Hi,
I have developed a news system .. I can select to insert an image into the
articles.
The function to extract the image calling them by the "imgid" of images
table is the following:
function getImm() {
?>
<select name="image">
<?
connect();
$sql = "select * from images";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)) {
printf("<option value=\"%s\">%s</option>", $row["id_img"],
$row["imgid"]);
}
?>
</select>
<?
I'd like to be able to see the image before to publish the piece of news.
I think about a window pop up with java in order to create a preview of the
image. The code follows:
<a href="../images/image.php?img=XXXXXXXXXX" target="_blank"
onClick="window.open(this.href, this.target, 'width=200,height=250');
return false;"><img SRC="../preview.jpg" ALT="Click to look at the picture
you are charging" border="0"></a>
The problem is that I can't pass the "imgid" to the href inspite of the
XXXXXXXX
May someone tell me how to pass the id number in order to show me the image
or how may I solve the trouble?
I hope someone will help me ...
Sincerely,
Alexander
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
