Code:
    $chkuserid = mysql_query("select UserID from users where Username = '".$_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.'">'.
  .'<a href="http://'.$image.'">View>'. <==== 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.