[SOLVED] Parse error

[supermerc]

im getting this error: Parse error: syntax error, unexpected $end in /home/xgame/public_html/random/edit.php on line 90

with this code

[code]<?php
//The following PHP script allows you to edit the
//contents of your MySQL table.
//Connecting to the MySQL database
require('dbconnect.php');

//Should we show a single item or a list?
if($_POST['edit']) {
//Simplifying the variables.

$id = $_POST['id'];
$title = $_POST['title'];
$author = $_POST['author'];
$date = $_POST['date'];
//trim() strips white space from the beginning and end of a line.
$date = trim($date);

$content = $_POST['content'];

//Checks for empty fields or invalid date.
if((empty($title)) OR (empty($author)) OR (empty($date)) OR (empty($content))) {
echo "<center><strong>Please fill in all fields!</strong></center>
";

} else {
//explode() separates the date by the '/' character and outputs it to an array.
$explode_date = explode('/', $date);
//checkdate() returns FALSE if the date is invalid.
$check_date = checkdate($explode_date[0], $explode_date[1], $explode_date[2]);
if($check_date == false) {

echo "<center><strong>Invalid date entered!</strong></center>
";
} else {
//htmlspecialchars() converts special characters into HTML entities.
$title = htmlspecialchars($title);
$author = htmlspecialchars($author);


//The MySQL query which will update the content in the table.
$query = "UPDATE news SET title = '$title', author = '$author', date = '$date', content = '$content' WHERE ID = '$id'";
//Execute the query.
$result = mysql_query($query) or die(mysql_error());
echo "<center><strong>News item modified!</strong></center>";

}
}
} elseif($_GET['action'] == "edit") {
//Display a single result.
$id = $_GET['id'];
//The MySQL query. Select all from the table news where the ID equals the id sent in URL.

$query = "SELECT * FROM news WHERE ID='$id'";
//Executing the query.
$result = mysql_query($query) or die(mysql_error());
//Displaying the results of the query.
while ($row = mysql_fetch_array($result)) {
//extract() takes an associative array and treats the keys as variable names and values as variable values.

extract($row);
?>
<form method="post" action="edit.php">
<table align="center">
<tr><td align="right">Title:</td><td><input type="text" name="title" value="<?php echo "$title"; ?>" maxlength="250" /></td></tr>
<tr><td align="right">Author:</td><td><input type="text" name="author" value="<?php echo "$author"; ?>" maxlength="250" /></td></tr>
<tr><td align="right">Date:</td><td><input type="text" name="date" value="<?php echo "$date"; ?>" maxlength="10" /></td></tr>
<tr><td align="right">Content:</td><td><textarea name="content" cols="50" rows="10"><?php echo "$content"; ?></textarea></td></tr>

<tr><td> </td><td><input type="hidden" name="id" value="<?php echo "$ID"; ?>" /><input type="submit" name="edit" value="Modify" /><input type="reset" name="reset" value="Reset" /></td></tr>
</table>
</form>
<?
}
} else {
//Since we're not displaying a single result,
//we're going to display a list of results.

//The MySQl query. Selects all from the table news.
$query = "SELECT * FROM news ORDER BY ID DESC";
//Execute the query.
$result = mysql_query($query) or die(mysql_error());
while ($row = mysql_fetch_array($result)) {
//extract() takes an associative array and treats the keys as variable names and values as variable values.

extract($row);
echo '<table><tr><td><strong><a href="edit.php?action=edit&id='.$ID.'">'.$title.'</a></strong></td></tr>
<tr><td><small>Written by '.$author.' on '.$date.'</small></td></tr>

<tr><td><strong><a href="delete.php?id='.$ID.'">DELETE</a></strong></td></tr>';
?>[/code]

I really cant find the issue

[kenrbnsn]

You're missing at least 2 closing "}" at the end of your script.  If you employ proper indentation in your script, this would have been much easier to find.

Ken