View Full Version : funny error

08-22-2007, 03:24 AM
heres the code

while($row = mysql_fetch_array($return)){
$num = count_com($row['blog.b_id']);
if($i % 2 == 0){
echo '<tr><td class="alt">'.$row['blog.b_id'].'</td><td class="alt">'.$row['blog.title'].'</td><td class="alt">'.$row['blogdate'].'</td><td class="alt">'.row['blog.author'].'</td><td class="alt">'.$num.'<td class="alt"><a class="link" href="post?id='.$row['blog.b_id'].'">Veiw Post</a></td></tr><td class="alt">edit</td><td class="alt">Delete</td></tr>';
echo '<tr><td>'.$row['blog.b_id'].'</td><td>'.$row['blog.title'].'</td><td>'.$row['blogdate'].'</td><td>'.row['blog.author'].'</td><td>'.$num.'<td><a class="link" href="post?id='.$row['blog.b_id'].'">Veiw Post</a></td></tr><td>edit</td><td>Delete</td></tr>';;

Parse error: syntax error, unexpected '[', expecting ',' or ';' in /home/richard/labs/admin/manage.php on line 27

this doesnt make sense i cant find it.

08-22-2007, 03:32 AM
One $row[...] variable in each of the echo statements is missing the $ on the front of the variable name.

08-22-2007, 03:43 AM
and you have two ; after echo in else

08-22-2007, 02:01 PM
Thanks, Let night coding gets to you. I ended up rewriting the whole thing differently.

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