image_gallery

<?php
                    $i=0;
                    while ($data = mysql_fetch_array($result)){
                    $result2=($i%2)?'#DFA09D':'white';

                            echo "<tr bgcolor='$result2'>";
                            echo '<td>'.$data['image'].'</td>';
                            echo '<td>'.$data['path'].'</td>';
                            echo '<td>'.<img src="photos/image".$i>.'</td>';
                            echo '<td><a href="admin.php?mode=delete&id='.$data['id'].'">Hapus</a> |<a href="input_image.php?id='.$data['id'].'">Edit</a></td>';
                        echo '</tr>';

                    $i++;
                    }
                    ?>

What is this:

Parse error: syntax error, unexpected '<' in C:\xampp\htdocs\Bread\administrator\image_gallery.php on line 153

            line: 153       echo '<td>'.<img src="photos/image".$i>.'</td>';

This line

echo '<td>'.<img src="photos/image".$i>.'</td>';

Should be

echo '<td><img src="photos/image"'.$i.'></td>';

OR

echo "<td><img src='photos/image$i'></td>";

echo "<td>"."<img src='photos/image'."$i>"."</td>"
try it like this .
you get the error your string concatenation is not correct.

error in "

 echo ('<td><img src="photos/image'.$i.' /></td>');

you have a missing " from your img tag

echo ('<td><img src="photos/image'.$i.' "/></td>');
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