msql_fetch_array and for loops

Anon

Hi, I'm extremely new with both MySQL and PHP, and I'm having a problem with my code. Hopefully someone here can point me in the right direction. I'm using some of my knowledge of C++, which may or may not be the right thing to do.

What I'm trying to do is display a certain number of news items (starting with the most recent), and display them in standard HTML tables. Here's my code:

if($show)
{
$total_news = mysql_query("SELECT COUNT() FROM news");
$allnews = mysql_query("SELECT FROM news, authors WHERE aid = authors.id");
for($i=$total_news; $i>$total_news-$show; $i--)
{
$news=mysql_fetch_array($allnews)
$subject=$news["subject"];
$summary=$news["summary"];
$fullnews=$news["fullnews"];
$date=$news["date"];
$time=$news["time"];
$username=$news["username"];

echo("<table width=\"100%\" border=\"0\" cellspacing=\"0\" cellpadding=\"2\" class=\"darkbd\">");
echo("<tr>");
echo("<td class=\"bluecn\"><font class=\"whitenmb\">Posted by: <a href=\"contact.php\">$username</a> on $date at $time</font></td>");
echo("</tr><tr>");
echo("<td class=\"ltblue\"><font class=\"blacksmb\">$subject</font></td>");
echo("</tr><tr>");
echo("<td class=\"ltblue\"><font class=\"blacksm\">$summary</font></td>");
echo("</tr>");
echo("</table>");
echo("<br>");
    }

}

One obvious problem is that even though the for loop moves backwards (starting with the total number of news items and decreasing from there), the mysql_fetch_array still starts at the first news item (instead of the last, where I need it to). However, that shouldn't cause an error, even though its incorrect. When I run the script, I get this error:

Parse error: parse error, unexpected T_VARIABLE in c:\path to news\news.php on line 59

Line 59 has: $subject=$news["subject"];
Hopefully someone can help me out here, since my tutorial doesn't really go this far into creating a news script. I thank you very much in advance.

scoppc

forgot to mysql_fetch_array the $total_news, I think.

Anon

Hmm...I didn't think you had to if the return value of the SELECT is just a number. Well, I'll try it.

Yeah, it didn't do anything.

scoppc

ic. Well why don't you try to use
ORDER BY somefield DESC

This way, you can also get the same results.
Give it a try.

Anon

Ah, thanks. That should do the trick.

Hmm...well, I tried it, and I got this:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\path to news\news.php on line 57

I have no idea why that would show up...

Anon

By the way, here's my new code:

if($show)
{
$done=0;
$allnews = mysql_query("SELECT * FROM news, authors WHERE aid = authors.id ORDER BY id DESC");
while($news = mysql_fetch_array($allnews) && $done<=$show)
{
$subject=$news["subject"];
$summary=$news["summary"];
$fullnews=$news["fullnews"];
$date=$news["date"];
$time=$news["time"];
$username=$news["username"];

echo("<table width=\"100%\" border=\"0\" cellspacing=\"0\" cellpadding=\"2\" class=\"darkbd\">");
cho("<tr>");
echo("<td class=\"bluecn\"><font class=\"whitenmb\">Posted by: <a href=\"contact.php\">$username</a> on $date at $time</font></td>");
echo("</tr><tr>");
echo("<td class=\"ltblue\"><font class=\"blacksmb\">$subject</font></td>");
echo("</tr><tr>");
echo("<td class=\"ltblue\"><font class=\"blacksm\">$summary</font></td>");
echo("</tr>");
echo("</table>");
echo("<br>");

$done++;
}
}

Which, again, gives the error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\path to news\news.php on line 56

And line 56 is of course the beginning of the while loop above. I have no idea why the argument $allnews isn't valid...

scoppc

Frankly, I never tried to use SQL join operation till now in my PHP application (I'm newbie =)).
I suggest you to try to execute the above SQL statement right from the mysql command line. Can you get any results?

Anon

Aha! Yeah, when I tried it, it told me "id" was ambiguous, after which I just changed it to "news.id" and it worked. Thank you so much for all your help.