Problem with PHP Code...

[petesmc]

Hi,

I'm having some problems with some code. Basically, it outputs 1 result and then gives this error:

Warning: Supplied argument is not a valid MySQL result resource 
in c:\phpdev\www\public\pcreview\new_list.php on line 13

The code is relatively simple:

<?php
$news = mysql_query("SELECT Name, AuthorID, Title, DateAdded, news.ID as nID, News FROM news, authors " .
"WHERE AuthorID=authors.ID ORDER BY DateAdded DESC");
if (!$news) {
print "Cannot query the database.<br>";
print mysql_error();
} else {
?>
<blockquote><ul>

<?php

while ($result = mysql_fetch_array($news)) {
  $id   = $result["nID"];
  $authorid = $result["AuthorID"];
  $news = $result["News"];
  $title = $result["Title"];
  $authorname = $result["Name"];
  $date = $result["DateAdded"];
  $title = stripslashes($title);
  $news = stripslashes($news);
  if($date != $olddate) {
      echo "<LI><font face=\"Verdana, Tahoma\" size=\"2\">$date</font><br>\n";

 }
   $olddate = $date;

     echo " - <font face=\"Verdana, Tahoma\" size=\"2\">$title by $authorname<br>$news<br>\n";


}
}
?>
</ul></blockquote>

Any Ideas what is causing the problem?
Thanks
-Peter

[JMulder]

Well, it's your line 13

$news = mysql_query("SELECT Name, AuthorID, Title, DateAdded, news.ID as nID, News FROM news, authors " .
"WHERE AuthorID=authors.ID ORDER BY DateAdded DESC"); 

Try changing it to:

$news = mysql_query("SELECT Name, AuthorID, Title, DateAdded, news.ID as nID, News FROM news, authors WHERE AuthorID=authors.ID ORDER BY DateAdded DESC"); 

Try that..?

[freddydoesphp]

That will change nothing.

Pete did you try copy and pasting the query to your mysql console to see whats being actually returned.

$sql = " SELECT Name, AuthorID, Title, DateAdded, news.ID as nID, News FROM news, authors WHERE AuthorID=authors.ID ORDER BY DateAdded DESC";
print $sql; 

[firepages]

your problem starts here..

if (!$news) {

whether the query returns any rows or not $news will always have a value (eg ,Resource id #2)...

and then the error is generated in the while loop on a call to an empty resultset (line 13 )

you can either use @ to suppress the errors , or use mysql_num_rows()

if(!@mysql_num_rows($news)){
print "Cannot query the database.<br>";
print mysql_error();
} else { .... etc 

[freddydoesphp]

Thus pasting the query into the mysql console will tell you why your qeury returned no rows.

[JMulder]

Originally posted by freddydoesphp
That will change nothing...

Believe it or not, but errors are always the things you least expect

Serious now, I've had so many wierd errors in my 'PHP Life'..you'll never know. But maybe you're right.

[freddydoesphp]

My point was that

"string string"

and

"string". ." string"

Will produce the same thing and using one over the other is not going to change anything.

[petesmc]

Hi,

JMulder: I did it like that because vBulletin doesn't wordwrap properly.

Freddy: Yes, I've tried it and there are two rows as expected.

Firepages: Thanks for the suggestion, although this doesn't apply as the there are results, as the first one is printed. Great program btw...(phpdev)

[freddydoesphp]

The problem lies in this line.

$news = $result["News"];
$news = stripslashes($news); 

You are assigning a vlaue to $news which is going to overwirte the query result stored in $news and thus break the loop after the first iteration. Either change the query result var or change the line above to use something other than $news.

[petesmc]

Didn't work...

Any other ideas?

-Peter

[petesmc]

It worked actually...i thought you meant something else to start off with.

Thanks allot!

-Peter

[freddydoesphp]

Pete I am 99% sure that its the problem, could you please post your new code with those changes, as I think you may have missed something.

[freddydoesphp]

/ignore my last.

[petesmc]

It's funny how one can miss the slightest thing and ruin their day...

Thanks again

-Peter