script che richiama photo

**Lino80** · 05-05-2010, 20:14

Ho questo script

Codice PHP:


<images>

<?

$dir = "slideshow/";



if (is_dir($dir)) {



    if ($dh = opendir($dir)) {

    

        while (($file = readdir($dh)) !== false) {



            $filetype = substr($file,-3);

            $filetype = strtolower($filetype);

                        

            if ($filetype == "jpg" || $filetype == "gif") { ?>

    <pic>

        <image>slideshow/<? echo $file;?></image>

        <caption><? echo $file;?></caption>

    </pic>

<?            

            }

        }

        closedir($dh);

    }

}

?>

</images>

dovrei sostituire la directory slideshow/ con admin/photos/uploads/thumbs/tn_". $row_photo['location']."

ho provato a fare così:

Codice PHP:


<images>

<?



    $sql = "select * from photos where ptid = ". $_REQUEST['id'] . " order by porder asc";

    $photo = mysql_query($sql,$myconn) or die(mysql_error()); 

    

    

     // start php to load javascript slide show

    $n = " ";

    

    $count = 0;

    while ($row_photo = mysql_fetch_assoc($photo)) {

        if ($count == 1) {

            $captions .=$n;

            $dir .=$n; 

        }

        else

            $count = $count + 1;

        $captions .= "'".$row_photo['caption']."'" ;

    

$dir = "'admin/photos/uploads/thumbs/tn_". $row_photo['location']."'";



if (is_dir($dir)) {



    if ($dh = opendir($dir)) {

    

        while (($file = readdir($dh)) !== false) {



            $filetype = substr($file,-3);

            $filetype = strtolower($filetype);

                        

            if ($filetype == "jpg" || $filetype == "gif") { ?>

    <pic>

        <image>slideshow/<? echo $file;?></image>

        <caption><? echo $file;?></caption>

    </pic>

<?            

            }

        }

        closedir($dh);

    }

}

?>

</images>

cosa sbaglio? si deve aggiungere la directory anche qui:

Codice PHP:


<image>slideshow/<? echo $file;?></image>

?

**Samleo** · 05-05-2010, 21:19

Prova così

Codice PHP:


<images>

<?

$dir = "admin/photos/uploads/thumbs/tn_". $row_photo['location']."/";



if (is_dir($dir)) {



    if ($dh = opendir($dir)) {

    

        while (($file = readdir($dh)) !== false) {



            $filetype = substr($file,-3);

            $filetype = strtolower($filetype);

                        

            if ($filetype == "jpg" || $filetype == "gif") { ?>

    <pic>

        <image>admin/photos/uploads/thumbs/tn_<? echo $row_photo['location']; ?>/<? echo $file;?></image>

        <caption><? echo $file;?></caption>

    </pic>

<?            

            }

        }

        closedir($dh);

    }

}

?>

</images>

**Lino80** · 05-05-2010, 21:52

ho provato così ma non va

Codice PHP:


<images>

<?



    $sql = "select * from photos where ptid = ". $_REQUEST['id'] . " order by porder asc";

    $photo = mysql_query($sql,$myconn) or die(mysql_error()); 

    

    

     // start php to load javascript slide show

    $n = " ";

    

    $count = 0;

    while ($row_photo = mysql_fetch_assoc($photo)) {

        if ($count == 1) {

            $captions .=$n;

            $dir .=$n; 

        }

        else

            $count = $count + 1;

        $captions .= "'".$row_photo['caption']."'" ;

    

$dir = "admin/photos/uploads/thumbs/tn_". $row_photo['location']."/";



if (is_dir($dir)) {



    if ($dh = opendir($dir)) {

    

        while (($file = readdir($dh)) !== false) {



            $filetype = substr($file,-3);

            $filetype = strtolower($filetype);

                        

            if ($filetype == "jpg" || $filetype == "gif") { ?>

    <pic>

        <image>admin/photos/uploads/thumbs/tn_<? echo $row_photo['location']; ?>/<? echo $file;?></image>

        <caption><? echo $file;?></caption>

    </pic>

    <?

    }else

        print    '<center>[img]admin/photos/uploads/thumbs/tn_noimage.jpg[/img]

';

        

?>

<?            

            }

        }

        closedir($dh);

    }

}

?> 

</images>

strano...cosa manca?

**Samleo** · 05-05-2010, 21:54

ma il risultato cosa è?

**Lino80** · 05-05-2010, 22:01

non visualizza le foto il player in flash..

lo script è questo: http://php.html.it/script/vedi/4308/...sh-slide-show/

pagina example.html

codice:

<!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>Slide Show Example</title>
</head>
<body>
<h2>DynamicFlash :: Slide Show</h2>
<blockquote>
  

http://code.google.com/p/dynamicflash</p>
</blockquote>
<div style="padding:5px;width:300px;background-color:#000000;text-align:center;">
<object classid="clsid:d27cdb6e-ae6d-11cf-96b8-444553540000"
 codebase="http://fpdownload.macromedia.com/pub/shockwave/cabs/flash/swflash.cab#version=8,0,0,0" name="Gallery"
 width="300" height="230" align="middle" id="Gallery">
  <param name="movie" value="gallery.swf"/>
  <param name="quality" value="high" />
  <param name="scale" value="noborder" />
  <param name="salign" value="lt" />
  <param name="bgcolor" value="#000000" />
  <embed src="gallery.swf" width="300" height="230" align="middle" quality="high" bgcolor="#000000"
 name="Gallery" scale="noborder" salign="lt" type="application/x-shockwave-flash"
 pluginspage="http://www.macromedia.com/go/getflashplayer" />
</object>
</div>


Use Nifty Corners to give it that nice round edge!</p>
</body>
</html>

images.php

Codice PHP:


<images>

<?

