Коментари ...

Player_pz

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Правя си аз едни коментари , но нещо не става ...

Код:
<?php

$view = mysql_fetch_array(mysql_query("select * from alle where  id=$view"));
if (empty ($view[id])) {
	print "Nema takova jivotno ...";
	exit;
}
if ($view[varified] == 'No') {
	print "Linka ne e potvyrden ..";
	exit;
}

print "Name: $view[name]<br>";
print "Opisanie: $view[descr]<br>";
print "Owner: $view[owner]<br>";

print "<b>Komentari</a></b>";
		 print" <form method=POST action=view.php?action=comment>
         <p>Komentar : <input type=text name=text size=30>
         <input type=submit value=изпрати name=send></p>
         </form> ";
		 print "<table border=1 width=500>
		 <tr><td>Date add</td><td>Name</td><td>Message</td></tr>";
	     $comments = mysql_query("select * from comments where link_id='$view[id]'  order by id asc");
	     while ($c = mysql_fetch_array($comments)) {
		 $user = mysql_fetch_array(mysql_query("select * from users where id='$c[user_id]' "));
		 print "<tr><td>$c[data]</td><td>$user[user] </td><td>$c[body]</td></tr>";
		
	     }
	     print "</table><BR><BR>";
		 
if ($action == comment ) {

 if (strlen($text) < 1 ) { 
print"Heheh , nemoje komentara da e tolkova malyk ....";
exit;

} 
$date1 = date("Y:m:d"); 
$date = date("H:i:s"); 
mysql_query("insert into comments (user_id, link_id, body, data ) values('$stat[id]', '$view[id]', '$text', ' $date , $date1' )") or die("Could not add reply.");
		 
		 
		 
	
	}
?>

Kато изпратя коментар ми показва :arrow:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in view.php on line 5
 
$view = mysql_fetch_array(mysql_query("select * from alle where id=$view"));

Дублираш $viel. Веднъз я ползваш за сравнение, и веднъж за заявката.
 
Скрипта, когато е без коментарите си работи , а когато ги сложа ми вади тази грешка ...
 

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