Field Offset Error: ahh!

[Anon]

Ok. I'm running a check on a database...of news that I have. The problem is, I want the LAST entries to print out first. So, what I did is do a select that ran the entries backwards by sorting by id descending. The problem comes when it's printing the for loop, on the 5th entry and beyond, it begins to print out this error:

Warning: MySQL: Bad field offset in /var/lib/apache/htdocs/phpscripts/index.php on line 99

But why is it printing that error if it's still printing the value of that field? I'm a tad confused. Can someone please help me?

Below is the code that I have. For those of you who know what's going on, it should be that hard to follow.

//Check on num of rows for the "for" loop...
$news_query = "SELECT * FROM news ORDER BY news_id DESC";

$news_result = mysql_db_query($DB,news_query)
$num_rows = mysql_num_rows($news_result);

//For query to run the latest 10 news posts
for( $i = 0; $i <= ($num_rows) - 1; ++$i )
{
$news_row = mysql_fetch_array$news_result);
mysql_fetch_field($news_result, $i);

?>

[Anon]

just a thought...

does mysql_fetch_field not just return info about a field type? if you only have 5 fields in your table then getting the sixth will produce this error message.

if you just want the data from the table, use the $news_row array you created

if you want the name of the field/whatever, use the mysql_fetch_field outside of the for loop

dom
🙂

[Anon]

Ahh no I figure it out. What I had was a for loop that ran through all of the current rows up to a max of 15. The problem was in the mysql_fetch_field ($sqlresult, $i);

I was trying to set the row I wanted to print out in the $i in taht fetch field, but it was already getting assigned elsewhere, so I accidentally pulled it out, refreshed the page, and tada, it works. Don't ask me ;-).