Seperate directory reading

**Peuplarchie** · Apr 21, 2009 17:38

Good day to you all,
I was wondering how to seperate the result of a direcory read.
Example, I know the files will be images.

How can I show the portrait first and then landscape second ?

here is my code :

PHP Code:




<html>
<head>

    <SCRIPT language="JavaScript">
    <!--
    if (document.images)
    {
      preload_image = new Image(25,25); 
<?PHP
//  directory name sent by url
$dir = $_GET['dir'];
$file = $_GET['file'];

//define the path as relative
$path = $dir."".$file;

//using the opendir function
$dir_handle = @opendir($path) or die("Unable to open $path");

echo "Directory Listing of $path<br/>";

//running the while loop
while ($file = readdir($dir_handle)) 
{
   if($file!="." && $file!="..")
    
      echo "preload_image.src=\"http://test.peuplies.info/PHP/Photos_gallery/".$path."/".$file.""; 

}

echo "} //-> </script> </head> <body><center>";



while ($file = readdir($dir_handle)) 
{
   if($file!="." && $file!="..")
    
      echo "<img src=\"".$path."/".$file."\" width=\"50px\">"; 

}







//closing the directory
closedir($dir_handle);


?>



</center>

</body>
</html>

Thanks!

**spiderling** · Apr 21, 2009 18:02

Try this.

PHP Code:


while (false !== ($file = readdir($handle))) {# notice this is different

   if ($file != '.' && $file != '..') {

    $imginfo = getimagesize($file);

    $ratio = $imginfo[0] / $imginfo[1];

    if ($ratio < 1) {

      $portrait[] = $file;

    } else {

      $landscape[] = $file;

    }

  }

}



$images = array_merge($portrait,$landscape);



foreach ($images as $image) {

  echo '<img src="'.$path.'/'.$image.'" width="50px">';

}

**Peuplarchie** · Apr 21, 2009 18:38

it don'twork, or I can't make it work.

here is a way I have tried :

PHP Code:






<html>

<head>



    <SCRIPT language="JavaScript">

    <!--

    if (document.images)

    {

      preload_image = new Image(25,25); 

<?PHP

//  directory name sent by url

$dir = $_GET['dir'];

$file = $_GET['file'];



//define the path as relative

$path = $dir."".$file;



//using the opendir function

$dir_handle = @opendir($path) or die("Unable to open $path");



echo "Directory Listing of $path<br/>";



//running the while loop

while ($file = readdir($dir_handle)) 

{

   if($file!="." && $file!="..")

    

      echo "preload_image.src=\"http://test.peuplies.info/PHP/Photos_gallery/".$path."/".$file.""; 



}



echo "} //-> </script> </head> <body><center>";







while (false !== ($file = readdir($dir_handle))) {# notice this is different



   if ($file != '.' && $file != '..') {



    $imginfo = getimagesize($file);



    $ratio = $imginfo[0] / $imginfo[1];



    if ($ratio < 1) {



      $portrait[] = $file;



    } else {



      $landscape[] = $file;



    }



  }



}







$images = array_merge($portrait,$landscape);







foreach ($images as $image) {



  echo '<img src="'.$path.'/'.$image.'" width="50px">';



} 









//closing the directory

closedir($dir_handle);





?>







</center>



</body>

</html>

thanks!

**spiderling** · Apr 21, 2009 18:49

Your $path needs to be the server path to the directory you are opening and reading, and not include a file in it.

Since you are receiving the directory and file values from the GET method, as well as if it was user input, you should validate and filter the variable values.

PHP Code:


$dir = $_GET['dir'];

$path = '/server/path/to/'.$dir;

**logic_earth** · Apr 21, 2009 18:53

*Wonders why one still uses opendir...

**Peuplarchie** · Apr 21, 2009 19:17

how do i find it ?

**spiderling** · Apr 21, 2009 19:20

@logic_earth

....because of the dir class?

**spiderling** · Apr 21, 2009 19:22

Originally Posted by Peuplarchie

how do i find it ?

PHP Code:


echo $_SERVER['DOCUMENT_ROOT'];

**logic_earth** · Apr 21, 2009 19:31

Originally Posted by spiderling

@logic_earth

....because of the dir class?

Nah glob, a lot nicer to work with.

**Peuplarchie** · Apr 21, 2009 19:37

can I just change the \ by / ?

**spiderling** · Apr 21, 2009 19:44

Originally Posted by logic_earth

Nah glob, a lot nicer to work with.

I'm always open to new and different ways. Thanks.

**spiderling** · Apr 21, 2009 19:50

Originally Posted by Peuplarchie

can I just change the \ by / ?

Huh?? Which ones are you referring to?

I noticed a problem with your double quotes not being escaped properly on the line below. As well $path would not work in this line. It would probably have to be directory ($dir).

PHP Code:


echo "preload_image.src=\"http://test.peuplies.info/PHP/Photos_gallery/".$dir."/".$file."\"";

or for it to be easier to read.

PHP Code:


echo 'preload_image.src="http://test.peuplies.info/PHP/Photos_gallery/'.$dir.'/'.$file.'"';

**Peuplarchie** · Apr 21, 2009 20:30

no error now, but it don't display anything ?

