I am doing the same with the following code
$rec1=@mysql_fetch_object($result);
$image=$rec1->iurl;
$alt=$rec->mnam;
$array=GetImageSize("image/$image");
echo "<img src='image/$image' alt=$alt border=0 align=left $array[2]>";
and it works a charm.
Malouin Design Graphique wrote:
> Hello,
>
> My problem is that I have in the table below an URL (see $indice_url)
> that points to the graphic (.gif) that has to be use by the the
> "$indice" variable. The graphics ( five different ones) located in
> the same folder as the script " data_out.php3"
>
> Any Guru out here could shred me the "Blue Light" on how to make the
> images to show?
>
> Thanks in advance,
>
> Best regards,
>
> Yves
>
>
>
> This is the table that feeds:
> --------------------------------
> CREATE TABLE indice (
> id int(11) DEFAULT '0' NOT NULL auto_increment,
> date text,
> indice text,
> indice_url varchar(255),
> PRIMARY KEY (id)
> );
> --------------------------------
>
>
> The file to get the data out:
> --------------------------------
> <?php
>
> /* data_out.php3 */
>
> $db = mysql_connect("www.server.com", "root", "password");
> mysql_select_db("db_name", $db);
> $sql = "select * from table_name order by 'date'";
> $sql = "select * from table_name";
> $result = mysql_query($sql);
> while ($row = mysql_fetch_array($result)) {
> print("<tr><td bgcolor=\"#003399\">");
> printf("<img src=\"$indice_url\">%s</td></tr>\n",
> $row["indice_url"]);
> }
> ?>
> --------------------------------
>
>
> The output that it gives me (in source mode):
> --------------------------------
> <table border="1" cellpadding="3" width="130">
> <tr><td bgcolor="#003399"><img src="">image_f03.gif</td></tr>
> <tr><td bgcolor="#003399"><img src="">image_f05.gif</td></tr>
> <tr><td bgcolor="#003399"><img src="">image_f01.gif</td></tr>
> </table>
> --------------------------------
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