$chkuserid = mysql_query("select UserID from users where Username = '".$_SESSION['Username']."'");
$row = mysql_fetch_array($chkuserid);
$finall = $row['UserID'];
$result = mysql_query("SELECT PhotoDesc, photoDateAdded, PhotoString from photo where photographerID='".$finall."'");
if (!$result) {
die("Database query failed: " . mysql_error());
}
//
// 4. Use returned data
//
// while ($row = mysql_fetch_array($result)) {
// echo $row[0]." ".$row[1]."<br />";
//}
?>
<?php
//
//Set up table headings
//
echo "<table border=5 align=center>";
echo "<tr>";
echo '<td>Photo Desc.</td>';
echo '<td>Date&Time Added</td>';
echo '<td>Photo</td>';
echo '<tr>';
//
//Display data returned by the query
//
$i=0;
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo '<td>'.mysql_result($result,$i,0).'</td>';
echo '<td>'.mysql_result($result,$i,1).'</td>';
$image = mysql_result($result,$i,2);
echo '<td>'.'<img src="http://'.$image.'">'.
[COLOR="#FF0000"].'<a href="http://'.$image.'">View>'.[/COLOR] <==== this is line 185
'</td>';
echo '</tr>';
$i++;
}
echo "</table>";
Parse error: syntax error, unexpected '.' in /srv/disk7/1259829/www/blablabla/imgup.php on line 185
I want to display an image and just below it a link to display it full size in seperate tab.
When I get rid of link code I can display the images but when I add link code I get this error message.