Top 10 Query

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I am trying to write a mySQL query to determine if the current word =
being displayed in the game is one of the top 10 most popular. I am =
trying to achieve this by creating a table that tracks how many times =
each word was accessed. A new row is created for each access to the =
word. The table structure is as follows:

CREATE TABLE IF NOT EXISTS `bible_word_scramble_usage` (
`reference` int(25) NOT NULL AUTO_INCREMENT,
`bible_dictionary_reference` int(4) NOT NULL,
`ip_address` varchar(20) NOT NULL,
`date_accessed` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE =
CURRENT_TIMESTAMP,
PRIMARY KEY (`reference`)
) ENGINE=3DMyISAM DEFAULT CHARSET=3Dlatin1 AUTO_INCREMENT=3D122 ;

The following is the SELECT query I need help tweaking. What I am =
trying to do is select the top 10 most popular words and then use a =
second select to see if the word being displayed is one of the top 10 =
(IE using the search results of the 10 top SELECT query).

The error this query is currently giving me is:=20

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 '=3D =
`top_ten`.`bible_dictionary_reference` =3D 1 LIMIT 1'

This is the query:

SELECT `top_ten`.`bible_dictionary_reference`=20
FROM (=20

SELECT `bible_dictionary_reference` , COUNT( `reference` ) AS word_usage
FROM `bible_word_scramble_usage`=20
GROUP BY `bible_dictionary_reference`=20
ORDER BY word_usage DESC=20
LIMIT 10=20
) AS top_ten
WHERE =3D `top_ten`.`bible_dictionary_reference` =3D1
LIMIT 1=20

Thank you for helping me. Ron

The Verse of the Day
â€=9CEncouragement from Godâ€=99s Wordâ€=9D
http://www.TheVerseOfTheDay.info=20

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Re: Top 10 Query

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I had an extra “=â€=9D sign by the WHERE, my mistake, I =
couldnâ€=99t see it originally. Ron

The Verse of the Day
â€=9CEncouragement from Godâ€=99s Wordâ€=9D
http://www.TheVerseOfTheDay.info=20


From: Ron Piggott=20
Sent: Saturday, January 08, 2011 7:42 AM
To: php-db@lists.php.net=20
Subject: Top 10 Query


I am trying to write a mySQL query to determine if the current word =
being displayed in the game is one of the top 10 most popular. I am =
trying to achieve this by creating a table that tracks how many times =
each word was accessed. A new row is created for each access to the =
word. The table structure is as follows:

CREATE TABLE IF NOT EXISTS `bible_word_scramble_usage` (
`reference` int(25) NOT NULL AUTO_INCREMENT,
`bible_dictionary_reference` int(4) NOT NULL,
`ip_address` varchar(20) NOT NULL,
`date_accessed` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE =
CURRENT_TIMESTAMP,
PRIMARY KEY (`reference`)
) ENGINE=3DMyISAM DEFAULT CHARSET=3Dlatin1 AUTO_INCREMENT=3D122 ;

The following is the SELECT query I need help tweaking. What I am =
trying to do is select the top 10 most popular words and then use a =
second select to see if the word being displayed is one of the top 10 =
(IE using the search results of the 10 top SELECT query).

The error this query is currently giving me is:=20

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 '=3D =
`top_ten`.`bible_dictionary_reference` =3D 1 LIMIT 1'

This is the query:

SELECT `top_ten`.`bible_dictionary_reference`=20
FROM (=20

SELECT `bible_dictionary_reference` , COUNT( `reference` ) AS word_usage
FROM `bible_word_scramble_usage`=20
GROUP BY `bible_dictionary_reference`=20
ORDER BY word_usage DESC=20
LIMIT 10=20
) AS top_ten
WHERE =3D `top_ten`.`bible_dictionary_reference` =3D1
LIMIT 1=20

Thank you for helping me. Ron

The Verse of the Day
â€=9CEncouragement from Godâ€=99s Wordâ€=9D
http://www.TheVerseOfTheDay.info=20

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