Help with images, please

**neonardo** · Oct 8, 2007 22:00

hello

i have a problem with images

i use the function mysql_num_rows for to obtein the images and, i want the show images in files of 5

my code is :

PHP Code:


<?php

extract($HTTP_GET_VARS);

require_once('funciones.php');

$lnk    = Cnn();

$sql    = "select * from imagenes3 where img_of_id=$of_id";

@$rs    = mysql_query($sql,$lnk);

$cant    = mysql_num_rows($rs);

if($cant==0){echo "&nbsp;";

}

if(isset($HTTP_GET_VARS['imagen_mostrar']))

{

    $show = $HTTP_GET_VARS['imagen_mostrar'];

}

else

{

    $show = mysql_result($rs,0,"img_id");

}

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml">

<head>

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

<title>Documento sin t&iacute;tulo</title>

<style type="text/css">

<!--

body {

    background-color: #EAEAEA;

}

-->

</style></head>



<body>

<table width="100%" border="0" cellspacing="0" cellpadding="0">

  <tr>

    <td colspan="5"><div align="center"><?php echo "<img border='3' src='imagen.php?img_id=$show&tamanio=600'>"; ?></div></td>

  </tr>

  <tr>

    <td>&nbsp;</td>

    <td>&nbsp;</td>

    <td>&nbsp;</td>

    <td>&nbsp;</td>

    <td>&nbsp;</td>

  </tr>

  

<?php

$filas = $cant/5;

if($filas<1){$filas=1;}

    for($i=1;$i<=$filas;$i++)

    {

        echo "  <tr>\n";

        for($i=1;$i<=$cant;$i++)

        {        

            $x+=1;



            if($x<=$cant)

            {

                $img_id = mysql_result($rs,$i-1,"img_id");

                echo "    <td width='20%'><div align='center'><a href='?of_id=$of_id&imagen_mostrar=$img_id' title='Click para ver...'><img border='3' src='imagen.php?img_id=$img_id&tamanio=100'></a></div></td>\n";

            }

            else

            {

                echo "    <td width='20%'><div align='center'>&nbsp;</div></td>\n";

            }

        }

        echo "  </tr>\n";

    }

?>

  

</table>

</body>

</html>

<?php

mysql_free_result($rs);

mysql_close($lnk);

?>

help me, please!!!!!
i dont know where is my fault!!!!

cheers

pd: sorry for my english, is very bad

**Raju Gautam** · Oct 8, 2007 22:33

Really i could not get what you want to do. If you mean that you want to get 5 images from the database, why don't you use while and you need to use mysql_fetch_array() function to loop. But i saw there in your code that you are probably wanting to get only one image from the database with one id.

**neonardo** · Oct 8, 2007 22:43

i want show the images in rows of 5
in this code, show the images in one row of n images

i dont know where is my fault in this code, if you show me, please

**Raju Gautam** · Oct 8, 2007 22:56

Oh that means you want to show 5 images in a row and others in next row if there are more than 5, right? How many images are being retrieved with that query for the time being?

**neonardo** · Oct 8, 2007 23:13

yep!!!!!!!!!!!!!!

show 5 images in a row and
other 5 images in other row, and next next next

in this code, show the n images in 1 only row

30 images for the time being!!!!

cheers

**Raju Gautam** · Oct 8, 2007 23:21

Ok do like this then. I changed your whole code like this:

Code php:

<?php
extract($HTTP_GET_VARS);
require_once('funciones.php');
$lnk	= Cnn();
$sql	= "SELECT * FROM imagenes3 WHERE img_of_id=" . $of_id;
$rs		= mysql_query($sql) or die(mysql_error());
$cant	= mysql_num_rows($rs);
 
if($cant >= 1){
	if(isset($HTTP_GET_VARS['imagen_mostrar']))
		$show = $HTTP_GET_VARS['imagen_mostrar'];
	else
		$show = mysql_result($rs, 0, "img_id");
}
else{
	echo "No images found.";
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Documento sin t&iacute;tulo</title>
<style type="text/css">
body {background-color: #EAEAEA;}
</style>
</head>
<body>
<table width="100%" border="0" cellspacing="0" cellpadding="0">
  <tr>
    <td colspan="5"><div align="center"><?php echo "<img border='3' src='imagen.php?img_id=$show&tamanio=600'>"; ?></div></td>
  </tr>
  <tr>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
    <td>&nbsp;</td>
  </tr>
<?php
$perrow 	= 5;
$cnter 		= 1;
if($cant >= 1){
	echo '<tr>\n';
	while($rows = mysql_fetch_array($rs)){
		$img_id = $rows['img_id'];
		echo '<td width="20%"><div align="center"><a href="?of_id=' . $of_id . '&imagen_mostrar=' . $img_id . '" title="Click para ver...">';
		echo '<img border="3" src="imagen.php?img_id=' . $img_id . '&tamanio=100"></a></div></td>\n';
		if(($cnter % $perrow) == 0){
			echo '</tr><tr>';
		}
		$cnter++;
	}
	echo '</tr>\n';
}
?>
</table>
</body>
</html>
<?php
mysql_free_result($rs);
mysql_close($lnk);
?>

**neonardo** · Oct 8, 2007 23:43

thank you so much!!!!!!!!!!!!!!

i have a little problem with this code

in one part, show

\n
\n
\n
\n
\n
\n

but, the problem is solveddddddddddddddd!!!!!!!!!!!!! :P

thank you again

cheers!!!

**Raju Gautam** · Oct 8, 2007 23:51

Oh yes if you use \n in single quote that won't work. So you can concatenate with double quote.

PHP Code:


echo 'content' . "\n";

**neonardo** · Oct 8, 2007 23:59

yep, but the problem is solved

thank you rajug for you time

other question

what do you advise me for to learn (better) php?

thanks!!!!!

**Raju Gautam** · Oct 9, 2007 00:03

If you are not the quite beginner for PHP at all, then the php manual is enough. Otherwise go for google and find some better site or buy some books from your local market.

**neonardo** · Oct 9, 2007 00:08

what books you advise me??????????

**Raju Gautam** · Oct 9, 2007 00:26

For the beginners :
http://www.wrox.com/WileyCDA/WroxTit...764557440.html

**neonardo** · Oct 9, 2007 00:30

thanks rajug again for you help and time

good night

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