Unknown column 'news_posts.ID' in 'on clause'

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Hi,

PHPFreeNews has been working beautifully for several months but is now giving the following error:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home2/entelsof/public_html/news/Inc/ListingFunctions.php on line 67
Query failed : Unknown column 'news_posts.ID' in 'on clause'

My ISP has just upgraded MySQL to 5.0 from 4.1 (without any warning). So I believe the problem is being cause by the new version of MySQL (Which has been experienced by others documented at
http://www.phpfreenews.co.uk/forums/viewtopic.php?t=343

I believe I am running PHPFreeNEws 1.55. Is there a current step by step list of instructions to fix this problem?

Many Thanks
Adrian

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Sorry I got a bit confused with the thread mentioned in my origina post (unsure if we had an official fix to the problem) but I went through and made the following changes and the problem seems to have gone away.

1. Edit your ListingFunctions.php. Find function BuildListingSQL. Change the first line from
Code:
$Query = "SELECT DISTINCT news_posts.*, news_users.FullName FROM news_posts, news_users"; 
to
Code:
$Query = "SELECT DISTINCT news_posts.*, news_users.FullName FROM (news_posts, news_users)"; 

2. in NewsList.php, line 28
Code:
$Query = "SELECT DISTINCT news_posts.*, news_users.FullName FROM news_posts, news_users"; 
becomes
Code:
$Query = "SELECT DISTINCT news_posts.*, news_users.FullName FROM (news_posts, news_users)"; 

3. In ViewAudit.php, line 83
Code:
$sql = "SELECT EventDateTime, Username, CatDesc, EventType, LinkedID, news_audit.Description FROM news_audit, news_audit_categories LEFT OUTER JOIN news_users ON news_audit.EventUserID=news_users.ID WHERE news_audit.EventCatID=news_audit_categories.ID "; 
becomes
Code:
$sql = "SELECT EventDateTime, Username, CatDesc, EventType, LinkedID, news_audit.Description FROM (news_audit, news_audit_categories) LEFT OUTER JOIN news_users ON news_audit.EventUserID=news_users.ID WHERE news_audit.EventCatID=news_audit_categories.ID "; 

4. In Images.php , prior to the line
Code:
$ResultSet = mysql_query("SELECT * FROM news_images ORDER BY ImageName ASC LIMIT $RecStart, $AdminImagesPerPage"); 

insert Code:
echo "SELECT * FROM news_images ORDER BY ImageName ASC LIMIT $RecStart, $AdminImagesPerPage"; 

e.g. the full code becomes:
Code:
echo "SELECT * FROM news_images ORDER BY ImageName ASC LIMIT $RecStart, $AdminImagesPerPage";
$ResultSet = mysql_query("SELECT * FROM news_images ORDER BY ImageName ASC LIMIT $RecStart, $AdminImagesPerPage"); 

Cheers Adrian