Displaying an image using echo command

Posted on 16th Feb 2014 07:03 pm by admin

I had done a query on my database, one of the fields being an image reference to a directory where an image is stored. e.g "/images/picture.jpg"

I want to display this image by using <img src= etc...> in the echo statement but am failing to get a result as i am unsure where to place " or '.

Can you help?

Heres my code:

while($row = mysql_fetch_array($result))

{

echo "<tr class='d0' valign='top'>";
echo "<td width='200'>" "<img src="'. $row['image_url'] .'.jpg>"" "</td>";
echo "<td class='large' width='300'>" . $row['title'] . "</td>";
echo "<td class='medium' width='100'>" . $row['date'] . "</td>";
echo "<td>" . $row['content'] . "</td>";

echo "</tr>";
}

I know ive probably made an obvious mistake but im new to PHP.

Thanks

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