PHP Code:










<html>

<head>



    <SCRIPT language="JavaScript">

    <!--

    if (document.images)

    {

      preload_image = new Image(25,25); 

<?PHP

//  directory name sent by url

$dir = $_GET['dir'];

$path = $_SERVER['DOCUMENT_ROOT'].''.$dir; 

$file = $_GET['file'];



//define the path as relative

$path = $dir."".$file;



//using the opendir function

$dir_handle = @opendir($path) or die("Unable to open $path");



echo "Directory Listing of $path<br/>";



//running the while loop

while ($file = readdir($dir_handle)) 

{

   if($file!="." && $file!="..")

    

      echo "preload_image.src=\"http://test.peuplies.info/PHP/Photos_gallery/".$path."/".$file.""; 



}



echo "} //-> </script> </head> <body><center>";





while (false !== ($file = readdir($dir_handle))) {# notice this is different



   if ($file != '.' && $file != '..') {



    $imginfo = getimagesize($file);



    $ratio = $imginfo[0] / $imginfo[1];



    if ($ratio < 1) {



      $portrait[] = $file;



    } else {



      $landscape[] = $file;



    }



  }





foreach ($portrait as $port) {



  echo '<img src="'.$path.'/'.$image.'" width="50px">';

  

}







foreach ($landscape as $land) {



  echo '<img src="'.$path.'/'.$image.'" width="50px">';



}



}





//closing the directory

closedir($dir_handle);



?>







</center>



</body>

</html>

**spiderling** · Apr 21, 2009 20:52

Let's see if the $file is getting put into the arrays.

PHP Code:


while (false !== ($file = readdir($dir_handle))) {# notice this is different

   if ($file != '.' && $file != '..') {

    $imginfo = getimagesize($file);

    $ratio = $imginfo[0] / $imginfo[1];



    if ($ratio < 1) {

#      $portrait[] = $file;

      print $file.' portrait file here<br>';

    } else {

#      $landscape[] = $file;

      print $file.' landscape file here<br>';

    }

  }

}



/*

$images = array_merge($portrait,$landscape);

foreach ($images as $image) {

  echo '<img src="'.$path.'/'.$image.'" width="50px">';

} 

*/

What does path echo as in this line?

PHP Code:


echo "Directory Listing of $path<br/>";

**Peuplarchie** · Apr 21, 2009 20:57

nothing so far, dont even echo "Directory..."

**Peuplarchie** · Apr 21, 2009 20:58

sorry :

Art/Panama_Feb_2007/Gamboa

**Peuplarchie** · Apr 21, 2009 21:06

If I copy your code :

Notice: Undefined variable: portrait in ....s\PHP\Photos_gallery\folder_view.php on line 62

Warning: Invalid argument supplied for foreach() in ....s\PHP\Photos_gallery\folder_view.php on line 62

Notice: Undefined variable: landscape in ...s\PHP\Photos_gallery\folder_view.php on line 70

Warning: Invalid argument supplied for foreach() in ....s\PHP\Photos_gallery\folder_view.php on line 70

**Peuplarchie** · Apr 21, 2009 21:06

PHP Code:





<html>
<head>

    <SCRIPT language="JavaScript">
    <!--
    if (document.images)
    {
      preload_image = new Image(25,25); 
<?PHP
//  directory name sent by url
$dir = $_GET['dir'];
$path = $_SERVER['DOCUMENT_ROOT'].''.$dir; 
$file = $_GET['file'];

//define the path as relative
$path = $dir."".$file;

//using the opendir function
$dir_handle = @opendir($path) or die("Unable to open $path");


//running the while loop
while ($file = readdir($dir_handle)) 
{
   if($file!="." && $file!="..")
    
      echo "preload_image.src=\"http://test.peuplies.info/PHP/Photos_gallery/".$path."/".$file.""; 

}

echo "} //-> </script> </head> <body>";
echo "Directory Listing of ".$path."<br/><center>";

while (false !== ($file = readdir($dir_handle))) {# notice this is different

   if ($file != '.' && $file != '..') {

    $imginfo = getimagesize($file);

    $ratio = $imginfo[0] / $imginfo[1];

    if ($ratio < 1) {

      $portrait[] = $file;

    } else {

    $landscape[] = $file;

    }

  }

}



foreach ($portrait as $port) {

  echo '<img src="'.$path.'/'.$image.'" width="50px">';
  
}



foreach ($landscape as $land) {

  echo '<img src="'.$path.'/'.$image.'" width="50px">';

}



//closing the directory
closedir($dir_handle);



?>



</center>

</body>
</html>

**spiderling** · Apr 21, 2009 21:19

You are assigning a value to $path twice.

PHP Code:


//  directory name sent by url

$dir = $_GET['dir'];

$path = $_SERVER['DOCUMENT_ROOT'].''.$dir; 

$file = $_GET['file'];



//define the path as relative

$path = $dir."".$file;

The path you use to open and read the directory needs to be the server path and cannot contain a file name.

PHP Code:


$server_path = $_SERVER['DOCUMENT_ROOT'].'/Art/Panama_Feb_2007/'.$dir;

opendir($server_path);

The path you use for HTML is different.

PHP Code:


$url_path = '/Art/Panama_Feb_2007/'.$dir;

echo '<img src="'.$url_path.'" border="0">';

Also, when you concatenate two variables you don't need the empty quotes in between.

PHP Code:


$variable = $value1.$value2;

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