrice, ItemDate I want to group first by the week day using the WeekDay function, Sum the item price for each Weekday and order next by the sum of the item price for each weekday in DESC order. I've tried several ways, but keep getting errors that say I'm using illegal group by functions, etc. Any suggestions? Lee Jenkins - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail <[EMAIL PROTECTED]> To unsubscribe, e-mail <[EMAIL PROTECTED]> Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
