tables in php with mysql

**davidjmorin** · 12-31-2007, 06:46 PM

i have this code and cannot for the life of me figure out how to make it so that i to only displays 4 pics per row. can someone please help. thanks

Code:

<?
if(isset($_GET['id']))
{
include 'config.php';
include 'opendb.php';

$id = $_GET['id'];
$query = "SELECT name, type, size, content FROM upload WHERE id = '$id'";
$result = mysql_query($query) or die('Error, query failed');
list($name, $type, $size, $content) = mysql_fetch_array($result);

header("Content-Disposition: attachment; filename=$name");
header("Content-length: $size");
header("Content-type: $type");
echo $content;

include 'closedb.php';
exit;
}

?>
<html>
<head>
<title>Download File From MySQL</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<?
include 'config.php';
include 'opendb.php';

$query = "SELECT id, name FROM upload";
$result = mysql_query($query) or die('Error, query failed');
if(mysql_num_rows($result) == 0)
{
	echo "Database is empty <br>";
} 
else
{
?>
<table border="1">
<?
	while(list($id, $name) = mysql_fetch_array($result))
	{
?>
<td>
        <a href="download.php?id=<?=$id;?>">
	<img src="download.php?id=<?=$id;?>" height="150" width="150"><?=$name;?></a>
</td>
<?
	}
?>
</table>
<?
}
include 'closedb.php';
?>
</body>
</html>

**GhettoT** · 01-02-2008, 08:26 AM

Without reposting your code, this is what I would do.

At the begging of the script add.

PHP Code:


$rowCount = 0;

Then, every time you echo a picture say,

PHP Code:


$rowCount ++;

Then in the same place you echo a row say,

PHP Code:


if($rowCount >= 4){

</tr></tr> //closing the first row, starting a new one

}

Quick fix, but it should work, good luck!

**Leafy** · 01-02-2008, 08:53 AM

First, an explanation of my changes.

1. Add a doctype
You need a doctype in your documents to tell the web browser what type of document you are giving them. This document looks like HTML 4.0 Transitional, so you must include this doctype. You may find a list of doctypes here. They must be before the <html> tag.
2. escape your database inputs.
Any user-derived data must go through mysql_real_escape_string to avoid SQL injections.

PHP Code:


<?

if(isset($_GET['id']))

{

include 'config.php';

include 'opendb.php';



$id = mysql_real_escape_string($_GET['id']);

$query = "SELECT name, type, size, content FROM upload WHERE id = '$id'";

$result = mysql_query($query) or die('Error, query failed');

list($name, $type, $size, $content) = mysql_fetch_array($result);



header("Content-Disposition: attachment; filename=$name");

header("Content-length: $size");

header("Content-type: $type");

echo $content;



include 'closedb.php';

exit;

}



?>

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"

   "http://www.w3.org/TR/html4/loose.dtd">

<html>

<head>

<title>Download File From MySQL</title>

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">

</head>



<body>

<div id="container">

<?

include 'config.php';

include 'opendb.php';



$query = "SELECT id, name FROM upload";

$result = mysql_query($query) or die('Error, query failed');

if(mysql_num_rows($result) == 0)

{

    echo "<p>Database is empty</p>";

}

else

{

?>

<table border="1">

<tr><td>Name</td><td>Preview</td><td>Download</td></tr>

<tr>

<?

    $rows = 0;

    while(list($id, $name) = mysql_fetch_array($result))

    {

        list($id,$name) = $arr;

        $rows++;

        echo (($rows % 4) == 0)?"</tr>":"";

        printf("<tr><td>%s</td><td><img src=\"%s\" height=\"150\" width=\"150\" /></td><td><a href=\"download.php?id=%s\">Download</a></td>\n",

            $name,

            "download.php?id=".$id,

            $id

        );

    }

?>

</tr>

</table>

</div>

<?

}

include 'closedb.php';

?>

</body>

</html>

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