[SOLVED] comment not working

[S A N T A]

Okay i am working in a blog from scratch and it is suppose to display previous blog entries and comments but it only shows the current blog entry and it says there is 1 comment but it doesn't display it....

 

i also get this error

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\index.php on line 47

 

ok here is the code

<?php

require("header.php");

$sql = "SELECT entries.*, categories.cat FROM entries, categories
WHERE entries.cat_id = categories.id
ORDER BY dateposted DESC
LIMIT 1;";
?>
<div id="main">
<?php
$result = mysql_query($sql)or die(mysql_error());
$row = mysql_fetch_assoc($result);
echo "<h2><a href='viewentry.php?id=" . $row['id'] . "'>" . $row['subject'] . "</a></h2><br />";
echo "<i>In <a href= viewcat.php?id=" . $row['cat_id'] ."'>" . $row['cat'] . "</a> - Posted on " . date("D jS F Y g.iA", strtotime($row['dateposted'])) . "</i>";
echo "<p>";
echo nl2br($row['body']);
echo "</p>";

echo "<p>";
$commsql = "SELECT name FROM comments WHERE blog_id = " . $row['id'] .
	" ORDER BY dateposted;";
$commresult = mysql_query($commsql);
$numrows_comm = mysql_num_rows($commresult);
if($numrows_comm == 0) {
echo "<p>No comments.</p>";
}
else {
echo "(<strong>" . $numrows_comm . "</strong>) comments : ";
$i = 1;
while($comrow = mysql_fetch_assoc($commresult)) {
	echo "<a href='viewentry.php?id=" . $row['id'] ."#comment" . $i . "'>" . $commrow['name'] . "</a> ";

	$i++;
	}
}
?>
</div>
<div id="bar">
<?php
echo "</p>";
$prevsql = "SELECT entries.*, categories.cat FROM entries, categories
WHERE entries.cat_id = categories.id
ORDER BY dateposted DESC
LIMIT 1, 5;";
$prevresult = mysql_query($presql);
$numrows_prev = mysql_num_rows($prevresult);

if($numrows_prev == 0) {
echo "<p>No previous entries.</p>";
}
else { 
echo "<ul>";
while($prevrow = mysql_fetch_assoc($prevresult)) {
	echo "<li><a href='viewentry.php?id=" . $prevrow['id'] . "'>" . $prevrow ['subject'] . "</a></li>";
	}
}
echo "</ul>";
?>
</div>
<?php
require("footer.php");
?>

 

Help appreciated

[BlueSkyIS]

i suggest that you always die after mysql_query() in case there are errors in your SQL:

 

$commresult = mysql_query($commsql) or die(mysql_error());

[DeanWhitehouse]

this might be it

$prevsql = "SELECT entries.*, categories.cat FROM entries, categories // $prevsql
WHERE entries.cat_id = categories.id
ORDER BY dateposted DESC
LIMIT 1, 5;";
$prevresult = mysql_query($presql); //$presql
$numrows_prev = mysql_num_rows($prevresult);

[S A N T A]

both of those did absolutely nothing except the die one

[DeanWhitehouse]

did you change what i said, or just copied and pasted the code?

if you just copied and pasted, it won't here is the code, with changes

$prevsql = "SELECT entries.*, categories.cat FROM entries, categories 
WHERE entries.cat_id = categories.id
ORDER BY dateposted DESC
LIMIT 1, 5;";
$prevresult = mysql_query($prevsql); 
$numrows_prev = mysql_num_rows($prevresult);

[S A N T A]

cool now i just need it to display the comments and thanks a bunch

[DeanWhitehouse]

Np, read the comments next time though, there were there for a reason

[DeanWhitehouse]

if it's solved, please click solved at the bottom of the page.

[S A N T A]

well half of its solved