MySQL: Column Contains Word From List of Words

I have a list of words. Lets say they are 'Apple', 'Orange', and 'Pear'. I have rows in the database like this:

------------------------------------------------
|author_id   |  content                        |
------------------------------------------------
| 54         |开发者_StackOverflow中文版 I ate an apple for breakfast.   |
| 63         | Going to the store.             |
| 12         | Should I wear the orange shirt? |
------------------------------------------------

I'm looking for a query on an InnoDB table that will return the 1st and 3rd row, because the content column contains one or more words from my list. I know I could query the table once for each word in my list, and use LIKE and the % wildcard character, but I'm wondering if there is a single query method for such a thing?

MySQL (I believe the 5.0 version) added the ability to use regular expressions in your SQL.

Check out: http://www.brainbell.com/tutorials/MySQL/Using_MySQL_Regular_Expressions.htm

SELECT author_id, content
FROM AuthorTableName
WHERE content REGEXP 'Apple|Orange|Pear'
ORDER BY author_id;

EDIT:

Something like this:

SELECT * FROM yourtable WHERE content LIKE '%apple%' OR content LIKE '%orange%'

You can loop your words to create WHERE clause conditions.

For Example:

$words = array( 'apple', 'orange' );
$whereClause = '';
foreach( $words as $word) {
   $whereClause .= ' content LIKE "%' . $word . '%" OR';
}

// Remove last 'OR'
$whereClause = substr($whereClause, 0, -2);

$sql = 'SELECT * FROM yourtable WHERE' . $whereClause;

echo $sql;

Output:

SELECT * FROM yourtable WHERE content LIKE "%apple%" OR content LIKE "%orange%"