php form, data behind form

[Porkie]

i have a form which i want to be able to put pictures behind, how can i do this ?,

 

Cheers in advance

[ialsoagree]

You mean, in the background of the form while people are filling it out? This would be a CSS question, but I believe you're looking for the background-image style property.

 

http://www.htmlref.com/

[Porkie]

the problem is a want to load multiple images from the database to show in the background behind my form

 

<style type="text/css">

<!--

body {

background-image: url(imagelocation);

}

-->

</style>

 

could i load multiple images using this function or would i need a different one?

 

cheers

[fantomel]

the problem is a want to load multiple images from the database to show in the background behind my form

 

<style type="text/css">

<!--

body {

background-image: url(imagelocation);

}

-->

</style>

 

could i load multiple images using this function or would i need a different one?

 

cheers

i think this might help you out.

click here for the tutorial

[Porkie]

i looked through and created a file like said on the link, where is the problem here?

 

<?php header("Content-type: text/css; charset: UTF-8");
error_reporting(E_ALL);
include('conn.php');
$sql = "SELECT * FROM fsdf_image";
$sql2 = mysql_query($sql);

echo '<table><tr>';
while ($row = mysql_fetch_array($sql2)) {
     echo '<td>';
     echo '<img src="/uploads/'.$row['image'].'/default.jpg"width=60" height="50" />';
}



?>  

error code

<table><tr><br />

<b>Warning</b>:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>/home/public_html/css/supercharged.php</b> on line <b>8</b><br />/quote]

 

any help will be greatly appreciated

 

[fantomel]

i looked through and created a file like said on the link, where is the problem here?

 

<?php header("Content-type: text/css; charset: UTF-8");
error_reporting(E_ALL);
include('conn.php');
$sql = "SELECT * FROM fsdf_image";
$sql2 = mysql_query($sql);

echo '<table><tr>';
while ($row = mysql_fetch_array($sql2)) {
     echo '<td>';
     echo '<img src="/uploads/'.$row['image'].'/default.jpg"width=60" height="50" />';
}



?>  

error code

<table><tr><br />

<b>Warning</b>:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>/home/public_html/css/supercharged.php</b> on line <b>8</b><br />/quote]

 

any help will be greatly appreciated

 

i don't see any error in the code have you checked if you have any data in the table? does the table exists?