Jump to content


Photo

built a news system for my website- need a little help, tho


  • Please log in to reply
1 reply to this topic

#1 sirstrongbad

sirstrongbad
  • New Members
  • Pip
  • Newbie
  • 1 posts

Posted 15 November 2003 - 10:55 PM

okay. the file in question is:

http://graphicintegr...l/list_news.php

i would like to display the number of comments to each particular article, but no matter how i set my SQL statement up, it always returns the same number for every article. as it is now, it is showing the total number of comments made.

i have two tables for this. news_comments and news_content. here\'s my SQL statement thus far:

 $total_comments = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM news_content con, news_comments com WHERE com.news_id = con.id"),0);

i keep looking at it and it makes sense. there may be 100+ comments with the same news_comments.news_id value referencing a particular news.id value, but it always shows the total.

help please? i\'m somewhat \"new\" to this whole php/mysql thing. i followed quite a few tutorials to get to where i have this thing now. thanks in advance![/code]
my sig is broken

#2 Barand

Barand
  • Moderators
  • Sen . ( ile || sei )
  • 18,021 posts

Posted 16 November 2003 - 06:19 PM

You need to specify which item. You are counting the total number of comments which have corresponding content items.

$conid = somevalue;

$total_comments = mysql_result(mysql_query(\"SELECT COUNT(*) as Num FROM news_content con, news_comments com WHERE com.news_id = con.id AND con.id = $conid \"),0);

or, for totals of each con.id :

$total_comments = mysql_result(mysql_query(\"SELECT con.id, COUNT(*) as Num FROM news_content con, news_comments com WHERE com.news_id = con.id GROUP BY con.id \"),0);
If you are still using mysql_ functions, STOP! Use mysqli_ or PDO. The longer you leave it the more you will have to rewrite.

Donations gratefully received






moon.png

|baaGrid| easy data tables - and more
|baaChart| easy line, column and pie charts




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users