Show image in or possibly out suggest

Question

Hi everybody,
I guess that since nobody answered my previous post that
maybe it's was to long to solve or to complex for a forum solution.
My other solution to my problem would be to show an image
selected in the "dropdown" from my autosuggest instantly.
I tried to show thumbnails in the menu as show in the example
but it loads all my picture (2400 right now)
I would like to choose a pic in my menu and the picture to
show on a separate layer in my page
i show you my code:
<div id="productSampleContains" class="container">
Entrez quelques lettres du produit
recherch&eacute;.<br />
<form method="get" action="phponlyworking.php">
<input type="text" id="blahblahblah" />
<div id="productMenuContains"
spry:region="dsProducts5">
<div spry:repeat="dsProducts5"
spry:suggest="{name}">{name}</div>
</div>
</form>
</div>
<script type="text/javascript">
var as5 = new
Spry.Widget.AutoSuggest("productSampleContains",
"productMenuContains", "dsProducts5", 'name', {containsString:
true,hoverSuggestClass:'highlight',minCharsType: 4});
</script>
And for my XML
<?php
$hostname_conn = "localhost";
$database_conn = "pubs_gm";
$username_conn = "root";
$password_conn = "";
$conn = mysql_pconnect($hostname_conn, $username_conn,
$password_conn) or trigger_error(mysql_error(),E_USER_ERROR);
?>
<?php
// Query the database and get all the records from the Images
table
mysql_select_db($database_conn, $conn);
$query_rsImages = "SELECT
`nom_image`,`ID_image`,`fichier_image`FROM tbl_images";
$rsImages = mysql_query($query_rsImages, $conn) or
die(mysql_error());
$row_rsImages = mysql_fetch_assoc($rsImages);
$totalRows_rsImages = mysql_num_rows($rsImages);
// Send the headers
header('Content-type: text/xml');
header('Pragma: public');
header('Cache-control: private');
header('Expires: -1');
?><?php echo('<?xml version="1.0"
encoding="iso-8859-1"?>'); ?>
<products>
<?php if ($totalRows_rsImages > 0) { // Show if
recordset not empty ?>
<?php do { ?>
<product>
<category><?php echo $row_rsImages['ID_image'];
?></category>
<name><?php echo $row_rsImages['nom_image'];
?></name>
<boximage><?php echo
$row_rsImages['fichier_image']; ?></boximage>
</product>
<?php } while ($row_rsImages =
mysql_fetch_assoc($rsImages)); ?>
<?php } // Show if recordset not empty ?>
</products>
<?php
mysql_free_result($rsImages);
?>
thanks for any help...have a nice week

Answer 1

Hi Deep,
An image type UDF only stores the image name in the database. When you add an image, the SBO client takes a copy of the image file and puts it in the Pictures folder (as defined in General Settings). Therefore, to use dynamic images in Crystal you will need to do the following:
1) Create a formula in Crystal that combines the image name from the UDF field and the path to the Pictures folder to create the full path to the image (note: your Pictures folder must use a UNC or mapped drive path rather than a local path). It's easiest just to hardcode the path to the Pictures folder in the report but you'll have to bear in mind that this setting is company database specific. If you want to be able to run the report for multiple SBO databases then you may need to use a formula to change the path (you could add the OADM table to the report but there is no suitable join between this table and other tables in the database. The report would work because there is only ever 1 record in OADM but it's not good report design).
2) Once you have your formula you can add a add a picture to the body of the report (at this point you should select a default image)
3) Right-click the picture on the layout and choose Format Graphic.
4) Select the Picture tab and click on the formula button to the right of the Graphic Location property
5) In the formula window set the return value to the formula you created in step 1.
6) At this point your image will dynamically change based on the UDF value. You may want to set a formula in the Suppress property of the image so that the default image is hidden if the UDF field is blank for a particular record.
Kind Regards,
Owen