[RESOLVED] Image from results in database

[fsumba]

I've created a simple array (I think that is what it is called) in Dreamweaver. I know how to display images from the db with a static page but I can't seem to figure out how to do it when the pages are dynamic (i.e. display one record at at time). The below script puts the image name rather than the image itself. (The field is called Pictures) I actually want the picture itself.

<p>There are <?php echo $totalRows_Recordset1 ?> total records in this database. </p>
<p><?php echo $row_Recordset1['last']; ?></p>
<p><?php echo $row_Recordset1['Pictures']; ?></p>

Thank you..

[1042]

fsumba wrote:

I've created a simple array (I think that is what it is called) in Dreamweaver. I know how to display images from the db with a static page but I can't seem to figure out how to do it when the pages are dynamic (i.e. display one record at at time). The below script puts the image name rather than the image itself. (The field is called Pictures) I actually want the picture itself.

<p>There are <?php echo $totalRows_Recordset1 ?> total records in this database. </p>
<p><?php echo $row_Recordset1['last']; ?></p>
<p><?php echo $row_Recordset1['Pictures']; ?></p>

Thank you..

Here is a way to do it if the file was uploaded to a folder;

<p>There are <?php echo $totalRows_Recordset1 ?> total records in this database. </p>
<p><?php echo $row_Recordset1['last']; ?></p>
<p><?
$newimage=$row_Recordset1['Pictures']; 
echo "<img src=\"pathtoserverfolder/$newimage\"  />";
?></p>

[fsumba]

1042 it worked perfectly. Thank you!

[1042]

fsumba wrote:

1042 it worked perfectly. Thank you!

You're welcome 🙂