PHP Formating Help

[artisticre]

The following code works. I just need to add the following in the echo part like:

<table border="1" width="36%">
<tr>
<td width="100%">$my_cards_row[greeting_card_name]</td>
</tr>
<tr>
<td width="100%">&lt;img border=0 height=100 width=100 alt='$my_cards_row[filename]' src='/images/greetingcards/$my_cards_row[filename]'></td>
</tr>
</table>

THANKS

<?php

$db = mysql_connect("xxx", "xxx", "xxx");
mysql_select_db("xxx",$db);

$result = mysql_query("SELECT packid, card_pack_name FROM cardpacks1 ORDER BY packid");
while ($myrow = mysql_fetch_array($result)) {
echo " <br>";
$cards_in_pack = mysql_query("SELECT distinct greeting_card_name, filename FROM greetingcards1 WHERE greetingcards1.packid = $myrow[packid]");
if (!empty($cards_in_pack)) {
while ($my_cards_row = mysql_fetch_array($cards_in_pack)) {
echo "

$my_cards_row[greeting_card_name]
<img border=0 height=100 width=100 alt='$my_cards_row[filename]' src='/images/greetingcards/$my_cards_row[filename]'>";
}
}
}
?>

[tsinka]

Hi,

I think you can't use arrays directly in strings so try the following

<?php 

$db = mysql_connect("xxx", "xxx", "xxx"); 
mysql_select_db("xxx",$db); 

$result = mysql_query("SELECT packid, card_pack_name FROM cardpacks1 ORDER BY packid"); 
while ($myrow = mysql_fetch_array($result)) { 
    echo "<br>\n"; 
    $cards_in_pack = mysql_query("SELECT distinct greeting_card_name, filename FROM greetingcards1 WHERE greetingcards1.packid = ".$myrow["packid"]); 
    if (!empty($cards_in_pack)) { 
        while ($my_cards_row = mysql_fetch_array($cards_in_pack)) { 
            $cardName = $my_cards_row["greeting_card_name"];
            $fileName = $my_cards_row["filename"];
            echo "$cardName<img border='0' height='100' width='100' alt='$fileName' src='/images/greetingcards/$fileName'>";
        } 
    } 
} 
?>

Thomas

[artisticre]

Thanks for the info but it doesn't answer the question on how to put a table in to format the data