$dir = "slideshow/";



if (is_dir($dir)) {



    if ($dh = opendir($dir)) {

    

        while (($file = readdir($dh)) !== false) {



            $filetype = substr($file,-3);

            $filetype = strtolower($filetype);

                        

            if ($filetype == "jpg" || $filetype == "gif") { ?>

    <pic>

        <image>slideshow/<? echo $file;?></image>

        <caption><? echo $file;?></caption>

    </pic>

<?            

            }

        }

        closedir($dh);

    }

}

?>

</images>

**Lino80** · 06-05-2010, 08:33

Devo sostituire la directory slideshow/ con admin/photos/uploads/thumbs/tn_". $row_photo['location']." a questo script chi mi dice gentilmente come fare?

grazie

**Lino80** · 06-05-2010, 12:59

ho provato così perchè erano sbagliati i percorsi.. ma non funziona neanche

Codice PHP:


<images>

<?



    $sql = "select * from photos where ptid = ". $_REQUEST['id'] . " order by porder asc";

    $photo = mysql_query($sql,$myconn) or die(mysql_error()); 

    

    

     // start php to load javascript slide show

    $n = " ";

    

    $count = 0;

    while ($row_photo = mysql_fetch_assoc($photo)) {

        if ($count == 1) {

            $captions .=$n;

            $dir .=$n; 

        }

        else

            $count = $count + 1;

        $captions .= "'".$row_photo['caption']."'" ;

    

$dir = "admin/photos/uploads/thumbs/";



if (is_dir($dir)) {



    if ($dh = opendir($dir)) {

    

        while (($row_photo = readdir($dh)) !== false) {



            $filetype = substr($row_photo,-3);

            $filetype = strtolower($filetype);

                        

            if ($filetype == "jpg" || $filetype == "gif") { ?>

    <pic>

        <image>admin/photos/uploads/thumbs/tn_<? echo $row_photo['location']; ?></image>

        <caption>tn_<? echo $row_photo['location']; ?></caption>

    </pic>

    <?

    }else

        print    '<center>[img]admin/photos/uploads/thumbs/tn_noimage.jpg[/img]

';

        

?>

<?            

            }

        }

        closedir($dh);

    }

}

?> 

</images>

**Lino80** · 06-05-2010, 16:51

così funziona:

Codice PHP:


<images>

<?



$dir = "admin/photos/uploads/thumbs/";



if (is_dir($dir)) {



    if ($dh = opendir($dir)) {

    

        while (($file = readdir($dh)) !== false) {



            $filetype = substr($file,-3);

            $filetype = strtolower($filetype);

                        

            if ($filetype == "jpg" || $filetype == "gif") { ?>

    <pic>

        <image>admin/photos/uploads/thumbs/<? echo $file;?></image>

        <caption><? echo $file;?></caption>

    </pic>

<?            

            }

        }

        closedir($dh);

    }

}

?>

</images>

se aggiungo la query

Codice PHP:


$sql = "select * from photos where ptid = ". $_REQUEST['id'] . " order by porder asc";

    $photo = mysql_query($sql,$myconn) or die(mysql_error());

già non funziona più

Codice PHP:


<images>

<?

$sql = "select * from photos where ptid = ". $_REQUEST['id'] . " order by porder asc";

    $photo = mysql_query($sql,$myconn) or die(mysql_error()); 



$dir = "admin/photos/uploads/thumbs/";



if (is_dir($dir)) {



    if ($dh = opendir($dir)) {

    

        while (($file = readdir($dh)) !== false) {



            $filetype = substr($file,-3);

            $filetype = strtolower($filetype);

                        

            if ($filetype == "jpg" || $filetype == "gif") { ?>

    <pic>

        <image>admin/photos/uploads/thumbs/<? echo $file;?></image>

        <caption><? echo $file;?></caption>

    </pic>

<?            

            }

        }

        closedir($dh);

    }

}

?>

</images>

come devo inserire la query per far si che mi carica le immagini riferite all'id della pagina?

la struttura della tabella photos è:

id ptype ptid caption porder location
28 ITEMS 56 2 1 33454.jpg

la pagina che aprirò per visualizzare lo slideshow è pagina.php?id=56

cosa manca allo script per funzionare a parte la query?:

Codice PHP:


$sql = "select * from photos where ptid = ". $_REQUEST['id'] . " order by porder asc";

    $photo = mysql_query($sql,$myconn) or die(mysql_error());

**Lino80** · 07-05-2010, 10:12

up

**Lino80** · 07-05-2010, 20:00

upp

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