Whats this error Mean

[dropfaith]

im lost

 

Notice: Undefined variable: Authors in /home/www/lawrenceguide.org/www/literature/author.php on line 60

 

Warning: in_array() [function.in-array]: Wrong datatype for second argument in /home/www/lawrenceguide.org/www/literature/author.php on line 60

 

<?php
// includes
include("../template/conf.php");
// open database connection
$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");
// select database
mysql_select_db($db) or die ("Unable to select database!");
$genres = array();

$q = "SELECT Author FROM story";
$r = mysql_query($q);

while ( $row = mysql_fetch_row($r) ):

$add = $row[0];

if ( !in_array($add, $Authors) ):

$Authors[] = $add;

Endif;

Endwhile;

$n = count($Authors);

for ($i = 0; $i < $n; $i++):


$Author = $Authors[$i];

$query  = "SELECT id FROM story WHERE Author='$Author' ORDER BY id DESC";
$result = mysql_query($query);

$num    = mysql_num_rows($result);

echo '<tr><td><a href="byauthor.php?Author=' . $Author . '">' . $Author . '</a> ( ' . $num . ' )</td></tr>';


Endfor;

?>

[F1Fan]

Declare $Authors as an array before your loop.

 

$Authors = array();

[DarkWater]

1. You don't have $Authors set, and you try to use it in in_array(), which causes BOTH errors.

 

2. You need to indent your code.  It helps.

 

3. It's often more readable to use { } instead of if: endif; and the like.

[dropfaith]

solved but i cant see the button to make this solved wierd

[F1Fan]

solved but i cant see the button to make this solved wierd

 

I have no idea what that means.

[DarkWater]

Read the post in PHPFreaks site board.  They upgraded to SMF2 and now Solved doesn't work.  They'll fix it eventually.