Women in Technology

Article:		Building a Simple Search Engine with PHP
Subject:		A more detailed query...
Date:		2003-12-23 06:52:17
From:		anonymous2
Hi Daniel, Thanks again for your fantastic tutorial. I was wondering how one might submit a query which exclueds certain words and includes others. I know that the following should work for finding several words: $result = mysql_query(" SELECT p.page_url AS url, COUNT(*) AS occurrences FROM page p, word w1, occurrence o1, word w2, occurrence o2 WHERE p.page_id = o1.page_id AND w1.word_id = o1.word_id AND w1.word_word = \"$keyword[1]\" AND w2.word_id = o2.word_id AND w2.word_word = \"$keyword[2]\" AND GROUP BY p.page_id ORDER BY occurrences DESC LIMIT $results" ); But how can I also tell it to NOT provide me with an article that contains a word I don't want as well? - Giff

Showing messages 1 through 2 of 2.

• A more detailed query...
  2003-12-23 22:43:15  anonymous2 [View]
  
  
  Hi Giff
  
  Thank you for that. However, there is still a minor defect. This is what came up when i pasted your code in. And keep in mind that i have not altered anything stated in the article.
  
  "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/jeo/dev/search/search.php on line 48
  query executed in 0.000 seconds."
  
  
  So does anyone have a suggestion?
  Please help!
  
  Thank you all for helping me so far.
  
  Best wishes for the festive season!
  
  Regards,
  Sean

• RE: A more detailed query...
  2003-12-23 07:15:55  dsolin [View]
  
  
  Hi Giff,
  
  I'm not sure if I got you right here, but couldn't you just use the != or NOT LIKE operators? This would result in something like:
  
  w2.word_word != \"$keyword[1]\"
  
  or:
  
  w2.word_word NOT LIKE \"$keyword[1]\"
  
  Hope that helps!
  
  -Daniel