If-then errors?

[Anon]

<?
$x = 0;
$query = "SELECT sid,title,postdate,username,post_type FROM gnus where username = '$username';

$mysql_result = mysql_query($query,$mysql_link);
while ($row = mysql_fetch_row($mysql_result))
{
$sid = $row[0];
$title = $row[1];
$postdate = $row[2];
$username = $row[3];
$posttype = $row[4];

$x++;
if($x%2 == 0) {
$bgcolor = '#FFFFF0';
} else {
$bgcolor = '#FFFFFF';
}
?>
<tr bgcolor = <? echo $bgcolor; ?>>
<td><input type = 'radio' name = 'curdel' value = '<? echo $sid; ?>'></td>
<td><?

#error begins here, line 46 from full src file
if ($posttype == 1) {
echo "<img src = 'news.gif'>";
}
else if ($posttype == 0) {
print("<img src = 'journal.gif'>");
}

?>

I have been over this and over this, and can't find any error. Am I missing something?

[zhabala]

Hello,

else if should be written together as
elseif

like this:

#error begins here, line 46 from full src file
if ($posttype == 1) {
echo "<img src = 'news.gif'>";
}
elseif ($posttype == 0) {
print("<img src = 'journal.gif'>");
}

or like this:

#error begins here, line 46 from full src file
if ($posttype == 1) {
echo "<img src = 'news.gif'>";
} else {
if ($posttype == 0) {
print("<img src = 'journal.gif'>");
}
}

[badman]

hi
what is the error you are getting?

From the code above i cannot see where the while{} has been closed.

send the entire code and then maybe we can get to it.